Systems of Equations, Geometry 方程组与几何

本页译自 LibreTexts · A First Course in Linear Algebra (Kuttler) 第 1 章「Systems of Equations」。公式经 MathJax 渲染，自定义宏已注入；正文与译文段段对照。

1.1 Geometry 几何

Outcomes. Relate the types of solution sets of a system of two (three) variables to the intersections of lines in a plane (the intersection of planes in three space).

As you may remember, linear equations like $2x+3y=6$ can be graphed as straight lines in the coordinate plane. We say that this equation is in two variables, in this case $x$ and $y$. Suppose you have two such equations, each of which can be graphed as a straight line, and consider the resulting graph of two lines. What would it mean if there exists a point of intersection between the two lines? This point, which lies on both graphs, gives $x$ and $y$ values for which both equations are true. In other words, this point gives the ordered pair $\left( x, y \right)$ that satisfy both equations. If the point $\left( x, y \right)$ is a point of intersection, we say that $\left( x, y \right)$ is a solution to the two equations. In linear algebra, we often are concerned with finding the solution(s) to a system of equations, if such solutions exist. First, we consider graphical representations of solutions and later we will consider the algebraic methods for finding solutions.

When looking for the intersection of two lines in a graph, several situations may arise. The following picture demonstrates the possible situations when considering two equations (two lines in the graph) involving two variables.

In the first diagram, there is a unique point of intersection, which means that there is only one (unique) solution to the two equations. In the second, there are no points of intersection and no solution. When no solution exists, this means that the two lines are parallel and they never intersect. The third situation which can occur, as demonstrated in diagram three, is that the two lines are really the same line. For example, $x+y=1$ and $2x+2y=2$ are equations which when graphed yield the same line. In this case there are infinitely many points which are solutions of these two equations, as every ordered pair which is on the graph of the line satisfies both equations. When considering linear systems of equations, there are always three types of solutions possible; exactly one (unique) solution, infinitely many solutions, or no solution.

Example 1.1.1: A Graphical Solution 图像解法

Use a graph to find the solution to the following system of equations $$\begin{array}{c} x+y=3 \\ y-x=5 \end{array}$$

Solution. Through graphing the above equations and identifying the point of intersection, we can find the solution(s). Remember that we must have either one solution, infinitely many, or no solutions at all. The following graph shows the two equations, as well as the intersection. Remember, the point of intersection represents the solution of the two equations, or the $\left( x,y\right)$ which satisfy both equations. In this case, there is one point of intersection at $\left( -1, 4 \right)$ which means we have one unique solution, $x = -1, y = 4$.

In the above example, we investigated the intersection point of two equations in two variables, $x$ and $y$. Now we will consider the graphical solutions of three equations in two variables.

Consider a system of three equations in two variables. Again, these equations can be graphed as straight lines in the plane, so that the resulting graph contains three straight lines. Recall the three possible types of solutions; no solution, one solution, and infinitely many solutions. There are now more complex ways of achieving these situations, due to the presence of the third line. For example, you can imagine the case of three intersecting lines having no common point of intersection. Perhaps you can also imagine three intersecting lines which do intersect at a single point. These two situations are illustrated below.

Consider the first picture above. While all three lines intersect with one another, there is no common point of intersection where all three lines meet at one point. Hence, there is no solution to the three equations. Remember, a solution is a point $\left( x, y \right)$ which satisfies all three equations. In the case of the second picture, the lines intersect at a common point. This means that there is one solution to the three equations whose graphs are the given lines. You should take a moment now to draw the graph of a system which results in three parallel lines. Next, try the graph of three identical lines. Which type of solution is represented in each of these graphs?

We have now considered the graphical solutions of systems of two equations in two variables, as well as three equations in two variables. However, there is no reason to limit our investigation to equations in two variables. We will now consider equations in three variables.

You may recall that equations in three variables, such as $2x+4y-5z=8$, form a plane. Above, we were looking for intersections of lines in order to identify any possible solutions. When graphically solving systems of equations in three variables, we look for intersections of planes. These points of intersection give the $\left( x, y, z \right)$ that satisfy all the equations in the system. What types of solutions are possible when working with three variables? Consider the following picture involving two planes, which are given by two equations in three variables.

Notice how these two planes intersect in a line. This means that the points $\left( x,y,z\right)$ on this line satisfy both equations in the system. Since the line contains infinitely many points, this system has infinitely many solutions.

It could also happen that the two planes fail to intersect. However, is it possible to have two planes intersect at a single point? Take a moment to attempt drawing this situation, and convince yourself that it is not possible! This means that when we have only two equations in three variables, there is no way to have a unique solution! Hence, the types of solutions possible for two equations in three variables are no solution or infinitely many solutions.

Now imagine adding a third plane. In other words, consider three equations in three variables. What types of solutions are now possible? Consider the following diagram.

In this diagram, there is no point which lies in all three planes. There is no intersection between all planes so there is no solution. The picture illustrates the situation in which the line of intersection of the new plane with one of the original planes forms a line parallel to the line of intersection of the first two planes. However, in three dimensions, it is possible for two lines to fail to intersect even though they are not parallel. Such lines are called skew lines.

Recall that when working with two equations in three variables, it was not possible to have a unique solution. Is it possible when considering three equations in three variables? In fact, it is possible, and we demonstrate this situation in the following picture.

In this case, the three planes have a single point of intersection. Can you think of other types of solutions possible? Another is that the three planes could intersect in a line, resulting in infinitely many solutions, as in the following diagram.

We have now seen how three equations in three variables can have no solution, a unique solution, or intersect in a line resulting in infinitely many solutions. It is also possible that the three equations graph the same plane, which also leads to infinitely many solutions.

You can see that when working with equations in three variables, there are many more ways to achieve the different types of solutions than when working with two variables. It may prove enlightening to spend time imagining (and drawing) many possible scenarios, and you should take some time to try a few.

You should also take some time to imagine (and draw) graphs of systems in more than three variables. Equations like $x+y-2z+4w=8$ with more than three variables are often called hyper-planes. You may soon realize that it is tricky to draw the graphs of hyper-planes! Through the tools of linear algebra, we can algebraically examine these types of systems which are difficult to graph. In the following section, we will consider these algebraic tools.

Exercises 习题

Exercise 1. Graphically, find the point $\left( x_{1},y_{1}\right)$ which lies on both lines, $x+3y=1$ and $4x-y=3$. That is, graph each line and see where they intersect.

Exercise 2. Graphically, find the point of intersection of the two lines, $3x+y=3$ and $x+2y=1$. That is, graph each line and see where they intersect.

Exercise 3. You have a system of $k$ equations in two variables, $k\geq 2$. Explain the geometric significance of

1. No solution.
2. A unique solution.
3. An infinite number of solutions.

1. 无解。
2. 唯一解。
3. 无穷多个解。

1.2 Algebraic Procedures 代数方法

Outcomes.

1. Use elementary operations to find the solution to a linear system of equations.
2. Find the row-echelon form and reduced row-echelon form of a matrix.
3. Determine whether a system of linear equations has no solution, a unique solution or an infinite number of solutions from its reduced row-echelon form.
4. Solve a system of equations using Gaussian Elimination and Gauss-Jordan Elimination.
5. Model a physical system with linear equations and then solve.

1. 使用初等运算求出线性方程组的解。
2. 求矩阵的行阶梯形和简化行阶梯形。
3. 由其简化行阶梯形判断一个线性方程组是无解、有唯一解还是有无穷多个解。
4. 使用高斯消元法和高斯-约当消元法求解方程组。
5. 用线性方程组为物理系统建模并求解。

We have taken an in depth look at graphical representations of systems of equations, as well as how to find possible solutions graphically. Our attention now turns to working with systems algebraically.

Definition 1.2.1: System of Linear Equations. A system of linear equations is a list of equations, $$\begin{array}{c} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}=b_{1} \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}=b_{2} \\ \vdots \\ a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}=b_{m} \end{array}$$ where $a_{ij}$ and $b_{j}$ are real numbers. The above is a system of $m$ equations in the $n$ variables, $x_{1},x_{2}\cdots ,x_{n}$. Written more simply in terms of summation notation, the above can be written in the form $\sum_{j=1}^{n}a_{ij}x_{j}=b_{i},$ for $i=1,2,3,\cdots ,m$.

The relative size of $m$ and $n$ is not important here. Notice that we have allowed $a_{ij}$ and $b_{j}$ to be any real number. We can also call these numbers scalars. We will use this term throughout the text, so keep in mind that the term scalar just means that we are working with real numbers.

Now, suppose we have a system where $b_{i} = 0$ for all $i$. In other words every equation equals $0$. This is a special type of system.

Definition 1.2.2: Homogeneous System of Equations. A system of equations is called homogeneous if each equation in the system is equal to $0$. A homogeneous system has the form $$\begin{array}{c} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}=0 \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}=0 \\ \vdots \\ a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}=0 \end{array}$$ where $a_{ij}$ are scalars and $x_{i}$ are variables.

Recall from the previous section that our goal when working with systems of linear equations was to find the point of intersection of the equations when graphed. In other words, we looked for the solutions to the system. We now wish to find these solutions algebraically. We want to find values for $x_{1},\cdots ,x_{n}$ which solve all of the equations. If such a set of values exists, we call $\left( x_{1},\cdots ,x_{n}\right)$ the solution set.

Recall the above discussions about the types of solutions possible. We will see that systems of linear equations will have one unique solution, infinitely many solutions, or no solution. Consider the following definition.

Definition 1.2.3: Consistent and Inconsistent Systems. A system of linear equations is called consistent if there exists at least one solution. It is called inconsistent if there is no solution.

If you think of each equation as a condition which must be satisfied by the variables, consistent would mean there is some choice of variables which can satisfy all the conditions. Inconsistent would mean there is no choice of the variables which can satisfy all of the conditions.

The following sections provide methods for determining if a system is consistent or inconsistent, and finding solutions if they exist.

Elementary Operations 初等运算

We begin this section with an example. Recall from Example 1.1.1 that the solution to the given system was $\left(x, y \right) = \left( -1, 4 \right)$.

Example 1.2.1: Verifying an Ordered Pair is a Solution. Algebraically verify that $\left(x, y \right) = \left( -1, 4 \right)$ is a solution to the following system of equations. $$\begin{array}{c} x+y=3 \\ y-x=5 \end{array}$$

Solution. By graphing these two equations and identifying the point of intersection, we previously found that $\left(x, y \right) = \left( -1, 4 \right)$ is the unique solution.

We can verify algebraically by substituting these values into the original equations, and ensuring that the equations hold. First, we substitute the values into the first equation and check that it equals $3$: $x + y = (-1)+(4) = 3$. This equals $3$ as needed, so we see that $\left( -1,4 \right)$ is a solution to the first equation. Substituting the values into the second equation yields $y - x = (4) - (-1) = 4 + 1 = 5$, which is true. For $\left( x,y\right) =\left( -1,4\right)$ each equation is true and therefore, this is a solution to the system.

Now, the interesting question is this: If you were not given these numbers to verify, how could you algebraically determine the solution? Linear algebra gives us the tools needed to answer this question. The following basic operations are important tools that we will utilize.

Definition 1.2.4: Elementary Operations. Elementary operations are those operations consisting of the following.

1. Interchange the order in which the equations are listed.
2. Multiply any equation by a nonzero number.
3. Replace any equation with itself added to a multiple of another equation.

1. 交换方程组中方程的排列顺序。
2. 用任意非零数乘某个方程。
3. 把某个方程换成它自身加上另一个方程的若干倍。

It is important to note that none of these operations will change the set of solutions of the system of equations. In fact, elementary operations are the key tool we use in linear algebra to find solutions to systems of equations.

Consider the following example.

Example 1.2.2: Effects of an Elementary Operation. Show that the system $\begin{array}{c} x+y=7 \\ 2x-y=8 \end{array}$ has the same solution as the system $\begin{array}{c} x+y=7 \\ -3y=-6 \end{array}$.

Solution. Notice that the second system has been obtained by taking the second equation of the first system and adding $-2$ times the first equation, as follows: $2x-y + (-2)(x+y) = 8 + (-2)(7)$. By simplifying, we obtain $-3y=-6$, which is the second equation in the second system. Now, from here we can solve for $y$ and see that $y=2$. Next, we substitute this value into the first equation as follows: $x+y=x+2=7$. Hence $x=5$ and so $\left( x,y\right) = \left(5,2 \right)$ is a solution to the second system. We want to check if $\left(5,2 \right)$ is also a solution to the first system. We check this by substituting $\left(x, y \right) = \left(5,2 \right)$ into the system and ensuring the equations are true: $$\begin{array}{c} x+y = \left(5 \right)+ \left( 2 \right) = 7 \\ 2x-y= 2 \left(5 \right) - \left( 2 \right) = 8 \end{array}$$ Hence, $\left(5,2 \right)$ is also a solution to the first system.

This example illustrates how an elementary operation applied to a system of two equations in two variables does not affect the solution set. However, a linear system may involve many equations and many variables and there is no reason to limit our study to small systems. For any size of system in any number of variables, the solution set is still the collection of solutions to the equations. In every case, the above operations of Definition 1.2.4 do not change the set of solutions to the system of linear equations.

In the following theorem, we use the notation $E_i$ to represent an equation, while $b_i$ denotes a constant.

Theorem 1.2.1: Elementary Operations and Solutions. Suppose you have a system of two linear equations $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$. Then the following systems have the same solution set as the above system:

1. $\begin{array}{c} E_{2}=b_{2} \\ E_{1}=b_{1} \end{array}$
2. $\begin{array}{c} E_{1}=b_{1} \\ kE_{2}=kb_{2} \end{array}$ for any scalar $k$, provided $k\neq0$.
3. $\begin{array}{c} E_{1}=b_{1} \\ E_{2}+kE_{1}=b_{2}+kb_{1} \end{array}$ for any scalar $k$ (including $k=0$).

1. $\begin{array}{c} E_{2}=b_{2} \\ E_{1}=b_{1} \end{array}$
2. $\begin{array}{c} E_{1}=b_{1} \\ kE_{2}=kb_{2} \end{array}$，其中 $k$ 为任意纯量，且 $k\neq0$。
3. $\begin{array}{c} E_{1}=b_{1} \\ E_{2}+kE_{1}=b_{2}+kb_{1} \end{array}$，其中 $k$ 为任意纯量（包括 $k=0$）。

Before we proceed with the proof of Theorem 1.2.1, let us consider this theorem in context of Example 1.2.2. Then, $\begin{array}{cc} E_{1} = x+y, & b_{1} = 7 \\ E_{2} = 2x-y, & b_{2} = 8 \end{array}$. Recall the elementary operations that we used to modify the system in the solution to the example. First, we added $\left( -2 \right)$ times the first equation to the second equation. In terms of Theorem 1.2.1, this action is given by $E_{2} + \left( -2 \right) E_{1} = b_{2} + \left( -2 \right)b_{1}$, or $2x-y + \left( -2 \right) \left(x+y \right) = 8 + \left( -2 \right) 7$. This gave us the second system in Example 1.2.2, given by $\begin{array}{c} E_{1} = b_{1} \\ E_{2} + \left( -2 \right) E_{1} = b_{2} + \left( -2 \right) b_{1} \end{array}$.

From this point, we were able to find the solution to the system. Theorem 1.2.1 tells us that the solution we found is in fact a solution to the original system.

We will now prove Theorem 1.2.1.

Proof.

1. The proof that the systems $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ and $\begin{array}{c} E_{2}=b_{2} \\ E_{1}=b_{1} \end{array}$ have the same solution set is as follows. Suppose that $\left( x_{1},\cdots ,x_{n}\right)$ is a solution to $E_{1}=b_{1},E_{2}=b_{2}$. We want to show that this is a solution to the second system above. This is clear, because the second system is the original system, but listed in a different order. Changing the order does not effect the solution set, so $\left( x_{1},\cdots ,x_{n}\right)$ is a solution to it.
2. Next we want to prove that the two systems $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ and $\begin{array}{c} E_{1}=b_{1} \\ kE_{2}=kb_{2} \end{array}$ have the same solution set, provided $k\neq 0$. Let $\left( x_{1},\cdots ,x_{n}\right)$ be a solution of $E_{1}=b_{1},E_{2}=b_{2}$. We want to show that it is a solution to $E_{1}=b_{1},kE_{2}=kb_{2}$. Notice that the only difference between these two systems is that the second involves multiplying the equation $E_{2}=b_{2}$ by the scalar $k$. Recall that when you multiply both sides of an equation by the same number, the sides are still equal to each other. Hence if $\left( x_{1},\cdots ,x_{n}\right)$ is a solution to $E_{2}=b_{2}$, then it will also be a solution to $kE_{2}=kb_{2}$. Hence, $\left( x_{1},\cdots ,x_{n}\right)$ is also a solution to the second system. Similarly, let $\left( x_{1},\cdots ,x_{n}\right)$ be a solution of $E_{1}=b_{1},kE_{2}=kb_{2}$. Then we can multiply the equation $kE_{2}=kb_{2}$ by the scalar $1/k$, which is possible only because we have required that $k\neq 0$. Just as above, this action preserves equality and we obtain the equation $E_{2} = b_{2}$. Hence $\left( x_{1},\cdots ,x_{n}\right)$ is also a solution to $E_{1}=b_{1},E_{2}=b_{2}$.
3. Finally, we will prove that the two systems $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ and $\begin{array}{c} E_{1}=b_{1} \\ E_{2}+kE_{1}=b_{2}+kb_{1} \end{array}$ have the same solution set. We will show that any solution of the first is also a solution of the second. Then, we will show that any solution of the second is also a solution of the first. Let $\left( x_{1},\cdots ,x_{n}\right)$ be a solution to $E_{1}=b_{1},E_{2}=b_{2}$. Then in particular it solves $E_{1} = b_{1}$. Hence, it solves the first equation in the second system. Similarly, it also solves $E_{2} = b_{2}$. By our proof of part 2, it also solves $kE_{1}=kb_{1}$. Notice that if we add $E_{2}$ and $kE_{1}$, this is equal to $b_{2} + kb_{1}$. Therefore, if $\left( x_{1},\cdots ,x_{n}\right)$ solves $E_{1}=b_{1},E_{2}=b_{2}$ it must also solve $E_{2}+kE_{1}=b_{2}+kb_{1}$. Now suppose $\left( x_{1},\cdots ,x_{n}\right)$ solves the system $E_{1}=b_{1}, E_{2}+kE_{1}=b_{2}+kb_{1}$. Then in particular it is a solution of $E_{1} = b_{1}$. Again by our proof of part 2, it is also a solution to $kE_{1}=kb_{1}$. Now if we subtract these equal quantities from both sides of $E_{2}+kE_{1}=b_{2}+kb_{1}$ we obtain $E_{2}=b_{2}$, which shows that the solution also satisfies $E_{1}=b_{1},E_{2}=b_{2}$.

1. 证明两个方程组 $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ 与 $\begin{array}{c} E_{2}=b_{2} \\ E_{1}=b_{1} \end{array}$ 有相同解集的方法如下。设 $\left( x_{1},\cdots ,x_{n}\right)$ 是 $E_{1}=b_{1},E_{2}=b_{2}$ 的解。我们要证明它也是第二个方程组的解。这是显然的，因为第二个方程组就是原方程组，只是列出顺序不同。改变顺序并不影响解集，所以 $\left( x_{1},\cdots ,x_{n}\right)$ 也是它的解。
2. 下面我们证明两个方程组 $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ 与 $\begin{array}{c} E_{1}=b_{1} \\ kE_{2}=kb_{2} \end{array}$ 在 $k\neq 0$ 时有相同的解集。设 $\left( x_{1},\cdots ,x_{n}\right)$ 是 $E_{1}=b_{1},E_{2}=b_{2}$ 的解，要证明它也是 $E_{1}=b_{1},kE_{2}=kb_{2}$ 的解。注意这两个方程组唯一的差别在于后者把方程 $E_{2}=b_{2}$ 乘以了纯量 $k$。回忆：方程两边同乘同一个数时，两边仍然相等。因此若 $\left( x_{1},\cdots ,x_{n}\right)$ 是 $E_{2}=b_{2}$ 的解，它也就必然是 $kE_{2}=kb_{2}$ 的解。于是 $\left( x_{1},\cdots ,x_{n}\right)$ 也是第二个方程组的解。 反过来，设 $\left( x_{1},\cdots ,x_{n}\right)$ 是 $E_{1}=b_{1},kE_{2}=kb_{2}$ 的解。那么我们可以把方程 $kE_{2}=kb_{2}$ 乘以纯量 $1/k$，这之所以可行，正是因为我们已经要求 $k\neq 0$。与上面一样，这一操作保持等式成立，得到方程 $E_{2} = b_{2}$。于是 $\left( x_{1},\cdots ,x_{n}\right)$ 也是 $E_{1}=b_{1},E_{2}=b_{2}$ 的解。
3. 最后，我们证明两个方程组 $\begin{array}{c} E_{1}=b_{1} \\ E_{2}=b_{2} \end{array}$ 与 $\begin{array}{c} E_{1}=b_{1} \\ E_{2}+kE_{1}=b_{2}+kb_{1} \end{array}$ 有相同的解集。我们将证明前者的任一解也是后者的解，再证明后者的任一解也是前者的解。设 $\left( x_{1},\cdots ,x_{n}\right)$ 是 $E_{1}=b_{1},E_{2}=b_{2}$ 的解，则它特别地解出 $E_{1} = b_{1}$，因而也解出第二个方程组中的第一个方程；同理它也解出 $E_{2} = b_{2}$。由第 2 部分的证明，它同样解出 $kE_{1}=kb_{1}$。注意如果把 $E_{2}$ 与 $kE_{1}$ 相加，就等于 $b_{2} + kb_{1}$。因此，若 $\left( x_{1},\cdots ,x_{n}\right)$ 解出 $E_{1}=b_{1},E_{2}=b_{2}$，则它必然也解出 $E_{2}+kE_{1}=b_{2}+kb_{1}$。 再设 $\left( x_{1},\cdots ,x_{n}\right)$ 解出方程组 $E_{1}=b_{1}, E_{2}+kE_{1}=b_{2}+kb_{1}$，则它特别地是 $E_{1} = b_{1}$ 的解。同样由第 2 部分的证明，它也是 $kE_{1}=kb_{1}$ 的解。现在从 $E_{2}+kE_{1}=b_{2}+kb_{1}$ 两边同时减去这些相等的量，就得到 $E_{2}=b_{2}$，证得该解也满足 $E_{1}=b_{1},E_{2}=b_{2}$。

Stated simply, the above theorem shows that the elementary operations do not change the solution set of a system of equations.

We will now look at an example of a system of three equations and three variables. Similarly to the previous examples, the goal is to find values for $x,y,z$ such that each of the given equations are satisfied when these values are substituted in.

Example 1.2.3: Solving a System of Equations with Elementary Operations. Find the solutions to the system, $$\begin{array}{c} x+3y+6z=25 \\ 2x+7y+14z=58 \\ 2y+5z=19 \end{array}$$

Solution. We can relate this system to Theorem 1.2.1 above. In this case, we have $$\begin{array}{c c} E_{1} = x + 3y + 6z, & b_{1} = 25 \\ E_{2} = 2x+7y+14z, & b_{2} = 58 \\ E_{3} = 2y+5z, & b_{3} = 19 \end{array}$$ Theorem 1.2.1 claims that if we do elementary operations on this system, we will not change the solution set. Therefore, we can solve this system using the elementary operations given in Definition 1.2.4. First, replace the second equation by $\left( -2\right)$ times the first equation added to the second. This yields the system $$\begin{array}{c} x+3y+6z=25 \\ y+2z=8 \\ 2y+5z=19 \end{array}$$ Now, replace the third equation with $\left( -2\right)$ times the second added to the third. This yields the system $$\begin{array}{c} x+3y+6z=25 \\ y+2z=8 \\ z=3 \end{array}$$ At this point, we can easily find the solution. Simply take $z=3$ and substitute this back into the previous equation to solve for $y$, and similarly to solve for $x$. $$\begin{array}{c} x + 3y + 6 \left(3 \right) = x + 3y + 18 = 25 \\ y + 2 \left(3 \right) = y + 6 = 8 \\ z = 3 \end{array}$$ The second equation is now $y+6=8$. You can see from this equation that $y = 2$. Therefore, we can substitute this value into the first equation as follows: $x + 3 \left(2 \right) + 18 = 25$. By simplifying this equation, we find that $x=1$. Hence, the solution to this system is $\left( x,y,z \right) = \left( 1,2,3 \right)$. This process is called back substitution.

Alternatively, in the third system you could have continued as follows. Add $\left( -2\right)$ times the third equation to the second and then add $\left( -6\right)$ times the second to the first. This yields $$\begin{array}{c} x+3y=7 \\ y=2 \\ z=3 \end{array}$$ Now add $\left( -3\right)$ times the second to the first. This yields $$\begin{array}{c} x=1 \\ y=2 \\ z=3 \end{array}$$ a system which has the same solution set as the original system. This avoided back substitution and led to the same solution set. It is your decision which you prefer to use, as both methods lead to the correct solution, $\left( x,y,z \right) = \left(1,2,3\right)$.

1.3 Gaussian Elimination 高斯消元法

The work we did in the previous section will always find the solution to the system. In this section, we will explore a less cumbersome way to find the solutions. First, we will represent a linear system with an augmented matrix. A matrix is simply a rectangular array of numbers. The size or dimension of a matrix is defined as $m\times n$ where $m$ is the number of rows and $n$ is the number of columns. In order to construct an augmented matrix from a linear system, we create a coefficient matrix from the coefficients of the variables in the system, as well as a constant matrix from the constants. The coefficients from one equation of the system create one row of the augmented matrix.

For example, consider the linear system from Example 1.2.3 $$\begin{array}{c} x+3y+6z=25 \\ 2x+7y+14z=58 \\ 2y+5z=19 \end{array}$$ This system can be written as an augmented matrix, as follows $$\left[ \begin{array}{rrr|r} 1 & 3 & 6 & 25 \\ 2 & 7 & 14 & 58 \\ 0 & 2 & 5 & 19 \end{array} \right]$$

Notice that it has exactly the same information as the original system. Here it is understood that the first column contains the coefficients from $x$ in each equation, in order, $\left[ \begin{array}{r} 1 \\ 2 \\ 0 \end{array} \right]$. Similarly, we create a column from the coefficients on $y$ in each equation, $\left[ \begin{array}{r} 3 \\ 7 \\ 2 \end{array} \right]$, and a column from the coefficients on $z$ in each equation, $\left[ \begin{array}{r} 6 \\ 14 \\ 5 \end{array} \right]$. For a system of more than three variables, we would continue in this way constructing a column for each variable. Similarly, for a system of less than three variables, we simply construct a column for each variable.

Finally, we construct a column from the constants of the equations, $\left[ \begin{array}{r} 25 \\ 58 \\ 19 \end{array} \right]$.

The rows of the augmented matrix correspond to the equations in the system. For example, the top row in the augmented matrix, $\left[ \begin{array}{rrrrr} 1 & 3 & 6 & | & 25 \end{array} \right]$, corresponds to the equation $x+3y+6z=25$.

Definition 1.3.1: Augmented Matrix of a Linear System. For a linear system of the form $$\begin{array}{c} a_{11}x_{1}+\cdots +a_{1n}x_{n}=b_{1} \\ \vdots \\ a_{m1}x_{1}+\cdots +a_{mn}x_{n}=b_{m} \end{array}$$ where the $x_{i}$ are variables and the $a_{ij}$ and $b_{i}$ are constants, the augmented matrix of this system is given by $$\left[ \begin{array}{rrr|r} a_{11} & \cdots & a_{1n} & b_{1} \\ \vdots & & \vdots & \vdots \\ a_{m1} & \cdots & a_{mn} & b_{m} \end{array} \right]$$

Now, consider elementary operations in the context of the augmented matrix. The elementary operations in Definition 1.2.4 can be used on the rows just as we used them on equations previously. Changes to a system of equations as a result of an elementary operation are equivalent to changes in the augmented matrix resulting from the corresponding row operation. Note that Theorem 1.2.1 implies that any elementary row operations used on an augmented matrix will not change the solution to the corresponding system of equations. We now formally define elementary row operations. These are the key tool we will use to find solutions to systems of equations.

Definition 1.3.2: Elementary Row Operations. The elementary row operations (also known as row operations) consist of the following

1. Switch two rows.
2. Multiply a row by a nonzero number.
3. Replace a row by any multiple of another row added to it.

1. 交换两行。
2. 用非零数乘某一行。
3. 把某一行换成它加上另一行的任意倍数。

Recall how we solved Example 1.2.3. We can do the exact same steps as above, except now in the context of an augmented matrix and using row operations. The augmented matrix of this system is $$\left[ \begin{array}{rrr|r} 1 & 3 & 6 & 25 \\ 2 & 7 & 14 & 58 \\ 0 & 2 & 5 & 19 \end{array} \right]$$ Thus the first step in solving the system would be to take $\left( -2\right)$ times the first row of the augmented matrix and add it to the second row, $$\left[ \begin{array}{rrr|r} 1 & 3 & 6 & 25 \\ 0 & 1 & 2 & 8 \\ 0 & 2 & 5 & 19 \end{array} \right]$$ Next take $\left( -2\right)$ times the second row and add to the third, $$\left[ \begin{array}{rrr|r} 1 & 3 & 6 & 25 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 1 & 3 \end{array} \right]$$ This augmented matrix corresponds to the system $\begin{array}{c} x+3y+6z=25 \\ y+2z=8 \\ z=3 \end{array}$, which is the same as the third system above. By back substitution you obtain the solution $x=1,y=2,$ and $z=3$.

Through a systematic procedure of row operations, we can simplify an augmented matrix and carry it to row-echelon form or reduced row-echelon form, which we define next. These forms are used to find the solutions of the system of equations corresponding to the augmented matrix.

In the following definitions, the term leading entry refers to the first nonzero entry of a row when scanning the row from left to right.

Definition 1.3.3: Row-Echelon Form. An augmented matrix is in row-echelon form if

1. All nonzero rows are above any rows of zeros.
2. Each leading entry of a row is in a column to the right of the leading entries of any row above it.
3. Each leading entry of a row is equal to $1$.

1. 所有非零行都在零行的上方。
2. 每一行的首项都位于它上方任何行的首项所在列的右侧一列及以后。
3. 每一行的首项都等于 $1$。

We also consider another reduced form of the augmented matrix which has one further condition.

Definition 1.3.4: Reduced Row-Echelon Form. An augmented matrix is in reduced row-echelon form if

1. All nonzero rows are above any rows of zeros.
2. Each leading entry of a row is in a column to the right of the leading entries of any rows above it.
3. Each leading entry of a row is equal to $1$.
4. All entries in a column above and below a leading entry are zero.

1. 所有非零行都在零行的上方。
2. 每一行的首项都位于它上方任何行的首项所在列的右侧一列及以后。
3. 每一行的首项都等于 $1$。
4. 首项所在列中位于其上、其下的所有元素都是零。

Notice that the first three conditions on a reduced row-echelon form matrix are the same as those for row-echelon form.

Hence, every reduced row-echelon form matrix is also in row-echelon form. The converse is not necessarily true; we cannot assume that every matrix in row-echelon form is also in reduced row-echelon form. However, it often happens that the row-echelon form is sufficient to provide information about the solution of a system.

The following examples describe matrices in these various forms. As an exercise, take the time to carefully verify that they are in the specified form.

Example 1.3.1: Not in Row-Echelon Form. The following augmented matrices are not in row-echelon form (and therefore also not in reduced row-echelon form). $$\left[ \begin{array}{rrr|r} 0 & 0 & 0 & 0 \\ 1 & 2 & 3 & 3 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right] ,\left[ \begin{array}{rr|r} 1 & 2 & 3 \\ 2 & 4 & -6 \\ 4 & 0 & 7 \end{array} \right] ,\left[ \begin{array}{rrr|r} 0 & 2 & 3 & 3 \\ 1 & 5 & 0 & 2 \\ 7 & 5 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{array} \right]$$

Example 1.3.2: Matrices in Row-Echelon Form. The following augmented matrices are in row-echelon form, but not in reduced row-echelon form. $$\left[ \begin{array}{rrrrr|r} 1 & 0 & 6 & 5 & 8 & 2 \\ 0 & 0 & 1 & 2 & 7 & 3 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] ,\left[ \begin{array}{rrr|r} 1 & 3 & 5 & 4 \\ 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right], \left[ \begin{array}{rrr|r} 1 & 0 & 6 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$$

Notice that we could apply further row operations to these matrices to carry them to reduced row-echelon form. Take the time to try that on your own. Consider the following matrices, which are in reduced row-echelon form.

Example 1.3.3: Matrices in Reduced Row-Echelon Form. The following augmented matrices are in reduced row-echelon form. $$\left[ \begin{array}{rrrrr|r} 1 & 0 & 0 & 5 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right] ,\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right] , \left[ \begin{array}{rrr|r} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 2 \end{array} \right]$$

One way in which the row-echelon form of a matrix is useful is in identifying the pivot positions and pivot columns of the matrix.

Definition 1.3.5: Pivot Position and Pivot Column. A pivot position in a matrix is the location of a leading entry in the row-echelon form of a matrix.

A pivot column is a column that contains a pivot position.

For example consider the following.

Example 1.3.4: Pivot Position. Let $A=\left[ \begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 3 & 2 & 1 & 6 \\ 4 & 4 & 4 & 10 \end{array} \right]$. Where are the pivot positions and pivot columns of the augmented matrix $A$?

Solution. The row-echelon form of this matrix is $\left[ \begin{array}{rrr|r} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & \frac{3}{2} \\ 0 & 0 & 0 & 0 \end{array} \right]$.

This is all we need in this example, but note that this matrix is not in reduced row-echelon form.

In order to identify the pivot positions in the original matrix, we look for the leading entries in the row-echelon form of the matrix. Here, the entry in the first row and first column, as well as the entry in the second row and second column are the leading entries. Hence, these locations are the pivot positions. We identify the pivot positions in the original matrix, as in the following: $\left[ \begin{array}{rrr|r} \fbox{1} & 2 & 3 & 4 \\ 3 & \fbox{2} & 1 & 6 \\ 4 & 4 & 4 & 10 \end{array} \right]$. Thus the pivot columns in the matrix are the first two columns.

The following is an algorithm for carrying a matrix to row-echelon form and reduced row-echelon form. You may wish to use this algorithm to carry the above matrix to row-echelon form or reduced row-echelon form yourself for practice.

Algorithm 1.3.1: Reduced Row-Echelon Form Algorithm. This algorithm provides a method for using row operations to take a matrix to its reduced row-echelon form. We begin with the matrix in its original form.

1. Starting from the left, find the first nonzero column. This is the first pivot column, and the position at the top of this column is the first pivot position. Switch rows if necessary to place a nonzero number in the first pivot position.
2. Use row operations to make the entries below the first pivot position (in the first pivot column) equal to zero.
3. Ignoring the row containing the first pivot position, repeat steps 1 and 2 with the remaining rows. Repeat the process until there are no more rows to modify.
4. Divide each nonzero row by the value of the leading entry, so that the leading entry becomes $1$. The matrix will then be in row-echelon form. The following step will carry the matrix from row-echelon form to reduced row-echelon form.
5. Moving from right to left, use row operations to create zeros in the entries of the pivot columns which are above the pivot positions. The result will be a matrix in reduced row-echelon form.

1. 从左端开始，找到第一个非零列。它就是第一个主元列，该列的顶端位置就是第一个主元位置。必要时交换行，使第一个主元位置出现非零数。
2. 用行运算把第一个主元位置下方（第一个主元列中）的元素化为零。
3. 忽略包含第一个主元位置的那一行，对其余各行重复第 1、2 步。一直重复，直到没有可再修改的行为止。
4. 把每个非零行除以首项的值，使首项变为 $1$。此时矩阵即处于行阶梯形。 下面这一步会把矩阵从行阶梯形化为简化行阶梯形。
5. 从右向左移动，用行运算在位于主元位置上方的主元列各元素处制造零。结果就是处于简化行阶梯形的矩阵。

Most often we will apply this algorithm to an augmented matrix in order to find the solution to a system of linear equations. However, we can use this algorithm to compute the reduced row-echelon form of any matrix which could be useful in other applications.

Consider the following example of Algorithm 1.3.1.

Example 1.3.5: Finding Row-Echelon Form and Reduced Row-Echelon Form of a Matrix. Let $A = \left[ \begin{array}{rrr} 0 & -5 & -4 \\ 1 & 4 & 3 \\ 5 & 10 & 7 \end{array} \right]$. Find the row-echelon form of $A$. Then complete the process until $A$ is in reduced row-echelon form.

Solution. In working through this example, we will use the steps outlined in Algorithm 1.3.1.

1. The first pivot column is the first column of the matrix, as this is the first nonzero column from the left. Hence the first pivot position is the one in the first row and first column. Switch the first two rows to obtain a nonzero entry in the first pivot position, outlined in a box below. $\left[ \begin{array}{rrr} \fbox{1} & 4 & 3 \\ 0 & -5 & -4 \\ 5 & 10 & 7 \end{array} \right]$
2. Step two involves creating zeros in the entries below the first pivot position. The first entry of the second row is already a zero. All we need to do is subtract $5$ times the first row from the third row. The resulting matrix is $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & -5 & -4 \\ 0 & 10 & 8 \end{array} \right]$
3. Now ignore the top row. Apply steps $1$ and $2$ to the smaller matrix $\left[ \begin{array}{rr} -5 & -4 \\ 10 & 8 \end{array} \right]$. In this matrix, the first column is a pivot column, and $-5$ is in the first pivot position. Therefore, we need to create a zero below it. To do this, add $2$ times the first row (of this matrix) to the second. The resulting matrix is $\left[ \begin{array}{rr} -5 & -4 \\ 0 & 0 \end{array} \right]$. Our original matrix now looks like $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & -5 & -4 \\ 0 & 0 & 0 \end{array} \right]$. We can see that there are no more rows to modify.
4. Now, we need to create leading $1$s in each row. The first row already has a leading $1$ so no work is needed here. Divide the second row by $-5$ to create a leading $1$. The resulting matrix is $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & 1 & \frac{4}{5} \\ 0 & 0 & 0 \end{array} \right]$. This matrix is now in row-echelon form.
5. Now create zeros in the entries above pivot positions in each column, in order to carry this matrix all the way to reduced row-echelon form. Notice that there is no pivot position in the third column so we do not need to create any zeros in this column! The column in which we need to create zeros is the second. To do so, subtract $4$ times the second row from the first row. The resulting matrix is $\left[ \begin{array}{rrr} 1 & 0 & - \frac{1}{5} \\ 0 & 1 & \frac{4}{5} \\ 0 & 0 & 0 \end{array} \right]$.

1. 第一个主元列就是矩阵的第一列，因为这是从左端算起的第一个非零列。因此第一个主元位置是第一行第一列。交换前两行，使第一个主元位置出现非零元（见下方方框）：$\left[ \begin{array}{rrr} \fbox{1} & 4 & 3 \\ 0 & -5 & -4 \\ 5 & 10 & 7 \end{array} \right]$
2. 第 2 步是在第一个主元位置下方各元素处制造零。第二行的第一个元素已经是零。我们只需从第三行减去第一行的 $5$ 倍即可。得到的矩阵是 $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & -5 & -4 \\ 0 & 10 & 8 \end{array} \right]$
3. 现在忽略最上面一行。把第 1、2 步应用于较小矩阵 $\left[ \begin{array}{rr} -5 & -4 \\ 10 & 8 \end{array} \right]$。在这个矩阵中，第一列是主元列，$-5$ 位于第一个主元位置，于是我们要在其下方制造零。为此，把（该矩阵的）第一行乘以 $2$ 加到第二行，得到矩阵 $\left[ \begin{array}{rr} -5 & -4 \\ 0 & 0 \end{array} \right]$。原矩阵现在看起来是 $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & -5 & -4 \\ 0 & 0 & 0 \end{array} \right]$。可以看到已没有需要再修改的行了。
4. 现在需要在每行制造首项 $1$。第一行已经有首项 $1$，所以此处无需操作。把第二行除以 $-5$ 制造首项 $1$，得到的矩阵是 $\left[ \begin{array}{rrr} 1 & 4 & 3 \\ 0 & 1 & \frac{4}{5} \\ 0 & 0 & 0 \end{array} \right]$。此时矩阵处于行阶梯形。
5. 现在在各主元列中位于主元位置上方的元素处制造零，以便把矩阵一路化为简化行阶梯形。注意第三列没有主元位置，因此不需要在此列制造任何零！需要制造零的列是第二列。为此，从第一行减去第二行的 $4$ 倍，得到的矩阵是 $\left[ \begin{array}{rrr} 1 & 0 & - \frac{1}{5} \\ 0 & 1 & \frac{4}{5} \\ 0 & 0 & 0 \end{array} \right]$。

This matrix is now in reduced row-echelon form.

The above algorithm gives you a simple way to obtain the row-echelon form and reduced row-echelon form of a matrix. The main idea is to do row operations in such a way as to end up with a matrix in row-echelon form or reduced row-echelon form. This process is important because the resulting matrix will allow you to describe the solutions to the corresponding linear system of equations in a meaningful way.

In the next example, we look at how to solve a system of equations using the corresponding augmented matrix.

Example 1.3.6: Finding the Solution to a System. Give the complete solution to the following system of equations $$\begin{array}{c} 2x+4y-3z=-1 \\ 5x+10y-7z=-2 \\ 3x+6y+5z=9 \end{array}$$

Solution. The augmented matrix for this system is $\left[ \begin{array}{rrr|r} 2 & 4 & -3 & -1 \\ 5 & 10 & -7 & -2 \\ 3 & 6 & 5 & 9 \end{array} \right]$.

In order to find the solution to this system, we wish to carry the augmented matrix to reduced row-echelon form. We will do so using Algorithm 1.3.1. Notice that the first column is nonzero, so this is our first pivot column. The first entry in the first row, $2$, is the first leading entry and it is in the first pivot position. We will use row operations to create zeros in the entries below the $2$. First, replace the second row with $-5$ times the first row plus $2$ times the second row. This yields $\left[ \begin{array}{rrr|r} 2 & 4 & -3 & -1 \\ 0 & 0 & 1 & 1 \\ 3 & 6 & 5 & 9 \end{array} \right]$. Now, replace the third row with $-3$ times the first row plus $2$ times the third row. This yields $\left[ \begin{array}{rrr|r} 2 & 4 & -3 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 21 \end{array} \right]$. Now the entries in the first column below the pivot position are zeros. We now look for the second pivot column, which in this case is column three. Here, the $1$ in the second row and third column is in the pivot position. We need to do just one row operation to create a zero below the $1$.

Taking $-1$ times the second row and adding it to the third row yields $\left[ \begin{array}{rrr|r} 2 & 4 & -3 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 20 \end{array} \right]$.

We could proceed with the algorithm to carry this matrix to row-echelon form or reduced row-echelon form. However, remember that we are looking for the solutions to the system of equations. Take another look at the third row of the matrix. Notice that it corresponds to the equation $0x+0y+0z=20$. There is no solution to this equation because for all $x,y,z$, the left side will equal $0$ and $0\neq 20.$ This shows there is no solution to the given system of equations. In other words, this system is inconsistent.

The following is another example of how to find the solution to a system of equations by carrying the corresponding augmented matrix to reduced row-echelon form.

Example 1.3.7: An Infinite Set of Solutions. Give the complete solution to the system of equations $$\begin{array}{c} 3x-y-5z=9 \\ y-10z=0 \\ -2x+y=-6 \end{array}$$

Solution. The augmented matrix of this system is $\left[ \begin{array}{rrr|r} 3 & -1 & -5 & 9 \\ 0 & 1 & -10 & 0 \\ -2 & 1 & 0 & -6 \end{array} \right]$. In order to find the solution to this system, we will carry the augmented matrix to reduced row-echelon form, using Algorithm 1.3.1. The first column is the first pivot column. We want to use row operations to create zeros beneath the first entry in this column, which is in the first pivot position. Replace the third row with $2$ times the first row added to $3$ times the third row. This gives $$\left[ \begin{array}{rrr|r} 3 & -1 & -5 & 9 \\ 0 & 1 & -10 & 0 \\ 0 & 1 & -10 & 0 \end{array} \right]$$

Now, we have created zeros beneath the $3$ in the first column, so we move on to the second pivot column (which is the second column) and repeat the procedure. Take $-1$ times the second row and add to the third row. $\left[ \begin{array}{rrr|r} 3 & -1 & -5 & 9 \\ 0 & 1 & -10 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$. The entry below the pivot position in the second column is now a zero. Notice that we have no more pivot columns because we have only two leading entries.

At this stage, we also want the leading entries to be equal to one. To do so, divide the first row by $3$. $\left[ \begin{array}{rrr|r} 1 & - \frac{1}{3} & - \frac{5}{3} & 3 \\ 0 & 1 & -10 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$.

This matrix is now in row-echelon form.

Let’s continue with row operations until the matrix is in reduced row-echelon form. This involves creating zeros above the pivot positions in each pivot column. This requires only one step, which is to add $\frac{1}{3}$ times the second row to the first row. $\left[ \begin{array}{rrr|r} 1 & 0 & -5 & 3 \\ 0 & 1 & -10 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$.

This is in reduced row-echelon form, which you should verify using Definition 1.3.4. The equations corresponding to this reduced row-echelon form are $\begin{array}{c} x - 5z=3 \\ y - 10z = 0 \end{array}$, or $\begin{array}{c} x=3+5z \\ y = 10z \end{array}$.

Observe that $z$ is not restrained by any equation. In fact, $z$ can equal any number. For example, we can let $z = t$, where we can choose $t$ to be any number. In this context $t$ is called a parameter. Therefore, the solution set of this system is $\begin{array}{c} x=3+5t \\ y=10t \\ z=t \end{array}$, where $t$ is arbitrary. The system has an infinite set of solutions which are given by these equations. For any value of $t$ we select, $x, y,$ and $z$ will be given by the above equations. For example, if we choose $t=4$ then the corresponding solution would be $\begin{array}{c} x = 3 + 5 (4) = 23 \\ y = 10(4)=40 \\ z=4 \end{array}$.

In Example 1.3.7 the solution involved one parameter. It may happen that the solution to a system involves more than one parameter, as shown in the following example.

Example 1.3.8: A Two Parameter Set of Solutions. Find the solution to the system $\begin{array}{c} x+2y-z+w=3 \\ x+y-z+w=1 \\ x+3y-z+w=5 \end{array}$.

Solution. The augmented matrix is $\left[ \begin{array}{rrrr|r} 1 & 2 & -1 & 1 & 3 \\ 1 & 1 & -1 & 1 & 1 \\ 1 & 3 & -1 & 1 & 5 \end{array} \right]$. We wish to carry this matrix to row-echelon form. Here, we will outline the row operations used. However, make sure that you understand the steps in terms of Algorithm 1.3.1.

Take $-1$ times the first row and add to the second. Then take $-1$ times the first row and add to the third. This yields $\left[ \begin{array}{rrrr|r} 1 & 2 & -1 & 1 & 3 \\ 0 & -1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 0 & 2 \end{array} \right]$.

Now add the second row to the third row and divide the second row by $-1$. $\left[ \begin{array}{rrrr|r} 1 & 2 & -1 & 1 & 3 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$.

This matrix is in row-echelon form and we can see that $x$ and $y$ correspond to pivot columns, while $z$ and $w$ do not. Therefore, we will assign parameters to the variables $z$ and $w$. Assign the parameter $s$ to $z$ and the parameter $t$ to $w.$ Then the first row yields the equation $x+2y-s+t=3$, while the second row yields the equation $y=2$. Since $y=2$, the first equation becomes $x+4-s+t=3$ showing that the solution is given by $\begin{array}{c} x=-1+s-t \\ y=2 \\ z=s \\ w=t \end{array}$. It is customary to write this solution in the form $$\left[ \begin{array}{c} x \\ y \\ z \\ w \end{array} \right] =\left[ \begin{array}{c} -1+s-t \\ 2 \\ s \\ t \end{array} \right].$$

This example shows a system of equations with an infinite solution set which depends on two parameters. It can be less confusing in the case of an infinite solution set to first place the augmented matrix in reduced row-echelon form rather than just row-echelon form before seeking to write down the description of the solution.

In the above steps, this means we don’t stop with the row-echelon form in equation above. Instead we first place it in reduced row-echelon form as follows. $\left[ \begin{array}{rrrr|r} 1 & 0 & -1 & 1 & -1 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$. Then the solution is $y=2$ from the second row and $x=-1+z-w$ from the first. Thus letting $z=s$ and $w=t,$ the solution is given by the equation above.

You can see here that there are two paths to the correct answer, which both yield the same answer. Hence, either approach may be used. The process which we first used in the above solution is called Gaussian Elimination. This process involves carrying the matrix to row-echelon form, converting back to equations, and using back substitution to find the solution. When you do row operations until you obtain reduced row-echelon form, the process is called Gauss-Jordan Elimination.

We have now found solutions for systems of equations with no solution and infinitely many solutions, with one parameter as well as two parameters. Recall the three types of solution sets which we discussed in the previous section; no solution, one solution, and infinitely many solutions. Each of these types of solutions could be identified from the graph of the system. It turns out that we can also identify the type of solution from the reduced row-echelon form of the augmented matrix.

• No Solution: In the case where the system of equations has no solution, the row-echelon form of the augmented matrix will have a row of the form $\left[ \begin{array}{rrrrr} 0 & 0 & 0 & | & 1 \end{array} \right]$. This row indicates that the system is inconsistent and has no solution.
• One Solution: In the case where the system of equations has one solution, every column of the coefficient matrix is a pivot column. The following is an example of an augmented matrix in reduced row-echelon form for a system of equations with one solution. $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \end{array} \right]$
• Infinitely Many Solutions: In the case where the system of equations has infinitely many solutions, the solution contains parameters. There will be columns of the coefficient matrix which are not pivot columns. The following are examples of augmented matrices in reduced row-echelon form for systems of equations with infinitely many solutions. $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array} \right]$ or $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & -3 \end{array} \right]$

• 无解：当方程组无解时，增广矩阵的行阶梯形会有一行形如 $\left[ \begin{array}{rrrrr} 0 & 0 & 0 & | & 1 \end{array} \right]$。这一行表明方程组不相容、没有解。
• 一个解：当方程组恰有一个解时，系数矩阵的每一列都是主元列。下面是一个含唯一解的方程组其增广矩阵处于简化行阶梯形的例子：$\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \end{array} \right]$
• 无穷多个解：当方程组有无穷多个解时，解中含有参数，系数矩阵中会有某些列不是主元列。下面是含无穷多个解的方程组其增广矩阵处于简化行阶梯形的例子：$\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 2 & -3 \\ 0 & 0 & 0 & 0 \end{array} \right]$ 或 $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 5 \\ 0 & 1 & 0 & -3 \end{array} \right]$

1.4 Uniqueness of the Reduced Row-Echelon Form 简化行阶梯形的唯一性

As we have seen in earlier sections, we know that every matrix can be brought into reduced row-echelon form by a sequence of elementary row operations. Here we will prove that the resulting matrix is unique; in other words, the resulting matrix in reduced row-echelon does not depend upon the particular sequence of elementary row operations or the order in which they were performed.

Let $A$ be the augmented matrix of a homogeneous system of linear equations in the variables $x_1, x_2, \cdots, x_n$ which is also in reduced row-echelon form. The matrix $A$ divides the set of variables in two different types. We say that $x_i$ is a basic variable whenever $A$ has a leading $1$ in column number $i$, in other words, when column $i$ is a pivot column. Otherwise we say that $x_i$ is a free variable.

Recall Example 1.3.8.

Example 1.4.1: Basic and Free Variables. Find the basic and free variables in the system $\begin{array}{c} x+2y-z+w=3 \\ x+y-z+w=1 \\ x+3y-z+w=5 \end{array}$.

Solution. Recall from the solution of Example 1.3.8 that the row-echelon form of the augmented matrix of this system is given by $$\left[ \begin{array}{rrrr|r} 1 & 2 & -1 & 1 & 3 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right].$$ You can see that columns $1$ and $2$ are pivot columns. These columns correspond to variables $x$ and $y$, making these the basic variables. Columns $3$ and $4$ are not pivot columns, which means that $z$ and $w$ are free variables.

We can write the solution to this system as $\begin{array}{c} x=-1+s-t \\ y=2 \\ z=s \\ w=t \end{array}$.

Here the free variables are written as parameters, and the basic variables are given by linear functions of these parameters.

In general, all solutions can be written in terms of the free variables. In such a description, the free variables can take any values (they become parameters), while the basic variables become simple linear functions of these parameters. Indeed, a basic variable $x_i$ is a linear function of only those free variables $x_j$ with $j>i$. This leads to the following observation.

Proposition 1.4.1: Basic and Free Variables. If $x_i$ is a basic variable of a homogeneous system of linear equations, then any solution of the system with $x_j=0$ for all those free variables $x_j$ with $j>i$ must also have $x_i=0$.

Using this proposition, we prove a lemma which will be used in the proof of the main result of this section below.

Lemma 1.4.1: Solutions and the Reduced Row-Echelon Form of a Matrix. Let $A$ and $B$ be two distinct augmented matrices for two homogeneous systems of $m$ equations in $n$ variables, such that $A$ and $B$ are each in reduced row-echelon form. Then, the two systems do not have exactly the same solutions.

Proof. With respect to the linear systems associated with the matrices $A$ and $B$, there are two cases to consider:

• Case $1$: the two systems have the same basic variables
• Case $2$: the two systems do not have the same basic variables

• 情形 $1$：两个方程组有相同的基本变量
• 情形 $2$：两个方程组没有相同的基本变量

In case $1$, the two matrices will have exactly the same pivot positions. However, since $A$ and $B$ are not identical, there is some row of $A$ which is different from the corresponding row of $B$ and yet the rows each have a pivot in the same column position. Let $i$ be the index of this column position. Since the matrices are in reduced row-echelon form, the two rows must differ at some entry in a column $j>i$. Let these entries be $a$ in $A$ and $b$ in $B$, where $a \neq b$. Since $A$ is in reduced row-echelon form, if $x_j$ were a basic variable for its linear system, we would have $a=0$. Similarly, if $x_j$ were a basic variable for the linear system of the matrix $B$, we would have $b=0$. Since $a$ and $b$ are unequal, they cannot both be equal to $0$, and hence $x_j$ cannot be a basic variable for both linear systems. However, since the systems have the same basic variables, $x_j$ must then be a free variable for each system. We now look at the solutions of the systems in which $x_j$ is set equal to $1$ and all other free variables are set equal to $0$. For this choice of parameters, the solution of the system for matrix $A$ has $x_j=-a$, while the solution of the system for matrix $B$ has $x_j=-b$, so that the two systems have different solutions.

In case $2$, there is a variable $x_i$ which is a basic variable for one matrix, let’s say $A$, and a free variable for the other matrix $B$. The system for matrix $B$ has a solution in which $x_i=1$ and $x_j=0$ for all other free variables $x_j$. However, by Proposition 1.4.1 this cannot be a solution of the system for the matrix $A$. This completes the proof of case $2$.

Now, we say that the matrix $B$ is equivalent to the matrix $A$ provided that $B$ can be obtained from $A$ by performing a sequence of elementary row operations beginning with $A$. The importance of this concept lies in the following result.

Theorem 1.4.1: Equivalent Matrices. The two linear systems of equations corresponding to two equivalent augmented matrices have exactly the same solutions.

Proof. The proof of this theorem is left as an exercise.

Now, we can use Lemma 1.4.1 and Theorem 1.4.1 to prove the main result of this section.

Theorem 1.4.2: Uniqueness of the Reduced Row-Echelon Form. Every matrix $A$ is equivalent to a unique matrix in reduced row-echelon form.

Proof. Let $A$ be an $m \times n$ matrix and let $B$ and $C$ be matrices in reduced row-echelon form, each equivalent to $A$. It suffices to show that $B=C$.

Let $A^{+}$ be the matrix $A$ augmented with a new rightmost column consisting entirely of zeros. Similarly, augment matrices $B$ and $C$ each with a rightmost column of zeros to obtain $B^{+}$ and $C^{+}$. Note that $B^{+}$ and $C^{+}$ are matrices in reduced row-echelon form which are obtained from $A^{+}$ by respectively applying the same sequence of elementary row operations which were used to obtain $B$ and $C$ from $A$.

Now, $A^{+}$, $B^{+}$, and $C^{+}$ can all be considered as augmented matrices of homogeneous linear systems in the variables $x_1, x_2, \cdots, x_n$. Because $B^{+}$ and $C^{+}$ are each equivalent to $A^{+}$, Theorem 1.4.1 ensures that all three homogeneous linear systems have exactly the same solutions. By Lemma 1.4.1 we conclude that $B^{+}=C^{+}$. By construction, we must also have $B=C$.

According to this theorem we can say that each matrix $A$ has a unique reduced row-echelon form.

1.5 Rank and Homogeneous Systems 秩与齐次方程组

There is a special type of system which requires additional study. This type of system is called a homogeneous system of equations, which we defined above in Definition 1.2.2. Our focus in this section is to consider what types of solutions are possible for a homogeneous system of equations.

Consider the following definition.

Definition 1.5.1: Trivial Solution. Consider the homogeneous system of equations given by $$\begin{array}{c} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}=0 \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}=0 \\ \vdots \\ a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}=0 \end{array}$$ Then, $x_{1} = 0, x_{2} = 0, \cdots, x_{n} =0$ is always a solution to this system. We call this the trivial solution.

If the system has a solution in which not all of the $x_1, \cdots, x_n$ are equal to zero, then we call this solution nontrivial. The trivial solution does not tell us much about the system, as it says that $0=0$! Therefore, when working with homogeneous systems of equations, we want to know when the system has a nontrivial solution.

Suppose we have a homogeneous system of $m$ equations, using $n$ variables, and suppose that $n > m$. In other words, there are more variables than equations. Then, it turns out that this system always has a nontrivial solution. Not only will the system have a nontrivial solution, but it also will have infinitely many solutions. It is also possible, but not required, to have a nontrivial solution if $n=m$ and $n

设我们有一个含 $n$ 个变量、$m$ 个方程的齐次方程组，且 $n > m$。换句话说，变量比方程多。事实证明，这个方程组总有非平凡解。方程组不仅会有非平凡解，而且它还会有无穷多个解。若 $n=m$ 或 $n

Consider the following example.

看下面的例子。

Example 1.5.1: Solutions to a Homogeneous System of Equations. Find the nontrivial solutions to the following homogeneous system of equations $$\begin{array}{c} 2x + y - z = 0 \\ x + 2y - 2z = 0 \end{array}.$$

示例 1.5.1：齐次方程组的解。求下面齐次方程组的非平凡解： $$\begin{array}{c} 2x + y - z = 0 \\ x + 2y - 2z = 0 \end{array}.$$

Solution. Notice that this system has $m = 2$ equations and $n = 3$ variables, so $n>m$. Therefore by our previous discussion, we expect this system to have infinitely many solutions.

解。注意这个方程组有 $m = 2$ 个方程、$n = 3$ 个变量，所以 $n>m$。因此，根据我们之前的讨论，可以预期这个方程组有无穷多个解。

The process we use to find the solutions for a homogeneous system of equations is the same process we used in the previous section. First, we construct the augmented matrix, given by $\left[ \begin{array}{rrr|r} 2 & 1 & -1 & 0 \\ 1 & 2 & -2 & 0 \end{array} \right]$. Then, we carry this matrix to its reduced row-echelon form, given below. $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{array} \right]$. The corresponding system of equations is $\begin{array}{c} x = 0 \\ y - z =0 \\ \end{array}$. Since $z$ is not restrained by any equation, we know that this variable will become our parameter. Let $z=t$ where $t$ is any number. Therefore, our solution has the form $\begin{array}{c} x = 0 \\ y = z = t \\ z = t \end{array}$. Hence this system has infinitely many solutions, with one parameter $t$.

我们用来求齐次方程组解的过程，与上一节所用的过程相同。首先构造增广矩阵，即 $\left[ \begin{array}{rrr|r} 2 & 1 & -1 & 0 \\ 1 & 2 & -2 & 0 \end{array} \right]$。然后把这个矩阵化为下面的简化行阶梯形：$\left[ \begin{array}{rrr|r} 1 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 \end{array} \right]$。对应的方程组是 $\begin{array}{c} x = 0 \\ y - z =0 \\ \end{array}$。由于 $z$ 不受任何方程约束，我们知道这个变量将成为我们的参数。令 $z=t$，其中 $t$ 为任意数。因此我们的解形如 $\begin{array}{c} x = 0 \\ y = z = t \\ z = t \end{array}$。于是这个方程组有无穷多个解，含一个参数 $t$。

Suppose we were to write the solution to the previous example in another form. Specifically, $\begin{array}{c} x = 0 \\ y = 0 + t \\ z = 0 + t \end{array}$ can be written as $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right].$$ Notice that we have constructed a column from the constants in the solution (all equal to $0$), as well as a column corresponding to the coefficients on $t$ in each equation. While we will discuss this form of solution more in further chapters, for now consider the column of coefficients of the parameter $t$. In this case, this is the column $\left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right]$.

设想我们用另一种形式来写上一示例的解。具体地说，$\begin{array}{c} x = 0 \\ y = 0 + t \\ z = 0 + t \end{array}$ 可以写成 $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right].$$ 注意我们由解中的常数（都等于 $0$）构造了一列，还由各方程中 $t$ 的系数构造了一列。虽然我们会在后续章节进一步讨论这种形式的解，但现在先考虑参数 $t$ 的系数列。本例中它就是列 $\left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right]$。

There is a special name for this column, which is basic solution. The basic solutions of a system are columns constructed from the coefficients on parameters in the solution. We often denote basic solutions by $X_1, X_2$ etc., depending on how many solutions occur. Therefore, Example 1.5.1 has the basic solution $X_1 = \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right]$.

这一列有一个专门的名称，叫做基础解（basic solution）。方程组的基础解是由解中参数的系数构造出来的列。我们常根据出现的数目，用 $X_1, X_2$ 等来记基础解。因此，示例 1.5.1 有基础解 $X_1 = \left[ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right]$。

We explore this further in the following example.

我们在下面的例子中进一步探讨这一点。

Example 1.5.2: Basic Solutions of a Homogeneous System. Consider the following homogeneous system of equations. $\begin{array}{c} x + 4y + 3z = 0 \\ 3x + 12y + 9z = 0 \end{array}$. Find the basic solutions to this system.

示例 1.5.2：齐次方程组的基础解。考虑下面的齐次方程组：$\begin{array}{c} x + 4y + 3z = 0 \\ 3x + 12y + 9z = 0 \end{array}$。求这个方程组的基础解。

Solution. The augmented matrix of this system and the resulting reduced row-echelon form are $\left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 3 & 12 & 9 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$. When written in equations, this system is given by $x + 4y +3z=0$. Notice that only $x$ corresponds to a pivot column. In this case, we will have two parameters, one for $y$ and one for $z$. Let $y = s$ and $z=t$ for any numbers $s$ and $t$. Then, our solution becomes $\begin{array}{c} x = -4s - 3t \\ y = s \\ z = t \end{array}$, which can be written as $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right] + s \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right].$$ You can see here that we have two columns of coefficients corresponding to parameters, specifically one for $s$ and one for $t$. Therefore, this system has two basic solutions! These are $X_1= \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right], X_2 = \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right]$.

解。这个方程组的增广矩阵及其化简得到的简化行阶梯形为 $\left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 3 & 12 & 9 & 0 \end{array} \right] \rightarrow \cdots \rightarrow \left[ \begin{array}{rrr|r} 1 & 4 & 3 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right]$。写成方程时，该方程组由 $x + 4y +3z=0$ 给出。注意只有 $x$ 对应一个主元列。这种情况下我们将有两个参数，一个用于 $y$，一个用于 $z$。对任意数 $s$ 和 $t$，令 $y = s$、$z=t$。于是解变成 $\begin{array}{c} x = -4s - 3t \\ y = s \\ z = t \end{array}$，可写成 $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right] + s \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + t \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right].$$ 这里可以看出，我们有两列对应参数的系数列，具体说一列对应 $s$、一列对应 $t$。因此这个方程组有两个基础解！它们是 $X_1= \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right]$、$X_2 = \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right]$。

We now present a new definition.

现在我们介绍一个新定义。

Definition 1.5.2: Linear Combination. Let $X_1,\cdots ,X_n,V$ be column matrices. Then $V$ is said to be a linear combination of the columns $X_1,\cdots , X_n$ if there exist scalars, $a_{1},\cdots ,a_{n}$ such that $V = a_1 X_1 + \cdots + a_n X_n$.

定义 1.5.2：线性组合。设 $X_1,\cdots ,X_n,V$ 是列矩阵。若存在纯量 $a_{1},\cdots ,a_{n}$，使得 $V = a_1 X_1 + \cdots + a_n X_n$，则称 $V$ 是各列 $X_1,\cdots , X_n$ 的一个线性组合。

A remarkable result of this section is that a linear combination of the basic solutions is again a solution to the system. Even more remarkable is that every solution can be written as a linear combination of these solutions. Therefore, if we take a linear combination of the two solutions to Example 1.5.2, this would also be a solution. For example, we could take the following linear combination $$3 \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + 2 \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right].$$ You should take a moment to verify that $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right]$$ is in fact a solution to the system in Example 1.5.2.

本节的一个引人注目的结果是：基础解的线性组合仍然是该方程组的解。更值得注意的是，每个解都可以写成这些解的线性组合。因此，如果我们取示例 1.5.2 中两个解的线性组合，那也会是一个解。例如，我们可以取如下的线性组合 $$3 \left[ \begin{array}{r} -4 \\ 1 \\ 0 \end{array} \right] + 2 \left[ \begin{array}{r} -3 \\ 0 \\ 1 \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right].$$ 你不妨花一点时间验证 $$\left[ \begin{array}{r} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{r} -18 \\ 3 \\ 2 \end{array} \right]$$ 事实上确实是示例 1.5.2 中方程组的解。

Another way in which we can find out more information about the solutions of a homogeneous system is to consider the rank of the associated coefficient matrix. We now define what is meant by the rank of a matrix.

要想了解齐次方程组解的更多信息，另一种途径是考虑相关复系数矩阵的秩（rank）。下面我们就来定义矩阵的秩。

Definition 1.5.3: Rank of a Matrix. Let $A$ be a matrix and consider any row-echelon form of $A$. Then, the number $r$ of leading entries of $A$ does not depend on the row-echelon form you choose, and is called the rank of $A$. We denote it by Rank($A$).

定义 1.5.3：矩阵的秩。设 $A$ 是一个矩阵，考虑 $A$ 的任意一个行阶梯形。那么 $A$ 的首项数目 $r$ 与你所选择的行阶梯形无关，称为 $A$ 的秩。我们用 Rank($A$) 记它。

Similarly, we could count the number of pivot positions (or pivot columns) to determine the rank of $A$.

类似地，我们也可以通过数主元位置（或主元列）的个数来确定 $A$ 的秩。

Example 1.5.3: Finding the Rank of a Matrix. Consider the matrix $\left[ \begin{array}{rrr} 1 & 2 & 3 \\ 1 & 5 & 9 \\ 2 & 4 & 6 \end{array} \right]$. What is its rank?

示例 1.5.3：求矩阵的秩。考虑矩阵 $\left[ \begin{array}{rrr} 1 & 2 & 3 \\ 1 & 5 & 9 \\ 2 & 4 & 6 \end{array} \right]$。它的秩是多少？

Solution. First, we need to find the reduced row-echelon form of $A$. Through the usual algorithm, we find that this is $\left[ \begin{array}{rrr} \fbox{1} & 0 & -1 \\ 0 & \fbox{1} & 2 \\ 0 & 0 & 0 \end{array} \right]$. Here we have two leading entries, or two pivot positions, shown above in boxes. The rank of $A$ is $r = 2.$

解。首先，我们需要求出 $A$ 的简化行阶梯形。通过通常的算法，我们求得结果是 $\left[ \begin{array}{rrr} \fbox{1} & 0 & -1 \\ 0 & \fbox{1} & 2 \\ 0 & 0 & 0 \end{array} \right]$。这里有三个首项（即两个主元位置），如上用方框标出。$A$ 的秩是 $r = 2.$

Notice that we would have achieved the same answer if we had found the row-echelon form of $A$ instead of the reduced row-echelon form.

注意，如果我们求的是 $A$ 的行阶梯形而不是简化行阶梯形，也会得到同样的答案。

Suppose we have a homogeneous system of $m$ equations in $n$ variables, and suppose that $n > m$. From our above discussion, we know that this system will have infinitely many solutions. If we consider the rank of the coefficient matrix of this system, we can find out even more about the solution. Note that we are looking at just the coefficient matrix, not the entire augmented matrix.

设我们有一个含 $n$ 个变量、$m$ 个方程的齐次方程组，且 $n > m$。由上面的讨论可知，该系统会有无穷多个解。如果我们考虑该系统系数矩阵的秩，还能进一步了解解的情况。注意我们只看系数矩阵，而不是整个增广矩阵。

Theorem 1.5.1: Rank and Solutions to a Homogeneous System. Let $A$ be the $m \times n$ coefficient matrix corresponding to a homogeneous system of equations, and suppose $A$ has rank $r$. Then, the solution to the corresponding system has $n-r$ parameters.

定理 1.5.1：秩与齐次方程组的解。设 $A$ 是对应于某个齐次方程组的 $m \times n$ 系数矩阵，并设 $A$ 的秩为 $r$。则该方程组的解含有 $n-r$ 个参数。

Consider our above Example 1.5.2 in the context of this theorem. The system in this example has $m = 2$ equations in $n = 3$ variables. First, because $n>m$, we know that the system has a nontrivial solution, and therefore infinitely many solutions. This tells us that the solution will contain at least one parameter. The rank of the coefficient matrix can tell us even more about the solution! The rank of the coefficient matrix of the system is $1$, as it has one leading entry in row-echelon form. Theorem 1.5.1 tells us that the solution will have $n-r = 3-1 = 2$ parameters. You can check that this is true in the solution to Example 1.5.2.

在这个定理的背景下，考虑我们上面的示例 1.5.2。该例中的方程组有 $m = 2$ 个方程、$n = 3$ 个变量。首先，因为 $n>m$，我们知道方程组有非平凡解，因此有无穷多个解。这告诉我们解将至少含有一个参数。系数矩阵的秩还能告诉我们更多的信息！该系统系数矩阵的秩是 $1$，因为它在行阶梯形中有一个首项。定理 1.5.1 告诉我们解将有 $n-r = 3-1 = 2$ 个参数。你可以验证这在示例 1.5.2 的解中确实成立。

Notice that if $n=m$ or $n

注意，若 $n=m$ 或 $n

We are not limited to homogeneous systems of equations here. The rank of a matrix can be used to learn about the solutions of any system of linear equations. In the previous section, we discussed that a system of equations can have no solution, a unique solution, or infinitely many solutions. Suppose the system is consistent, whether it is homogeneous or not. The following theorem tells us how we can use the rank to learn about the type of solution we have.

这里我们不局限于齐次方程组。矩阵的秩可以用来了解任何线性方程组的解的情况。上一节我们讨论过，方程组可以无解、有唯一解或有无穷多个解。设方程组是相容的（无论是否齐次）。下面的定理告诉我们如何利用秩来了解我们所处的是哪种类型的解。

Theorem 1.5.2: Rank and Solutions to a Consistent System of Equations. Let $A$ be the $m \times \left( n+1 \right)$ augmented matrix corresponding to a consistent system of equations in $n$ variables, and suppose $A$ has rank $r$. Then

定理 1.5.2：秩与相容方程组的解。设 $A$ 是对应于含 $n$ 个变量的相容方程组的 $m \times \left( n+1 \right)$ 增广矩阵，并设 $A$ 的秩为 $r$。则

1. the system has a unique solution if $r = n$
2. the system has infinitely many solutions if $r < n$

1. 若 $r = n$，则方程组有唯一解；
2. 若 $r < n$，则方程组有无穷多个解。

We will not present a formal proof of this, but consider the following discussions.

我们不会给出正式的证明，但请看下面的讨论。

1. No Solution The above theorem assumes that the system is consistent, that is, that it has a solution. It turns out that it is possible for the augmented matrix of a system with no solution to have any rank $r$ as long as $r>1$. Therefore, we must know that the system is consistent in order to use this theorem!
2. Unique Solution Suppose $r=n$. Then, there is a pivot position in every column of the coefficient matrix of $A$. Hence, there is a unique solution.
3. Infinitely Many Solutions Suppose $r

1. 无解 上面的定理假定方程组是相容的，也就是说它有解。事实证明，一个无解的方程组，其增广矩阵的秩可以是任何满足 $r>1$ 的 $r$。因此，我们必须先知道方程组是相容的，才能使用这个定理！
2. 唯一解 设 $r=n$。则 $A$ 的系数矩阵每一列都有一个主元位置，因此有唯一解。
3. 无穷多个解 设 $r

1.6 Balancing Chemical Reactions 化学反应的配平

The tools of linear algebra can also be used in the subject area of Chemistry, specifically for balancing chemical reactions.

线性代数的工具也可以用于化学领域，特别是用于配平化学反应。

Consider the chemical reaction $SnO_{2}+H_{2}\rightarrow Sn+H_{2}O$ Here the elements involved are tin ($Sn$), oxygen ($O$), and hydrogen ($H$). A chemical reaction occurs and the result is a combination of tin ($Sn$) and water ($H_{2}O$). When considering chemical reactions, we want to investigate how much of each element we began with and how much of each element is involved in the result.

考虑化学反应 $SnO_{2}+H_{2}\rightarrow Sn+H_{2}O$。这里涉及的元素是锡（$Sn$）、氧（$O$）和氢（$H$）。发生化学反应后，结果是锡（$Sn$）与水的组合。在研究化学反应时，我们想弄清反应开始时每种元素各有多少，以及结果中又各涉及多少每种元素。

An important theory we will use here is the mass balance theory. It tells us that we cannot create or delete elements within a chemical reaction. For example, in the above expression, we must have the same number of oxygen, tin, and hydrogen on both sides of the reaction. Notice that this is not currently the case. For example, there are two oxygen atoms on the left and only one on the right. In order to fix this, we want to find numbers $x,y,z,w$ such that $xSnO_{2}+yH_{2}\rightarrow zSn+wH_{2}O$ where both sides of the reaction have the same number of atoms of the various elements.

这里我们要用到的一个重要理论是质量平衡理论。它告诉我们，在化学反应中不能创造或消灭元素。例如，在上面的表达式中，反应两边的氧、锡、氢的个数必须相同。注意目前并非如此。例如，左边有两个氧原子，而右边只有一个。为了解决这个问题，我们希望找到数 $x,y,z,w$，使得 $xSnO_{2}+yH_{2}\rightarrow zSn+wH_{2}O$，其中反应两边各种元素的原子个数相同。

This is a familiar problem. We can solve it by setting up a system of equations in the variables $x,y,z,w$. Thus you need $\begin{array}{cl} Sn: & x=z \\ O: & 2x=w \\ H: & 2y=2w \end{array}$

这是一个熟悉的问题。我们可以通过建立含变量 $x,y,z,w$ 的方程组来求解。因此需要 $\begin{array}{cl} Sn: & x=z \\ O: & 2x=w \\ H: & 2y=2w \end{array}$

We can rewrite these equations as $\begin{array}{cl} Sn: & x - z = 0 \\ O: & 2x - w = 0 \\ H: & 2y - 2w = 0 \end{array}$

我们可以把这些方程改写成 $\begin{array}{cl} Sn: & x - z = 0 \\ O: & 2x - w = 0 \\ H: & 2y - 2w = 0 \end{array}$

The augmented matrix for this system of equations is given by $\left[ \begin{array}{rrrr|r} 1 & 0 & -1 & 0 & 0 \\ 2 & 0 & 0 & -1 & 0 \\ 0 & 2 & 0 & -2 & 0 \end{array} \right]$

这个方程组的增广矩阵为 $\left[ \begin{array}{rrrr|r} 1 & 0 & -1 & 0 & 0 \\ 2 & 0 & 0 & -1 & 0 \\ 0 & 2 & 0 & -2 & 0 \end{array} \right]$

The reduced row-echelon form of this matrix is $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -\frac{1}{2} & 0 \end{array} \right]$

这个矩阵的简化行阶梯形为 $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -\frac{1}{2} & 0 \end{array} \right]$

The solution is given by $\begin{array}{c} x - \frac{1}{2} w = 0 \\ y - w = 0 \\ z - \frac{1}{2}w = 0 \end{array}$

解为 $\begin{array}{c} x - \frac{1}{2} w = 0 \\ y - w = 0 \\ z - \frac{1}{2}w = 0 \end{array}$

which we can write as $\begin{array}{c} x = \frac{1}{2} t \\ y = t \\ z = \frac{1}{2}t \\ w = t \end{array}$

我们可以把它写成 $\begin{array}{c} x = \frac{1}{2} t \\ y = t \\ z = \frac{1}{2}t \\ w = t \end{array}$

For example, let $w=2$ and this would yield $x=1,y=2,$ and $z=1.$ We can put these values back into the expression for the reaction which yields $SnO_{2}+2H_{2}\rightarrow Sn+2H_{2}O$ Observe that each side of the expression contains the same number of atoms of each element. This means that it preserves the total number of atoms, as required, and so the chemical reaction is balanced.

例如，令 $w=2$，则得到 $x=1,y=2$ 和 $z=1$。我们可以把这些值代回反应的表达式，得到 $SnO_{2}+2H_{2}\rightarrow Sn+2H_{2}O$。可以观察到表达式的每一边都含有每种元素的相同原子个数。这意味着它保持了原子总数不变，满足要求，因此这个化学反应配平了。

Example 1.6.1: Balancing a Chemical Reaction 配平一个化学反应

Potassium is denoted by $K,$ oxygen by $O,$ phosphorus by $P$ and hydrogen by $H$. Consider the reaction given by $KOH+H_{3}PO_{4}\rightarrow K_{3}PO_{4}+H_{2}O$

钾记作 $K$，氧记作 $O$，磷记作 $P$，氢记作 $H$。考虑如下的反应：$KOH+H_{3}PO_{4}\rightarrow K_{3}PO_{4}+H_{2}O$

Balance this chemical reaction.

配平这个化学反应。

Solution. We will use the same procedure as above to solve this problem. We need to find values for $x,y,z,w$ such that $xKOH+yH_{3}PO_{4}\rightarrow zK_{3}PO_{4}+wH_{2}O$ preserves the total number of atoms of each element.

解。我们用与上面相同的步骤来解决这个问题。需要找到 $x,y,z,w$ 的值，使得 $xKOH+yH_{3}PO_{4}\rightarrow zK_{3}PO_{4}+wH_{2}O$ 保持每种元素的总原子数不变。

Finding these values can be done by finding the solution to the following system of equations. $\begin{array}{cl} K: & x=3z \\ O: & x+4y=4z+w \\ H: & x+3y=2w \\ P: & y=z \end{array}$

只需求出如下方程组的解，即可找到这些值。$\begin{array}{cl} K: & x=3z \\ O: & x+4y=4z+w \\ H: & x+3y=2w \\ P: & y=z \end{array}$

The augmented matrix for this system is $\left[ \begin{array}{rrrr|r} 1 & 0 & -3 & 0 & 0 \\ 1 & 4 & -4 & -1 & 0 \\ 1 & 3 & 0 & -2 & 0 \\ 0 & 1 & -1 & 0 & 0 \end{array} \right]$ and the reduced row-echelon form is $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -\frac{1}{3} & 0 \\ 0 & 0 & 1 & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

这个方程组的增广矩阵为 $\left[ \begin{array}{rrrr|r} 1 & 0 & -3 & 0 & 0 \\ 1 & 4 & -4 & -1 & 0 \\ 1 & 3 & 0 & -2 & 0 \\ 0 & 1 & -1 & 0 & 0 \end{array} \right]$，其简化行阶梯形为 $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & -\frac{1}{3} & 0 \\ 0 & 0 & 1 & -\frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

The solution is given by $\begin{array}{c} x - w = 0 \\ y - \frac{1}{3}w = 0 \\ z - \frac{1}{3}w = 0 \end{array}$ which can be written as $\begin{array}{c} x = t \\ y = \frac{1}{3}t \\ z = \frac{1}{3}t \\ w = t \end{array}$

解为 $\begin{array}{c} x - w = 0 \\ y - \frac{1}{3}w = 0 \\ z - \frac{1}{3}w = 0 \end{array}$，可写成 $\begin{array}{c} x = t \\ y = \frac{1}{3}t \\ z = \frac{1}{3}t \\ w = t \end{array}$

Choose a value for $t$, say $3$. Then $w=3$ and this yields $x=3,y=1,z=1.$ It follows that the balanced reaction is given by $3KOH+1H_{3}PO_{4}\rightarrow 1K_{3}PO_{4}+3H_{2}O$ Note that this results in the same number of atoms on both sides.

为 $t$ 取一个值，例如取 $3$。则 $w=3$，由此得到 $x=3,y=1,z=1$。于是配平后的反应为 $3KOH+1H_{3}PO_{4}\rightarrow 1K_{3}PO_{4}+3H_{2}O$。注意这使两边的原子个数相同。

Of course these numbers you are finding would typically be the number of moles of the molecules on each side. Thus three moles of $KOH$ added to one mole of $H_{3}PO_{4}$ yields one mole of $K_{3}PO_{4}$ and three moles of $H_{2}O$.

当然，你找到的这些数通常就是每一侧分子的摩尔数。因此 $3$ 摩尔 $KOH$ 加入 $1$ 摩尔 $H_{3}PO_{4}$，会生成 $1$ 摩尔 $K_{3}PO_{4}$ 和 $3$ 摩尔 $H_{2}O$。

1.7 Dimensionless Variables 无量纲变量

This section shows how solving systems of equations can be used to determine appropriate dimensionless variables. It is only an introduction to this topic and considers a specific example of a simple airplane wing shown below. We assume for simplicity that it is a flat plane at an angle to the wind which is blowing against it with speed $V$ as shown.

本节说明如何用解方程组的方法来确定合适的无量纲变量。这只是对该主题的一个介绍，它考察下面所示的一个简单飞机机翼的具体例子。为简单起见，我们假设它是一个与风成一定角度的平面，风以速度 $V$ 迎面吹来，如图所示。

The angle $\theta$ is called the angle of incidence, $B$ is the span of the wing and $A$ is called the chord. Denote by $l$ the lift. Then this should depend on various quantities like $\theta ,V,B,A$ and so forth. Here is a table which indicates various quantities on which it is reasonable to expect $l$ to depend.

角 $\theta$ 称为入射角，$B$ 是机翼的翼展，$A$ 称为弦长。用 $l$ 表示升力。那么升力应依赖于各种量，如 $\theta ,V,B,A$ 等等。下面是一张表，列出了合理地预期 $l$ 所依赖的各种量。

Table 1.7.1 · 表 1.7.1

变量 Variable	符号 Symbol	单位 Units
弦长 chord	$A$	$m$
翼展 span	$B$	$m$
入射角 angle of incidence	$\theta$	$m^{0}kg^{0}\sec^{0}$
风速 speed of wind	$V$	$m\,\sec^{-1}$
声速 speed of sound	$V_{0}$	$m\,\sec^{-1}$
空气密度 density of air	$\rho$	$kg\,m^{-3}$
黏度 viscosity	$\mu$	$kg\,\sec^{-1}m^{-1}$
升力 lift	$l$	$kg\,\sec^{-2}m$

Here $m$ denotes meters, $\sec$ refers to seconds and $kg$ refers to kilograms. All of these are likely familiar except for $\mu$, which we will discuss in further detail now.

这里 $m$ 表示米，$\sec$ 指秒，$kg$ 指千克。这些量你可能都很熟悉，唯独 $\mu$ 可能不熟，我们现在就来更详细地讨论它。

Viscosity is a measure of how much internal friction is experienced when the fluid moves. It is roughly a measure of how “sticky” the fluid is. Consider a piece of area parallel to the direction of motion of the fluid. To say that the viscosity is large is to say that the tangential force applied to this area must be large in order to achieve a given change in speed of the fluid in a direction normal to the tangential force. Thus $\mu \left( \text{area}\right) \left( \text{velocity gradient}\right) =\text{ tangential force}$ Hence $\left( \text{units on }\mu \right) m^{2}\left( \frac{m}{\sec\, m}\right) =kg\sec ^{-2}m$ Thus the units on $\mu$ are $kg\sec ^{-1}m^{-1}$ as claimed above.

黏度是流体运动时内部摩擦大小的度量，粗略地讲就是流体有多“黏”。考虑一块与流体运动方向平行的面积。说黏度很大，就是说施加在这块面积上的切向力必须很大，才能在垂直于切向力的方向上使流体速度产生给定的变化。因此 $\mu \left( \text{面积}\right) \left( \text{速度梯度}\right) =\text{切向力}$。于是 $\left( \mu \text{的单位}\right) m^{2}\left( \frac{m}{\sec\, m}\right) =kg\sec ^{-2}m$。因此 $\mu$ 的单位为 $kg\sec ^{-1}m^{-1}$，正如上面所述。

Returning to our original discussion, you may think that we would want $l=f\left( A,B,\theta ,V,V_{0},\rho ,\mu \right)$ This is very cumbersome because it depends on seven variables. Also, it is likely that without much care, a change in the units such as going from meters to feet would result in an incorrect value for $l$. The way to get around this problem is to look for $l$ as a function of dimensionless variables multiplied by something which has units of force. It is helpful because first of all, you will likely have fewer independent variables and secondly, you could expect the formula to hold independent of the way of specifying length, mass and so forth. One looks for $l=f\left( g_{1},\cdots ,g_{k}\right) \rho V^{2}AB$ where the units on $\rho V^{2}AB$ are $\frac{kg}{m^{3}}\left( \frac{m}{\sec }\right) ^{2}m^{2}=\frac{kg\times m}{ \sec ^{2}}$ which are the units of force. Each of these $g_{i}$ is of the form $A^{x_{1}}B^{x_{2}}\theta ^{x_{3}}V^{x_{4}}V_{0}^{x_{5}}\rho ^{x_{6}}\mu ^{x_{7}}$ and each $g_{i}$ is independent of the dimensions. That is, this expression must not depend on meters, kilograms, seconds, etc. Thus, placing in the units for each of these quantities, one needs $m^{x_{1}}m^{x_{2}}\left( m^{x_{4}}\sec ^{-x_{4}}\right) \left( m^{x_{5}}\sec ^{-x_{5}}\right) \left( kgm^{-3}\right) ^{x_{6}}\left( kg\sec ^{-1}m^{-1}\right) ^{x_{7}}=m^{0}kg^{0}\sec ^{0}$ Notice that there are no units on $\theta$ because it is just the radian measure of an angle. Hence its dimensions consist of length divided by length, thus it is dimensionless. Then this leads to the following equations for the $x_{i}$.

回到最初的讨论，你可能认为我们想要 $l=f\left( A,B,\theta ,V,V_{0},\rho ,\mu \right)$。这非常繁琐，因为它依赖七个变量。而且，如果不加小心，更改单位（例如从米改成英尺）很可能会得到错误的 $l$ 值。绕过这个问题的方法，是把 $l$ 看作无量纲变量的函数再乘以某个具有力单位的量。这样做是有益的，因为首先你可以拥有更少的独立变量，其次你可以期望该公式与长度、质量等的指定方式无关。我们寻找 $l=f\left( g_{1},\cdots ,g_{k}\right) \rho V^{2}AB$，其中 $\rho V^{2}AB$ 的单位为 $\frac{kg}{m^{3}}\left( \frac{m}{\sec }\right) ^{2}m^{2}=\frac{kg\times m}{ \sec ^{2}}$，这正是力的单位。这里的每一个 $g_{i}$ 都具有形式 $A^{x_{1}}B^{x_{2}}\theta ^{x_{3}}V^{x_{4}}V_{0}^{x_{5}}\rho ^{x_{6}}\mu ^{x_{7}}$，并且每个 $g_{i}$ 都与量纲无关。也就是说，这个表达式不能依赖于米、千克、秒等。因此，把它们各自的单位代入这些量，就需要 $m^{x_{1}}m^{x_{2}}\left( m^{x_{4}}\sec ^{-x_{4}}\right) \left( m^{x_{5}}\sec ^{-x_{5}}\right) \left( kgm^{-3}\right) ^{x_{6}}\left( kg\sec ^{-1}m^{-1}\right) ^{x_{7}}=m^{0}kg^{0}\sec ^{0}$。注意 $\theta$ 没有单位，因为它只是角的弧度度量。因此它的量纲是长度除以长度，所以它是无量纲的。于是，这就导出了关于 $x_{i}$ 的下述方程。

$$\begin{array}{cc} m: & x_{1}+x_{2}+x_{4}+x_{5}-3x_{6}-x_{7}=0 \\ \sec: & -x_{4}-x_{5}-x_{7}=0 \\ kg: & x_{6}+x_{7}=0 \end{array}$$

The augmented matrix for this system is

这个方程组的增广矩阵为

$\left[ \begin{array}{rrrrrrr|r} 1 & 1 & 0 & 1 & 1 & -3 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right]$ The reduced row-echelon form is given by

$\left[ \begin{array}{rrrrrrr|r} 1 & 1 & 0 & 1 & 1 & -3 & -1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right]$，其简化行阶梯形为

$\left[ \begin{array}{rrrrrrr|r} 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right]$ and so the solutions are of the form $\begin{aligned} x_{1} &= -x_{2}-x_{7} \\ x_{3} &= x_{3} \\ x_{4} &= -x_{5}-x_{7} \\ x_{6} &= -x_{7}\end{aligned}$ Thus, in terms of vectors, the solution is $\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \\ x_{7} \end{array} \right] =\left[ \begin{array}{c} -x_{2}-x_{7} \\ x_{2} \\ x_{3} \\ -x_{5}-x_{7} \\ x_{5} \\ -x_{7} \\ x_{7} \end{array} \right]$ Thus the free variables are $x_{2},x_{3},x_{5},x_{7}.$ By assigning values to these, we can obtain dimensionless variables by placing the values obtained for the $x_{i}$ in the above equation. For example, let $x_{2}=1$ and all the rest of the free variables are 0. This yields $x_{1}=-1,x_{2}=1,x_{3}=0,x_{4}=0,x_{5}=0,x_{6}=0,x_{7}=0$ The dimensionless variable is then $A^{-1}B^{1}$. This is the ratio between the span and the chord. It is called the aspect ratio, denoted as $AR$. Next let $x_{3}=1$ and all others equal zero. This gives for a dimensionless quantity the angle $\theta$. Next let $x_{5}=1$ and all others equal zero. This gives $x_{1}=0,x_{2}=0,x_{3}=0,x_{4}=-1,x_{5}=1,x_{6}=0,x_{7}=0$ Then the dimensionless variable is $V^{-1}V_{0}^{1}.$ However, it is written as $V/V_{0}$. This is called the Mach number $\mathcal{M}$. Finally, let $x_{7}=1$ and all the other free variables equal 0. Then $x_{1}=-1,x_{2}=0,x_{3}=0,x_{4}=-1,x_{5}=0,x_{6}=-1,x_{7}=1$ then the dimensionless variable which results from this is $A^{-1}V^{-1}\rho ^{-1}\mu .$ It is customary to write it as $Re=\left( AV\rho \right) /\mu$. This one is called the Reynold’s number. It is the one which involves viscosity. Thus we would look for $l=f\left(Re,AR,\theta ,\mathcal{M}\right) kg\times m/\sec ^{2}$ This is quite interesting because it is easy to vary $Re$ by simply adjusting the velocity or $A$ but it is hard to vary things like $\mu$ or $\rho$. Note that all the quantities are easy to adjust. Now this could be used, along with wind tunnel experiments to get a formula for the lift which would be reasonable. You could also consider more variables and more complicated situations in the same way.

$\left[ \begin{array}{rrrrrrr|r} 1 & 1 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array} \right]$，因此解具有形式 $\begin{aligned} x_{1} &= -x_{2}-x_{7} \\ x_{3} &= x_{3} \\ x_{4} &= -x_{5}-x_{7} \\ x_{6} &= -x_{7}\end{aligned}$。于是用向量表示时，解为 $\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \\ x_{5} \\ x_{6} \\ x_{7} \end{array} \right] =\left[ \begin{array}{c} -x_{2}-x_{7} \\ x_{2} \\ x_{3} \\ -x_{5}-x_{7} \\ x_{5} \\ -x_{7} \\ x_{7} \end{array} \right]$。于是自由变量为 $x_{2},x_{3},x_{5},x_{7}$。通过为这些变量赋值，并把所得 $x_{i}$ 的值代入上述等式，就可以得到无量纲变量。例如，令 $x_{2}=1$，其余自由变量都为 $0$。这得到 $x_{1}=-1,x_{2}=1,x_{3}=0,x_{4}=0,x_{5}=0,x_{6}=0,x_{7}=0$，于是无量纲变量为 $A^{-1}B^{1}$。这是翼展与弦长之比，称为展弦比，记作 $AR$。接下来令 $x_{3}=1$，其他都为 $0$，这就给出一无量纲量：角 $\theta$。再令 $x_{5}=1$，其他都为 $0$，得到 $x_{1}=0,x_{2}=0,x_{3}=0,x_{4}=-1,x_{5}=1,x_{6}=0,x_{7}=0$，则无量纲变量为 $V^{-1}V_{0}^{1}$，不过通常写成 $V/V_{0}$。这称为马赫数 $\mathcal{M}$。最后，令 $x_{7}=1$，其余自由变量都为 $0$，则 $x_{1}=-1,x_{2}=0,x_{3}=0,x_{4}=-1,x_{5}=0,x_{6}=-1,x_{7}=1$，由此得到的无量纲变量为 $A^{-1}V^{-1}\rho ^{-1}\mu$，习惯上写成 $Re=\left( AV\rho \right) /\mu$。这个称之为雷诺数，它是与黏度有关的那个量。于是我们要寻找 $l=f\left(Re,AR,\theta ,\mathcal{M}\right) kg\times m/\sec ^{2}$。这相当有趣，因为只需调整速度或 $A$，就很容易改变 $Re$，但要改变像 $\mu$ 或 $\rho$ 这样的量却很难。注意所有这些量都容易调整。现在，这就可以与风洞实验一起使用，从而得到可用的升力公式。你也可以用同样的方法考察更多变量和更复杂的情形。

1.8 An Application to Resistor Networks 电阻网络的应用

The tools of linear algebra can be used to study the application of resistor networks. An example of an electrical circuit is below.

线性代数的工具可以用来研究电阻网络这一应用。下面是一个电路示例。

The jagged lines () denote resistors and the numbers next to them give their resistance in ohms, written as $\Omega$. The voltage source () causes the current to flow in the direction from the shorter of the two lines toward the longer (as indicated by the arrow). The current for a circuit is labeled $I_k$.

锯齿状线段（电阻器的符号）表示电阻，它们旁边的数字给出以欧姆计的电阻值，记作 $\Omega$。电压源（符号）使电流从两条线中较短的一条流向较长的一条（如箭头所示）。电路中某电流记作 $I_k$。

In the above figure, the current $I_1$ has been labeled with an arrow in the counter clockwise direction. This is an entirely arbitrary decision and we could have chosen to label the current in the clockwise direction. With our choice of direction here, we define a positive current to flow in the counter clockwise direction and a negative current to flow in the clockwise direction.

在上图中，电流 $I_1$ 被标上逆时针方向的箭头。这完全是个任意的决定，我们也可以选择把电流标成顺时针方向。采用这里的流向选择，我们把沿逆时针方向流动的电流定义为正的，把沿顺时针方向流动的电流定义为负的。

The goal of this section is to use the values of resistors and voltage sources in a circuit to determine the current. An essential theorem for this application is Kirchhoff’s law.

本节的目标是利用电路中电阻和电压源的值来确定电流。这一应用的重要定理是基尔霍夫定律。

Theorem 1.8.1: Kirchhoff’s Law 基尔霍夫定律

The sum of the resistance ($R$) times the amps ($I$) in the counter clockwise direction around a loop equals the sum of the voltage sources ($V$) in the same direction around the loop.

沿回路逆时针方向，各电阻（$R$）乘以电流（$I$）之和，等于沿同一方向的各路电压源（$V$）之和。

Kirchhoff’s law allows us to set up a system of linear equations and solve for any unknown variables. When setting up this system, it is important to trace the circuit in the counter clockwise direction. If a resistor or voltage source is crossed against this direction, the related term must be given a negative sign.

基尔霍夫定律使我们能够建立线性方程组并求解任何未知变量。在建立该方程组时，按逆时针方向沿电路走一圈很重要。如果以相反方向经过某个电阻或电压源，相关项就必须取负号。

We will explore this in the next example where we determine the value of the current in the initial diagram.

我们将在下一个示例中探讨这一点，在那里我们求出初始图中电流的值。

Example 1.8.1: Solving for Current 求电流

Applying Kirchhoff’s Law to the diagram below, determine the value for $I_1$.

对下图运用基尔霍夫定律，确定 $I_1$ 的值。

Solution. Begin in the bottom left corner, and trace the circuit in the counter clockwise direction. At the first resistor, multiplying resistance and current gives $2I_1$. Continuing in this way through all three resistors gives $2I_1 + 4I_1 + 2 I_1$. This must equal the voltage source in the same direction. Notice that the direction of the voltage source matches the counter clockwise direction specified, so the voltage is positive.

解。从左下角开始，沿逆时针方向沿电路走一圈。经过第一个电阻时，把电阻与电流相乘得 $2I_1$。以此类推通过三个电阻，得到 $2I_1 + 4I_1 + 2 I_1$。这必须等于同一方向的电压源。注意电压源的方向与所指定的逆时针方向一致，所以电压为正。

Therefore the equation and solution are given by $\begin{aligned} 2I_1 + 4I_1 + 2 I_1 &= 18 \\ 8I_1 &= 18 \\ I_1 &= \frac{9}{4} A\end{aligned}$

因此方程和解为 $\begin{aligned} 2I_1 + 4I_1 + 2 I_1 &= 18 \\ 8I_1 &= 18 \\ I_1 &= \frac{9}{4} A\end{aligned}$

Since the answer is positive, this confirms that the current flows counter clockwise.

因为答案是正的，这确认电流沿逆时针方向流动。

Example 1.8.2: Solving for Current 求电流

Applying Kirchhoff’s Law to the diagram below, determine the value for $I_1$.

对下图运用基尔霍夫定律，确定 $I_1$ 的值。

Solution. Begin in the top left corner this time, and trace the circuit in the counter clockwise direction. At the first resistor, multiplying resistance and current gives $4I_1$. Continuing in this way through the four resistors gives $4I_1 + 6I_1 + 1 I_1 + 3I_1$. This must equal the voltage source in the same direction. Notice that the direction of the voltage source is opposite to the counter clockwise direction, so the voltage is negative.

解。这次从左上角开始，沿逆时针方向沿电路走一圈。经过第一个电阻时，把电阻与电流相乘得 $4I_1$。以此类推通过四个电阻，得到 $4I_1 + 6I_1 + 1 I_1 + 3I_1$。这必须等于同一方向的电压源。注意电压源的方向与逆时针方向相反，所以电压为负。

Therefore the equation and solution are given by $\begin{aligned} 4I_1 + 6I_1 + 1 I_1 + 3I_1 &= -27 \\ 14I_1 &= -27 \\ I_1 &= -\frac{27}{14} A\end{aligned}$

因此方程和解为 $\begin{aligned} 4I_1 + 6I_1 + 1 I_1 + 3I_1 &= -27 \\ 14I_1 &= -27 \\ I_1 &= -\frac{27}{14} A\end{aligned}$

Since the answer is negative, this tells us that the current flows clockwise.

因为答案是负的，这告诉我们电流沿顺时针方向流动。

A more complicated example follows. Two of the circuits below may be familiar; they were examined in the examples above. However as they are now part of a larger system of circuits, the answers will differ.

下面是一个更复杂的例子。下面四个电路中的两个可能很眼熟，它们在上面几个例子中研究过。不过，由于现在它们是更大电路系统的一部分，答案会不同。

Example 1.8.3: Unknown Currents 未知电流

The diagram below consists of four circuits. The current ($I_k$) in the four circuits is denoted by $I_{1},I_{2},I_{3},I_{4}$. Using Kirchhoff’s Law, write an equation for each circuit and solve for each current.

下图由四个电路组成。四个电路中的电流（$I_k$）记为 $I_{1},I_{2},I_{3},I_{4}$。利用基尔霍夫定律，为每个电路写出一个方程并求出各电流。

Solution. The circuits are given in the following diagram.

解。各电路如下图所示。

Starting with the top left circuit, multiply the resistance by the amps and sum the resulting products. Specifically, consider the resistor labeled $2\,\Omega$ that is part of the circuits of $I_1$ and $I_2$. Notice that current $I_2$ runs through this in a positive (counter clockwise) direction, and $I_1$ runs through in the opposite (negative) direction. The product of resistance and amps is then $2 (I_2 - I_1) = 2I_2 - 2I_1$. Continue in this way for each resistor, and set the sum of the products equal to the voltage source to write the equation: $2I_{2}-2I_{1}+4I_{2}-4I_{3}+2I_{2}=18$ The above process is used on each of the other three circuits, and the resulting equations are:

从左上角电路开始，把电阻乘以电流并把所得乘积相加。具体来说，考虑标着 $2\,\Omega$ 的、属于 $I_1$ 和 $I_2$ 电路的电阻。注意电流 $I_2$ 沿正的（逆时针）方向流过它，而 $I_1$ 沿相反的（负）方向流过它。那么电阻与电流的乘积为 $2 (I_2 - I_1) = 2I_2 - 2I_1$。对每个电阻都这样处理，并把各乘积之和等于电压源，从而写出方程：$2I_{2}-2I_{1}+4I_{2}-4I_{3}+2I_{2}=18$。对其他三个电路也运用上面的过程，所得的方程为：

Upper right circuit: $4I_{3} - 4I_{2} + 6I_{3} - 6I_{4} + I_{3} + 3I_{3} = -27$ Lower right circuit: $3I_{4} + 2I_{4} + 6I_{4} - 6I_{3} + I_{4} - I_{1} = 0$ Lower left circuit: $5I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-23$

右上电路：$4I_{3} - 4I_{2} + 6I_{3} - 6I_{4} + I_{3} + 3I_{3} = -27$；右下电路：$3I_{4} + 2I_{4} + 6I_{4} - 6I_{3} + I_{4} - I_{1} = 0$；左下电路：$5I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-23$

Notice that the voltage for the upper right and lower left circuits are negative due to the clockwise direction they indicate.

注意右上电路和左下电路的电压为负，因为它们所指示的是顺时针方向。

The resulting system of four equations in four unknowns is $\begin{aligned} 2I_{2}-2I_{1}+4I_{2}-4I_{3}+2I_{2}&=18 \\ 4I_{3} - 4I_{2} + 6I_{3} - 6I_{4} + I_{3} + I_{3} &= -27 \\ 3I_{4} + 2I_{4} + 6I_{4} - 6I_{3} + I_{4} - I_{1} &= 0 \\ 5I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}&= -23\end{aligned}$ Simplifying and rearranging with variables in order, we have: $\begin{aligned} -2I_{1}+8I_{2}-4I_{3}&=18 \\ - 4I_{2} + 14I_{3} - 6I_{4} &= -27 \\ -I_{1} - 6I_{3} + 12I_{4} &= 0 \\ 8I_{1}-2I_{2} - I_{4} &= -23\end{aligned}$ The augmented matrix is $\left[ \begin{array}{rrrr|r} -2 & 8 & -4 & 0 & 18 \\ 0 & -4 & 14 & -6 & -27 \\ -1 & 0 & -6 & 12 & 0 \\ 8 & -2 & 0 & -1 & -23 \end{array} \right]$

所得到的四元方程组为 $\begin{aligned} 2I_{2}-2I_{1}+4I_{2}-4I_{3}+2I_{2}&=18 \\ 4I_{3} - 4I_{2} + 6I_{3} - 6I_{4} + I_{3} + I_{3} &= -27 \\ 3I_{4} + 2I_{4} + 6I_{4} - 6I_{3} + I_{4} - I_{1} &= 0 \\ 5I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}&= -23\end{aligned}$。按变量顺序化简重排后，我们得到：$\begin{aligned} -2I_{1}+8I_{2}-4I_{3}&=18 \\ - 4I_{2} + 14I_{3} - 6I_{4} &= -27 \\ -I_{1} - 6I_{3} + 12I_{4} &= 0 \\ 8I_{1}-2I_{2} - I_{4} &= -23\end{aligned}$。增广矩阵为 $\left[ \begin{array}{rrrr|r} -2 & 8 & -4 & 0 & 18 \\ 0 & -4 & 14 & -6 & -27 \\ -1 & 0 & -6 & 12 & 0 \\ 8 & -2 & 0 & -1 & -23 \end{array} \right]$

The solution to this matrix is $\begin{aligned} I_{1} &= -3 A \\ I_{2} &= \frac{1}{4} A \\ I_{3} &= -\frac{5}{2} A \\ I_{4} &= -\frac{3}{2} A\end{aligned}$

该矩阵的解为 $\begin{aligned} I_{1} &= -3 A \\ I_{2} &= \frac{1}{4} A \\ I_{3} &= -\frac{5}{2} A \\ I_{4} &= -\frac{3}{2} A\end{aligned}$

This tells us that currents $I_1, I_3,$ and $I_4$ travel clockwise while $I_2$ travels counter clockwise.

这告诉我们，电流 $I_1, I_3$ 和 $I_4$ 沿顺时针方向流动，而 $I_2$ 沿逆时针方向流动。

1.9 Exercises 习题

Exercise 1. Graphically, find the point $\left( x_{1},y_{1}\right)$ which lies on both lines, $x+3y=1$ and $4x-y=3.$ That is, graph each line and see where they intersect.

习题 1。用图像法求出同时位于直线 $x+3y=1$ 与 $4x-y=3$ 上的点 $\left( x_{1},y_{1}\right)$。也就是说，画出各直线，看它们相交于何处。

Answer $\begin{array}{c} x+3y=1 \\ 4x-y=3 \end{array}$, Solution is: $\left[ x=\frac{10}{13},y=\frac{1}{13}\right]$.

答案 $\begin{array}{c} x+3y=1 \\ 4x-y=3 \end{array}$，解为：$\left[ x=\frac{10}{13},y=\frac{1}{13}\right]$。

Exercise 2. Graphically, find the point of intersection of the two lines $3x+y=3$ and $x+2y=1.$ That is, graph each line and see where they intersect.

习题 2。用图像法求出两条直线 $3x+y=3$ 与 $x+2y=1$ 的交点。也就是说，画出各直线，看它们相交于何处。

Answer $\begin{array}{c} 3x+y=3 \\ x+2y=1 \end{array}$, Solution is: $\left[ x=1,y=0\right]$

答案 $\begin{array}{c} 3x+y=3 \\ x+2y=1 \end{array}$，解为：$\left[ x=1,y=0\right]$

Exercise 3. Do the three lines, $x + 2y = 1$, $2x - y = 1$, and $4x + 3y = 3$ have a common point of intersection? If so, find the point and if not, tell why they don’t have such a common point of intersection.

习题 3。三条直线 $x + 2y = 1$、$2x - y = 1$ 和 $4x + 3y = 3$ 是否有一个公共交点？如果有，求出该点；如果没有，请说明为什么它们没有这样的公共交点。

Answer $\begin{array}{c}x+2y=1 \\ 2x-y=1 \\ 4x+3y=3\end{array}$, Solution is: $\left[ x=\frac{3}{5},y=\frac{1}{5}\right]$

答案 $\begin{array}{c}x+2y=1 \\ 2x-y=1 \\ 4x+3y=3\end{array}$，解为：$\left[ x=\frac{3}{5},y=\frac{1}{5}\right]$

Exercise 4. Do the three planes, $x+y-3z = 2$, $2x+y+z = 1$, and $3x+2y-2z = 0$ have a common point of intersection? If so, find one and if not, tell why there is no such point.

习题 4。三个平面 $x+y-3z = 2$、$2x+y+z = 1$ 和 $3x+2y-2z = 0$ 是否有公共交点？如果有，求出一个；如果没有，说明为什么不存在这样的点。

Answer No solution exists. You can see this by writing the augmented matrix and doing row operations. $\left[\begin{array}{cccc}1&1&-3&2 \\ 2&1&1&1 \\ 3&2&-2&0\end{array}\right]$, row echelon form: $\left[\begin{array}{cccc}1&0&4&0 \\ 0&1&-7&0 \\ 0&0&0&1\end{array}\right]$. Thus one of the equations says $0=1$ in an equivalent system of equations.

答案 不存在解。写出增广矩阵并做行运算便可看出这一点。$\left[\begin{array}{cccc}1&1&-3&2 \\ 2&1&1&1 \\ 3&2&-2&0\end{array}\right]$，行阶梯形为 $\left[\begin{array}{cccc}1&0&4&0 \\ 0&1&-7&0 \\ 0&0&0&1\end{array}\right]$。于是在一个等价的方程组中有一个方程说 $0=1$。

Exercise 5. Four times the weight of Gaston is $150$ pounds more than the weight of Ichabod. Four times the weight of Ichabod is $660$ pounds less than seventeen times the weight of Gaston. Four times the weight of Gaston plus the weight of Siegfried equals $290$ pounds. Brunhilde would balance all three of the others. Find the weights of the four people.

习题 5。加斯通体重的四倍比伊卡博德的体重多 $150$ 磅。伊卡博德体重的四倍比加斯通体重的十七倍少 $660$ 磅。加斯通体重的四倍加上齐格弗里德的体重等于 $290$ 磅。布伦希尔的体重恰好等于其他三人体重之和。求这四个人的体重。

Answer $\begin{array}{c}4g-I=150 \\ 4I-17g=-660 \\ 4g+s=290 \\ g+I+s-b=0\end{array}$, Solution is: $\left\{ g=60, I=90, b=200, s=50\right\}$

答案 $\begin{array}{c}4g-I=150 \\ 4I-17g=-660 \\ 4g+s=290 \\ g+I+s-b=0\end{array}$，解为：$\left\{ g=60, I=90, b=200, s=50\right\}$

Exercise 6. Consider the following augmented matrix in which $\ast$ denotes an arbitrary number and $\blacksquare$ denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? $\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & \blacksquare & \ast & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]$

习题 6。考虑下面的增广矩阵，其中 $\ast$ 表示任意数，$\blacksquare$ 表示非零数。判断所给增广矩阵是否相容。若相容，解是否唯一？$\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & \blacksquare & \ast & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]$

Answer The solution exists but is not unique.

答案 解存在但不唯一。

Exercise 7. Consider the following augmented matrix in which $\ast$ denotes an arbitrary number and $\blacksquare$ denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? $\left[ \begin{array}{ccc|c} \blacksquare & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast \\ 0 & 0 & \blacksquare & \ast \end{array} \right]$

习题 7。考虑下面的增广矩阵，其中 $\ast$ 表示任意数，$\blacksquare$ 表示非零数。判断所给增广矩阵是否相容。若相容，解是否唯一？$\left[ \begin{array}{ccc|c} \blacksquare & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast \\ 0 & 0 & \blacksquare & \ast \end{array} \right]$

Answer A solution exists and is unique.

答案 存在解且解唯一。

Exercise 8. Consider the following augmented matrix in which $\ast$ denotes an arbitrary number and $\blacksquare$ denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? $\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & 0 & \ast & 0 & \ast \\ 0 & 0 & 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]$

习题 8。考虑下面的增广矩阵，其中 $\ast$ 表示任意数，$\blacksquare$ 表示非零数。判断所给增广矩阵是否相容。若相容，解是否唯一？$\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & 0 & \ast & 0 & \ast \\ 0 & 0 & 0 & \blacksquare & \ast & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & \ast \end{array} \right]$

Exercise 9. Consider the following augmented matrix in which $\ast$ denotes an arbitrary number and $\blacksquare$ denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique? $\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & 0 \\ 0 & 0 & 0 & 0 & \ast & \blacksquare \end{array} \right]$

习题 9。考虑下面的增广矩阵，其中 $\ast$ 表示任意数，$\blacksquare$ 表示非零数。判断所给增广矩阵是否相容。若相容，解是否唯一？$\left[ \begin{array}{ccccc|c} \blacksquare & \ast & \ast & \ast & \ast & \ast \\ 0 & \blacksquare & \ast & \ast & 0 & \ast \\ 0 & 0 & 0 & 0 & \blacksquare & 0 \\ 0 & 0 & 0 & 0 & \ast & \blacksquare \end{array} \right]$

Answer There might be a solution. If so, there are infinitely many.

答案 可能存在解；若存在，则有无穷多个解。

Exercise 10. Suppose a system of equations has fewer equations than variables. Will such a system necessarily be consistent? If so, explain why and if not, give an example which is not consistent.

习题 10。假设某个方程组的方程个数少于变量个数。这样的方程组一定相容吗？如果一定，请解释原因；如果不一定，请给出一个不相容的例子。

Answer No. Consider $x+y+z=2$ and $x+y+z=1.$

答案 不一定。考虑 $x+y+z=2$ 和 $x+y+z=1$。

Exercise 11. If a system of equations has more equations than variables, can it have a solution? If so, give an example and if not, tell why not.

习题 11。如果某个方程组的方程个数多于变量个数，它可能有解吗？如果有，请举一个例子；如果没有，说明为什么。

Answer These can have a solution. For example, $x+y=1,2x+2y=2,3x+3y=3$ even has an infinite set of solutions.

答案 这样的方程组可以有解。例如 $x+y=1,2x+2y=2,3x+3y=3$ 甚至有无穷多组解。

Exercise 12. Find $h$ such that $\left[ \begin{array}{rr|r} 2 & h & 4 \\ 3 & 6 & 7 \end{array} \right]$ is the augmented matrix of an inconsistent system.

习题 12。求 $h$，使得 $\left[ \begin{array}{rr|r} 2 & h & 4 \\ 3 & 6 & 7 \end{array} \right]$ 是一个不相容方程组的增广矩阵。

Answer $h=4$

答案 $h=4$

Exercise 13. Find $h$ such that $\left[ \begin{array}{rr|r} 1 & h & 3 \\ 2 & 4 & 6 \end{array} \right]$ is the augmented matrix of a consistent system.

习题 13。求 $h$，使得 $\left[ \begin{array}{rr|r} 1 & h & 3 \\ 2 & 4 & 6 \end{array} \right]$ 是一个相容方程组的增广矩阵。

Answer Any $h$ will work.

答案 任何 $h$ 都行。

Exercise 14. Find $h$ such that $\left[ \begin{array}{rr|r} 1 & 1 & 4 \\ 3 & h & 12 \end{array} \right]$ is the augmented matrix of a consistent system.

习题 14。求 $h$，使得 $\left[ \begin{array}{rr|r} 1 & 1 & 4 \\ 3 & h & 12 \end{array} \right]$ 是一个相容方程组的增广矩阵。

Answer Any $h$ will work.

答案 任何 $h$ 都行。

Exercise 15. Choose $h$ and $k$ such that the augmented matrix shown has each of the following:

习题 15。选择 $h$ 和 $k$，使得所给增广矩阵分别具有下列每种性质：

1. one solution
2. no solution
3. infinitely many solutions

1. 唯一解
2. 无解
3. 无穷多个解

$$\left[ \begin{array}{rr|r} 1 & h & 2 \\ 2 & 4 & k \end{array} \right]$$

Answer If $h\neq 2$ there will be a unique solution for any $k$. If $h=2$ and $k\neq 4,$ there are no solutions. If $h=2$ and $k=4,$ then there are infinitely many solutions.

答案 若 $h\neq 2$，则对任意 $k$ 都有唯一解。若 $h=2$ 且 $k\neq 4$，则无解。若 $h=2$ 且 $k=4$，则有无穷多个解。

Exercise 16. Choose $h$ and $k$ such that the augmented matrix shown has each of the following:

习题 16。选择 $h$ 和 $k$，使得所给增广矩阵分别具有下列每种性质：

1. one solution
2. no solution
3. infinitely many solutions

1. 唯一解
2. 无解
3. 无穷多个解

$$\left[ \begin{array}{rr|r} 1 & 2 & 2 \\ 2 & h & k \end{array} \right]$$

Answer If $h\neq 4,$ then there is exactly one solution. If $h=4$ and $k\neq 4,$ then there are no solutions. If $h=4$ and $k=4,$ then there are infinitely many solutions.

答案 若 $h\neq 4$，则恰有一个解。若 $h=4$ 且 $k\neq 4$，则无解。若 $h=4$ 且 $k=4$，则有无穷多个解。

Exercise 17. Determine if the system is consistent. If so, is the solution unique? $\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=1 \\ 2x+y-z=1 \\ 4x+2y+z=5 \end{array}$

习题 17。判断方程组是否相容。若相容，解是否唯一？$\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=1 \\ 2x+y-z=1 \\ 4x+2y+z=5 \end{array}$

Answer There is no solution. The system is inconsistent. You can see this from the augmented matrix. $\left[ \begin{array}{rrrrr} 1 & 2 & 1 & -1 & 2 \\ 1 & -1 & 1 & 1 & 1 \\ 2 & 1 & -1 & 0 & 1 \\ 4 & 2 & 1 & 0 & 5 \end{array} \right]$, reduced row-echelon form: $\left[ \begin{array}{rrrrr} 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & 0 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right] .$

答案 无解，方程组不相容。你可以从增广矩阵看出这一点。$\left[ \begin{array}{rrrrr} 1 & 2 & 1 & -1 & 2 \\ 1 & -1 & 1 & 1 & 1 \\ 2 & 1 & -1 & 0 & 1 \\ 4 & 2 & 1 & 0 & 5 \end{array} \right]$，简化行阶梯形为 $\left[ \begin{array}{rrrrr} 1 & 0 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & 0 & -\frac{2}{3} & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array} \right]$。

Exercise 18. Determine if the system is consistent. If so, is the solution unique? $\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=0 \\ 2x+y-z=1 \\ 4x+2y+z=3 \end{array}$

习题 18。判断方程组是否相容。若相容，解是否唯一？$\begin{array}{c} x+2y+z-w=2 \\ x-y+z+w=0 \\ 2x+y-z=1 \\ 4x+2y+z=3 \end{array}$

Answer Solution is: $\left[ w=\frac{3}{2}y-1,\ x=\frac{2}{3}-\frac{1}{2}y,\ z=\frac{1}{3}\right]$

答案 解为：$\left[ w=\frac{3}{2}y-1,\ x=\frac{2}{3}-\frac{1}{2}y,\ z=\frac{1}{3}\right]$

Exercise 19. Determine which matrices are in reduced row-echelon form.

习题 19。判断下列哪些矩阵属于简化行阶梯形。

1. $\left[ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 7 \end{array} \right]$
2. $\left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right]$
3. $\left[ \begin{array}{rrrrrr} 1 & 1 & 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 2 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 3 \end{array} \right]$

1. $\left[ \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 7 \end{array} \right]$
2. $\left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right]$
3. $\left[ \begin{array}{rrrrrr} 1 & 1 & 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 2 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 3 \end{array} \right]$

Answer

1. This one is not.
2. This one is.
3. This one is.

答案

1. 这一个不是。
2. 这一个属于。
3. 这一个属于。

Exercise 20. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 2 & -1 & 3 & -1 \\ 1 & 0 & 2 & 1 \\ 1 & -1 & 1 & -2 \end{array} \right]$

习题 20。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 2 & -1 & 3 & -1 \\ 1 & 0 & 2 & 1 \\ 1 & -1 & 1 & -2 \end{array} \right]$

Exercise 21. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & -1 \end{array} \right]$

习题 21。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 0 & 0 & -1 & -1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & -1 \end{array} \right]$

Exercise 22. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 3 & -6 & -7 & -8 \\ 1 & -2 & -2 & -2 \\ 1 & -2 & -3 & -4 \end{array} \right]$

习题 22。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 3 & -6 & -7 & -8 \\ 1 & -2 & -2 & -2 \\ 1 & -2 & -3 & -4 \end{array} \right]$

Exercise 23. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 2 & 4 & 5 & 15 \\ 1 & 2 & 3 & 9 \\ 1 & 2 & 2 & 6 \end{array} \right]$

习题 23。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 2 & 4 & 5 & 15 \\ 1 & 2 & 3 & 9 \\ 1 & 2 & 2 & 6 \end{array} \right]$

Exercise 24. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 4 & -1 & 7 & 10 \\ 1 & 0 & 3 & 3 \\ 1 & -1 & -2 & 1 \end{array} \right]$

习题 24。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 4 & -1 & 7 & 10 \\ 1 & 0 & 3 & 3 \\ 1 & -1 & -2 & 1 \end{array} \right]$

Exercise 25. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} 3 & 5 & -4 & 2 \\ 1 & 2 & -1 & 1 \\ 1 & 1 & -2 & 0 \end{array} \right]$

习题 25。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} 3 & 5 & -4 & 2 \\ 1 & 2 & -1 & 1 \\ 1 & 1 & -2 & 0 \end{array} \right]$

Exercise 26. Row reduce the following matrix to obtain the row-echelon form. Then continue to obtain the reduced row-echelon form. $\left[ \begin{array}{rrrr} -2 & 3 & -8 & 7 \\ 1 & -2 & 5 & -5 \\ 1 & -3 & 7 & -8 \end{array} \right]$

习题 26。对下列矩阵做行化简得到行阶梯形，然后继续化简得到简化行阶梯形。$\left[ \begin{array}{rrrr} -2 & 3 & -8 & 7 \\ 1 & -2 & 5 & -5 \\ 1 & -3 & 7 & -8 \end{array} \right]$

Exercise 27. Find the solution of the system whose augmented matrix is $\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 1 & 3 & 4 & 2 \\ 1 & 0 & 2 & 1 \end{array} \right]$

习题 27。求增广矩阵为 $\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 1 & 3 & 4 & 2 \\ 1 & 0 & 2 & 1 \end{array} \right]$ 的方程组的解。

Exercise 28. Find the solution of the system whose augmented matrix is $\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 2 & 0 & 1 & 1 \\ 3 & 2 & 1 & 3 \end{array} \right]$

习题 28。求增广矩阵为 $\left[ \begin{array}{rrr|r} 1 & 2 & 0 & 2 \\ 2 & 0 & 1 & 1 \\ 3 & 2 & 1 & 3 \end{array} \right]$ 的方程组的解。

Answer The reduced row-echelon form is $\left[ \begin{array}{rrr|r} 1 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{1}{4} &\frac{3}{4} \\ 0 & 0 & 0 & 0 \end{array} \right] .$ Therefore, the solution is of the form $z=t,\ y=\frac{3}{4}+\frac{1}{4}t,\ x=\frac{1}{2}-\frac{1}{2}t$ where $t\in \mathbb{R}$.

答案 简化行阶梯形为 $\left[ \begin{array}{rrr|r} 1 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 1 & -\frac{1}{4} &\frac{3}{4} \\ 0 & 0 & 0 & 0 \end{array} \right]$。因此解具有形式 $z=t,\ y=\frac{3}{4}+\frac{1}{4}t,\ x=\frac{1}{2}-\frac{1}{2}t$，其中 $t\in \mathbb{R}$。

Exercise 29. Find the solution of the system whose augmented matrix is $\left[ \begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 1 & 0 & 4 & 2 \end{array} \right]$

习题 29。求增广矩阵为 $\left[ \begin{array}{rrr|r} 1 & 1 & 0 & 1 \\ 1 & 0 & 4 & 2 \end{array} \right]$ 的方程组的解。

Answer The reduced row-echelon form is $\left[ \begin{array}{rrr|r} 1 & 0 & 4 & 2 \\ 0 & 1 & -4 & -1 \end{array} \right]$ and so the solution is $z=t,\ y=4t,\ x=2-4t.$

答案 简化行阶梯形为 $\left[ \begin{array}{rrr|r} 1 & 0 & 4 & 2 \\ 0 & 1 & -4 & -1 \end{array} \right]$，因此解为 $z=t,\ y=4t,\ x=2-4t$。

Exercise 30. Find the solution of the system whose augmented matrix is $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 1 & 2 & 0 & 0 & 1 & 3 \\ 1 & 0 & 1 & 0 & 2 & 2 \end{array} \right]$

习题 30。求增广矩阵为 $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 1 & 2 & 0 & 0 & 1 & 3 \\ 1 & 0 & 1 & 0 & 2 & 2 \end{array} \right]$ 的方程组的解。

Answer The reduced row-echelon form is $\left[ \begin{array}{rrrrr|r} 1 & 0 & 0 & 0 & 9 & 3 \\ 0 & 1 & 0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0 & -7 & -1 \\ 0 & 0 & 0 & 1 & 6 & 1 \end{array} \right]$ and so $x_{5}=t,\ x_{4}=1-6t,\ x_{3}=-1+7t,\ x_{2}=4t,\ x_{1}=3-9t$.

答案 简化行阶梯形为 $\left[ \begin{array}{rrrrr|r} 1 & 0 & 0 & 0 & 9 & 3 \\ 0 & 1 & 0 & 0 & -4 & 0 \\ 0 & 0 & 1 & 0 & -7 & -1 \\ 0 & 0 & 0 & 1 & 6 & 1 \end{array} \right]$，因此 $x_{5}=t,\ x_{4}=1-6t,\ x_{3}=-1+7t,\ x_{2}=4t,\ x_{1}=3-9t$。

Exercise 31. Find the solution of the system whose augmented matrix is $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 0 & 2 & 0 & 0 & 1 & 3 \\ 1 & -1 & 2 & 2 & 2 & 0 \end{array} \right]$

习题 31。求增广矩阵为 $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 1 & 1 & 2 \\ 0 & 1 & 0 & 1 & 2 & 1 \\ 0 & 2 & 0 & 0 & 1 & 3 \\ 1 & -1 & 2 & 2 & 2 & 0 \end{array} \right]$ 的方程组的解。

Answer The reduced row-echelon form is $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 0 & -\frac{1}{2} & \frac{5}{2} \\ 0 & 1 & 0 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 1 & \frac{3}{2} & -\frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]$. Therefore, let $x_{5}=t,\ x_{3}=s.$ Then the other variables are given by $x_{4}=-\frac{1}{2}-\frac{3}{2}t,\ x_{2}=\frac{3}{2}-\frac{1}{2}t,\ x_{1}=\frac{5}{2}+\frac{1}{2}t-2s.$

答案 简化行阶梯形为 $\left[ \begin{array}{rrrrr|r} 1 & 0 & 2 & 0 & -\frac{1}{2} & \frac{5}{2} \\ 0 & 1 & 0 & 0 & \frac{1}{2} & \frac{3}{2} \\ 0 & 0 & 0 & 1 & \frac{3}{2} & -\frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right]$。因此令 $x_{5}=t,\ x_{3}=s$，则其余变量为 $x_{4}=-\frac{1}{2}-\frac{3}{2}t,\ x_{2}=\frac{3}{2}-\frac{1}{2}t,\ x_{1}=\frac{5}{2}+\frac{1}{2}t-2s$。

Exercise 32. Find the solution to the system of equations, $7x+14y+15z=22,$ $2x+4y+3z=5,$ and $3x+6y+10z=13.$

习题 32。求方程组 $7x+14y+15z=22$、$2x+4y+3z=5$ 和 $3x+6y+10z=13$ 的解。

Answer Solution is: $\left[ x=1-2t,\ z=1,\ y=t\right]$

答案 解为：$\left[ x=1-2t,\ z=1,\ y=t\right]$

Exercise 33. Find the solution to the system of equations, $3x-y+4z=6,$ $y+8z=0,$ and $-2x+y=-4.$

习题 33。求方程组 $3x-y+4z=6$、$y+8z=0$ 和 $-2x+y=-4$ 的解。

Answer Solution is: $\left[ x=2-4t,\ y=-8t,\ z=t\right]$

答案 解为：$\left[ x=2-4t,\ y=-8t,\ z=t\right]$

Exercise 34. Find the solution to the system of equations, $9x-2y+4z=-17,$ $13x-3y+6z=-25,$ and $-2x-z=3.$

习题 34。求方程组 $9x-2y+4z=-17$、$13x-3y+6z=-25$ 和 $-2x-z=3$ 的解。

Answer Solution is: $\left[x=-1,y=2,z=-1\right]$

答案 解为：$\left[x=-1,y=2,z=-1\right]$

Exercise 35. Find the solution to the system of equations, $65x+84y+16z=546,$ $81x+105y+20z=682,$ and $84x+110y+21z=713.$

习题 35。求方程组 $65x+84y+16z=546$、$81x+105y+20z=682$ 和 $84x+110y+21z=713$ 的解。

Answer Solution is: $\left[ x=2,y=4,z=5\right]$

答案 解为：$\left[ x=2,y=4,z=5\right]$

Exercise 36. Find the solution to the system of equations, $8x+2y+3z=-3,\ 8x+3y+3z=-1,$ and $4x+y+3z=-9.$

习题 36。求方程组 $8x+2y+3z=-3$、$8x+3y+3z=-1$ 和 $4x+y+3z=-9$ 的解。

Answer Solution is: $\left[ x=1,y=2,z=-5\right]$

答案 解为：$\left[ x=1,y=2,z=-5\right]$

Exercise 37. Find the solution to the system of equations, $-8x+2y+5z=18,\ -8x+3y+5z=13,$ and $-4x+y+5z=19.$

习题 37。求方程组 $-8x+2y+5z=18$、$-8x+3y+5z=13$ 和 $-4x+y+5z=19$ 的解。

Answer Solution is: $\left[x=-1,y=-5,z=4\right]$

答案 解为：$\left[x=-1,y=-5,z=4\right]$

Exercise 38. Find the solution to the system of equations, $3x-y-2z=3,$ $y-4z=0,$ and $-2x+y=-2.$

习题 38。求方程组 $3x-y-2z=3$、$y-4z=0$ 和 $-2x+y=-2$ 的解。

Answer Solution is: $\left[ x=2t+1,\ y=4t,\ z=t\right]$

答案 解为：$\left[ x=2t+1,\ y=4t,\ z=t\right]$

Exercise 39. Find the solution to the system of equations, $-9x+15y=66,\ -11x+18y=79$, $-x+y=4$, and $z=3$.

习题 39。求方程组 $-9x+15y=66$、$-11x+18y=79$、$-x+y=4$ 和 $z=3$ 的解。

Answer Solution is: $\left[x=1,y=5,z=3\right]$

答案 解为：$\left[x=1,y=5,z=3\right]$

Exercise 40. Find the solution to the system of equations, $-19x+8y=-108,$ $-71x+30y=-404,$ $-2x+y=-12,$ $4x+z=14.$

习题 40。求方程组 $-19x+8y=-108$、$-71x+30y=-404$、$-2x+y=-12$ 和 $4x+z=14$ 的解。

Answer Solution is: $\left[ x=4,y=-4,z=-2\right]$

答案 解为：$\left[ x=4,y=-4,z=-2\right]$

Exercise 41. Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain.

习题 41。假设某个方程组的方程个数少于变量个数，并且你已求出这个方程组的一个解。你的这个解有没有可能是唯一解？请解释。

Answer No. Consider $x+y+z=2$ and $x+y+z=1.$

答案 不可能。考虑 $x+y+z=2$ 和 $x+y+z=1$。

Exercise 42. Suppose a system of linear equations has a $2\times 4$ augmented matrix and the last column is a pivot column. Could the system of linear equations be consistent? Explain.

习题 42。假设一个线性方程组的增广矩阵是 $2\times 4$ 的，并且最后一列是主元列。这个线性方程组可能相容吗？请解释。

Answer No. This would lead to $0=1.$

答案 不可能。这会导致 $0=1$。

Exercise 43. Suppose the coefficient matrix of a system of $n$ equations with $n$ variables has the property that every column is a pivot column. Does it follow that the system of equations must have a solution? If so, must the solution be unique? Explain.

习题 43。假设一个含 $n$ 个变量、$n$ 个方程的方程组的系数矩阵具有每一列都是主元列的性质。能否推出该方程组一定有解？如果有，解是否必须唯一？请解释。

Answer Yes. It has a unique solution.

答案 能。它有唯一解。

Exercise 44. Suppose there is a unique solution to a system of linear equations. What must be true of the pivot columns in the augmented matrix?

习题 44。假设某个线性方程组有唯一解。那么在增广矩阵中，主元列必须满足什么条件？

Answer The last column must not be a pivot column. The remaining columns must each be pivot columns.

答案 最后一列不能是主元列，其余各列必须每一列都是主元列。

Exercise 45. The steady state temperature, $u$, of a plate solves Laplace’s equation, $\Delta u=0.$ One way to approximate the solution is to divide the plate into a square mesh and require the temperature at each node to equal the average of the temperature at the four adjacent nodes. In the following picture, the numbers represent the observed temperature at the indicated nodes. Find the temperature at the interior nodes, indicated by $x,y,z,$ and $w$. One of the equations is $z=\frac{1}{4}\left( 10+0+w+x\right)$.

习题 45。一块板的稳态温度 $u$ 满足拉普拉斯方程 $\Delta u=0$。近似求解的一种方法，是把板划分成正方形网格，并要求每个节点的温度等于其相邻四个节点温度的平均值。在下图中，数字表示所标注节点处观测到的温度。求由 $x,y,z$ 和 $w$ 标注的内部节点的温度。其中一个方程是 $z=\frac{1}{4}\left( 10+0+w+x\right)$。

Answer You need $\begin{array}{c} \frac{1}{4}\left( 20+30+w+x\right) - y=0 \\ \frac{1}{4}\left( y+30+0+z\right) - w=0 \\ \frac{1}{4}\left( 20+y+z+10\right) - x=0 \\ \frac{1}{4}\left( x+w+0+10\right) - z=0 \end{array}$, Solution is: $\left[ w=15,x=15,y=20,z=10\right] .$

答案 你需要 $\begin{array}{c} \frac{1}{4}\left( 20+30+w+x\right) - y=0 \\ \frac{1}{4}\left( y+30+0+z\right) - w=0 \\ \frac{1}{4}\left( 20+y+z+10\right) - x=0 \\ \frac{1}{4}\left( x+w+0+10\right) - z=0 \end{array}$，解为：$\left[ w=15,x=15,y=20,z=10\right]$。

Exercise 46. Find the rank of the following matrix. $\left[ \begin{array}{rrrr} 4 & -16 & -1 & -5 \\ 1 & -4 & 0 & -1 \\ 1 & -4 & -1 & -2 \end{array} \right]$

习题 46。求下列矩阵的秩。$\left[ \begin{array}{rrrr} 4 & -16 & -1 & -5 \\ 1 & -4 & 0 & -1 \\ 1 & -4 & -1 & -2 \end{array} \right]$

Exercise 47. Find the rank of the following matrix. $\left[ \begin{array}{rrrr} 3 & 6 & 5 & 12 \\ 1 & 2 & 2 & 5 \\ 1 & 2 & 1 & 2 \end{array} \right]$

习题 47。求下列矩阵的秩。$\left[ \begin{array}{rrrr} 3 & 6 & 5 & 12 \\ 1 & 2 & 2 & 5 \\ 1 & 2 & 1 & 2 \end{array} \right]$

Exercise 48. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 3 \\ 1 & 4 & 1 & 0 & -8 \\ 1 & 4 & 0 & 1 & 2 \\ -1 & -4 & 0 & -1 & -2 \end{array} \right]$

习题 48。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 3 \\ 1 & 4 & 1 & 0 & -8 \\ 1 & 4 & 0 & 1 & 2 \\ -1 & -4 & 0 & -1 & -2 \end{array} \right]$

Exercise 49. Find the rank of the following matrix. $\left[ \begin{array}{rrrr} 4 & -4 & 3 & -9 \\ 1 & -1 & 1 & -2 \\ 1 & -1 & 0 & -3 \end{array} \right]$

习题 49。求下列矩阵的秩。$\left[ \begin{array}{rrrr} 4 & -4 & 3 & -9 \\ 1 & -1 & 1 & -2 \\ 1 & -1 & 0 & -3 \end{array} \right]$

Exercise 50. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} 2 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 7 \\ 1 & 0 & 0 & 1 & 7 \end{array} \right]$

习题 50。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} 2 & 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 7 \\ 1 & 0 & 0 & 1 & 7 \end{array} \right]$

Exercise 51. Find the rank of the following matrix. $\left[ \begin{array}{rrr} 4 & 15 & 29 \\ 1 & 4 & 8 \\ 1 & 3 & 5 \\ 3 & 9 & 15 \end{array} \right]$

习题 51。求下列矩阵的秩。$\left[ \begin{array}{rrr} 4 & 15 & 29 \\ 1 & 4 & 8 \\ 1 & 3 & 5 \\ 3 & 9 & 15 \end{array} \right]$

Exercise 52. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 1 \\ 1 & 2 & 3 & -2 & -18 \\ 1 & 2 & 2 & -1 & -11 \\ -1 & -2 & -2 & 1 & 11 \end{array} \right]$

习题 52。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} 0 & 0 & -1 & 0 & 1 \\ 1 & 2 & 3 & -2 & -18 \\ 1 & 2 & 2 & -1 & -11 \\ -1 & -2 & -2 & 1 & 11 \end{array} \right]$

Exercise 53. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} 1 & -2 & 0 & 3 & 11 \\ 1 & -2 & 0 & 4 & 15 \\ 1 & -2 & 0 & 3 & 11 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

习题 53。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} 1 & -2 & 0 & 3 & 11 \\ 1 & -2 & 0 & 4 & 15 \\ 1 & -2 & 0 & 3 & 11 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

Exercise 54. Find the rank of the following matrix. $\left[ \begin{array}{rrr} -2 & -3 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ -3 & 0 & -3 \end{array} \right]$

习题 54。求下列矩阵的秩。$\left[ \begin{array}{rrr} -2 & -3 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \\ -3 & 0 & -3 \end{array} \right]$

Exercise 55. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} 4 & 4 & 20 & -1 & 17 \\ 1 & 1 & 5 & 0 & 5 \\ 1 & 1 & 5 & -1 & 2 \\ 3 & 3 & 15 & -3 & 6 \end{array} \right]$

习题 55。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} 4 & 4 & 20 & -1 & 17 \\ 1 & 1 & 5 & 0 & 5 \\ 1 & 1 & 5 & -1 & 2 \\ 3 & 3 & 15 & -3 & 6 \end{array} \right]$

Exercise 56. Find the rank of the following matrix. $\left[ \begin{array}{rrrrr} -1 & 3 & 4 & -3 & 8 \\ 1 & -3 & -4 & 2 & -5 \\ 1 & -3 & -4 & 1 & -2 \\ -2 & 6 & 8 & -2 & 4 \end{array} \right]$

习题 56。求下列矩阵的秩。$\left[ \begin{array}{rrrrr} -1 & 3 & 4 & -3 & 8 \\ 1 & -3 & -4 & 2 & -5 \\ 1 & -3 & -4 & 1 & -2 \\ -2 & 6 & 8 & -2 & 4 \end{array} \right]$

Exercise 57. Suppose $A$ is an $m\times n$ matrix. Explain why the rank of $A$ is always no larger than $\min \left( m,n\right) .$

习题 57。假设 $A$ 是一个 $m\times n$ 矩阵。解释为什么 $A$ 的秩总是不超过 $\min \left( m,n\right)$。

Answer It is because you cannot have more than $\min \left( m,n\right)$ nonzero rows in the reduced row-echelon form. Recall that the number of pivot columns is the same as the number of nonzero rows from the description of this reduced row-echelon form.

答案 因为在简化行阶梯形中，非零行的个数不可能超过 $\min \left( m,n\right)$。回忆一下：由简化行阶梯形的描述可知，主元列的个数与非零行的个数相同。

Exercise 58. State whether each of the following sets of data are possible for the matrix equation $AX=B$. If possible, describe the solution set. That is, tell whether there exists a unique solution, no solution or infinitely many solutions. Here, $\left[ A \|B \right]$ denotes the augmented matrix.

习题 58。指出下列各组数据对矩阵方程 $AX=B$ 来说是否可能。如果可能，描述其解集。也就是说，说明是存在唯一解、无解还是无穷多个解。这里 $\left[ A \|B \right]$ 表示增广矩阵。

1. $A$ is a $5\times 6$ matrix, $rank\left( A\right) =4$ and $rank\left[ A\|B \right] =4.$
2. $A$ is a $3\times 4$ matrix, $rank\left( A\right) =3$ and $rank\left[ A\|B\right] =2.$
3. $A$ is a $4\times 2$ matrix, $rank\left( A\right) =4$ and $rank\left[ A\|B \right] =4.$
4. $A$ is a $5\times 5$ matrix, $rank\left( A\right) =4$ and $rank\left[ A\|B \right] =5.$
5. $A$ is a $4\times 2$ matrix, $rank\left( A\right) =2$ and $rank\left[ A\|B \right] =2$。

1. $A$ 是一个 $5\times 6$ 矩阵，$rank\left( A\right) =4$，且 $rank\left[ A\|B \right] =4$。
2. $A$ 是一个 $3\times 4$ 矩阵，$rank\left( A\right) =3$，且 $rank\left[ A\|B\right] =2$。
3. $A$ 是一个 $4\times 2$ 矩阵，$rank\left( A\right) =4$，且 $rank\left[ A\|B \right] =4$。
4. $A$ 是一个 $5\times 5$ 矩阵，$rank\left( A\right) =4$，且 $rank\left[ A\|B \right] =5$。
5. $A$ 是一个 $4\times 2$ 矩阵，$rank\left( A\right) =2$，且 $rank\left[ A\|B \right] =2$。

Answer

1. This says $B$ is in the span of four of the columns. Thus the columns are not independent. Infinite solution set.
2. This surely can’t happen. If you add in another column, the rank does not get smaller.
3. This says $B$ is in the span of the columns and the columns must be independent. You can’t have the rank equal $4$ if you only have two columns.
4. This says $B$ is not in the span of the columns. In this case, there is no solution to the system of equations represented by the augmented matrix.
5. In this case, there is a unique solution since the columns of $A$ are independent.

答案

1. 这说明 $B$ 位于其中四列的张成空间中，因而各列不是线性无关的。解集是无穷的。
2. 这种情况肯定不会发生。如果再加一列，秩不会变小。
3. 这说明 $B$ 位于各列的张成空间中，且各列必须线性无关。如果只有两列，秩不可能等于 $4$。
4. 这说明 $B$ 不在各列的张成空间中。此时，增广矩阵所表示的方程组无解。
5. 在这种情况下，因为 $A$ 的各列线性无关，所以有唯一解。

Exercise 59. Consider the system $-5x+2y-z=0$ and $-5x-2y-z=0.$ Both equations equal zero and so $-5x+2y-z=-5x-2y-z$ which is equivalent to $y=0.$ Does it follow that $x$ and $z$ can equal anything? Notice that when $x=1$, $z=-4,$ and $y=0$ are plugged in to the equations, the equations do not equal $0$. Why?

习题 59。考虑方程组 $-5x+2y-z=0$ 和 $-5x-2y-z=0$。两个方程都等于 $0$，因此 $-5x+2y-z=-5x-2y-z$，这等价于 $y=0$。能否由此推出 $x$ 和 $z$ 可以等于任何值？注意当把 $x=1,z=-4,y=0$ 代入方程时，方程并不等于 $0$。为什么？

Answer These are not legitimate row operations. They do not preserve the solution set of the system.

答案 这些并不是合法的行运算。它们不能保持方程组的解集不变。

Exercise 60. Balance the following chemical reactions.

习题 60。配平下列化学反应。

1. $KNO_{3}+H_{2}CO_{3}\rightarrow K_{2}CO_{3}+HNO_{3}$
2. $AgI+Na_{2}S\rightarrow Ag_{2}S+NaI$
3. $Ba_{3}N_{2}+H_{2}O\rightarrow Ba\left( OH\right) _{2}+NH_{3}$
4. $CaCl_{2}+Na_{3}PO_{4}\rightarrow Ca_{3}\left( PO_{4}\right) _{2}+NaCl$

1. $KNO_{3}+H_{2}CO_{3}\rightarrow K_{2}CO_{3}+HNO_{3}$
2. $AgI+Na_{2}S\rightarrow Ag_{2}S+NaI$
3. $Ba_{3}N_{2}+H_{2}O\rightarrow Ba\left( OH\right) _{2}+NH_{3}$
4. $CaCl_{2}+Na_{3}PO_{4}\rightarrow Ca_{3}\left( PO_{4}\right) _{2}+NaCl$

Exercise 61. In the section on dimensionless variables it was observed that $\rho V^{2}AB$ has the units of force. Describe a systematic way to obtain such combinations of the variables which will yield something which has the units of force.

习题 61。在无量纲变量一节中已观察到，$\rho V^{2}AB$ 具有力的单位。描述一种系统的方法，用来获得这样的变量组合，使其结果具有力的单位。

Exercise 62. Consider the following diagram of four circuits.

习题 62。考虑下面的四个电路图。

The current in amps in the four circuits is denoted by $I_{1},I_{2},I_{3},I_{4}$ and it is understood that the motion is in the counter clockwise direction. If $I_{k}$ ends up being negative, then it just means the current flows in the clockwise direction.

四个电路中的电流（安培）记为 $I_{1},I_{2},I_{3},I_{4}$，并约定流动方向为逆时针方向。若 $I_{k}$ 最终为负，则只表示电流沿顺时针方向流动。

In the above diagram, the top left circuit should give the equation $2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5$ For the circuit on the lower left, you should have $4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10$ Write equations for each of the other two circuits and then give a solution to the resulting system of equations.

在上图中，左上电路应给出方程 $2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5$。对于左下电路，你应该有 $4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10$。请为另外两个电路各写出方程，然后给出所得方程组的解。

Answer The other two equations are $\begin{aligned} 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2} &=-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1} &=0\end{aligned}$ Then the system is $\begin{array}{c} 2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5 \\ 4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10 \\ 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2}=-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1}=0 \end{array}$ The solution is: $\begin{aligned} I_{1}&= -\frac{750}{373} \\ I_{2}&= -\frac{1421}{1119} \\ I_{3}&= -\frac{3061}{1119} \\ I_{4}&= -\frac{1718}{1119}\end{aligned}$

答案 另外两个方程为 $\begin{aligned} 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2} &=-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1} &=0\end{aligned}$。于是方程组为 $\begin{array}{c} 2I_{2}-2I_{1}+5I_{2}-5I_{3}+3I_{2}=5 \\ 4I_{1}+I_{1}-I_{4}+2I_{1}-2I_{2}=-10 \\ 6I_{3}-6I_{4}+I_{3}+I_{3}+5I_{3}-5I_{2}=-20 \\ 2I_{4}+3I_{4}+6I_{4}-6I_{3}+I_{4}-I_{1}=0 \end{array}$。解为：$\begin{aligned} I_{1}&= -\frac{750}{373} \\ I_{2}&= -\frac{1421}{1119} \\ I_{3}&= -\frac{3061}{1119} \\ I_{4}&= -\frac{1718}{1119}\end{aligned}$

Exercise 63. Consider the following diagram of three circuits.

习题 63。考虑下面的三个电路图。

The current in amps in the three circuits is denoted by $I_{1},I_{2},I_{3}$ and it is understood that the motion is in the counter clockwise direction. If $I_{k}$ ends up being negative, then it just means the current flows in the clockwise direction.

三个电路中的电流（安培）记为 $I_{1},I_{2},I_{3}$，并约定流动方向为逆时针方向。若 $I_{k}$ 最终为负，则只表示电流沿顺时针方向流动。

Find $I_{1},I_{2},I_{3}$.

求 $I_{1},I_{2},I_{3}$。

Answer You have $\begin{aligned} 2I_{1}+5I_{1}+3I_{1}-5I_{2} &= 10 \\ I_{2}- I_{3} +3I_{2}+7I_{2}+5I_{2}-5I_{1} &=-12 \\ 2I_{3}+4I_{3}+4I_{3}+I_{3}-I_{2} &= 0\end{aligned}$ Simplifying this yields $\begin{aligned} 10I_{1}-5I_{2} &= 10 \\ -5I_{1} + 16I_{2}- I_{3} &=-12 \\ -I_{2} + 11I_{3} &=0\end{aligned}$ The solution is given by $I_{1}=\frac{218}{295},I_{2}=-\frac{154}{295},I_{3}=-\frac{14}{295}$

答案 你有 $\begin{aligned} 2I_{1}+5I_{1}+3I_{1}-5I_{2} &= 10 \\ I_{2}- I_{3} +3I_{2}+7I_{2}+5I_{2}-5I_{1} &=-12 \\ 2I_{3}+4I_{3}+4I_{3}+I_{3}-I_{2} &= 0\end{aligned}$。化简后得到 $\begin{aligned} 10I_{1}-5I_{2} &= 10 \\ -5I_{1} + 16I_{2}- I_{3} &=-12 \\ -I_{2} + 11I_{3} &=0\end{aligned}$。解为 $I_{1}=\frac{218}{295},I_{2}=-\frac{154}{295},I_{3}=-\frac{14}{295}$。