Determinants 行列式 行列式

Since $2(j-i)+1$ is an odd number $(-1)^{2(j-i)+1}=-1$ and we have that $\det A=-\det B$.

(2) This is like (1)… but much easier. Assume that (2) is true for all $n-1\times n-1$ matrices. We have that $a_{ji}=k b_{ji}$ for $1\leq j\leq n$. In particular $a_{1i}=kb_{1i}$, and for $l\neq i$ matrix $A(l)$ is obtained from $B(l)$ by multiplying one of its rows by $k$. Therefore $\mathrm{cof}(A)_{1l}=k\mathrm{cof}(B)_{1l}$ for $l\neq i$, and for all $l$ we have $a_{1l} \mathrm{cof}(A)_{1l}=k b_{1l}\mathrm{cof}(B)_{1l}$. By the equation above, we have $\det A=k\det B$.

(3) This is a consequence of (1). If two rows of $A$ are identical, then $A$ is equal to the matrix obtained by interchanging those two rows and therefore by (1) $\det A=-\det A$. This implies $\det A=0$.

(4) Assume (4) is true for all $n-1\times n-1$ matrices and fix $A$ and $B$ such that $A$ is obtained by multiplying $i$th row of $B$ by $k$ and adding it to $j$th row of $B$ ($i\neq j$). Then $\det A=\det B$. If $k=0$ then $A=B$ and there is nothing to prove, so we may assume $k\neq 0$.

Let $C$ be the matrix obtained by replacing the $j$th row of $B$ by the $i$th row of $B$ multiplied by $k$. By Lemma 2, we have that $\det A=\det B+\det C $ and we 'only' need to show that $\det C=0$. But $i$th and $j$th rows of $C$ are proportional. If $D$ is obtained by multiplying the $j$th row of $C$ by $\frac 1k$ then by (2) we have $\det C=\frac 1k\det D$ (recall that $k\neq 0$!). But $i$th and $j$th rows of $D$ are identical, hence by (3) we have $\det D=0$ and therefore $\det C=0$.

定理 9 Theorem 9

Let $A$ and $B$ be two $n\times n$ matrices. Then $\det \left( AB\right) =\det \left( A\right) \det \left( B\right) $.

Proof: If $A$ is an elementary matrix of either type, then multiplying by $A$ on the left has the same effect as performing the corresponding elementary row operation. Therefore the equality $\det (AB) =\det A\det B$ in this case follows by Example 8 and Theorem 8.

If $C$ is the reduced row-echelon form of $A$ then we can write $A=E_1\cdot E_2\cdot\dots\cdot E_m\cdot C$ for some elementary matrices $E_1,\dots, E_m$.

Now we consider two cases.

Assume first that $C=I$. Then $A=E_1\cdot E_2\cdot \dots\cdot E_m$ and $AB= E_1\cdot E_2\cdot \dots\cdot E_m B$. By applying the above equality $m$ times, and then $m-1$ times, we have that $\begin{aligned} \det AB&=\det E_1\det E_2\cdot \det E_m\cdot \det B \\ &=\det (E_1\cdot E_2\cdot\dots\cdot E_m) \det B \\ &=\det A\det B. \end{aligned}$

Now assume $C\neq I$. Since it is in reduced row-echelon form, its last row consists of zeros and by (4) of Example 8 the last row of $CB$ consists of zeros. By Lemma 1 we have $\det C=\det (CB)=0$ and therefore $\det A=\det (E_1\cdot E_2\cdot E_m)\cdot \det (C) = \det (E_1\cdot E_2\cdot E_m)\cdot 0=0 $ and also $\det AB=\det (E_1\cdot E_2\cdot E_m)\cdot \det (C B) =\det (E_1\cdot E_2\cdot\dots\cdot E_m) \cdot 0 =0 $ hence $\det AB=0=\det A \det B$.

The same 'machine' used in the previous proof will be used again.

定理 10 Theorem 10

Let $A$ be a matrix where $A^T$ is the transpose of $A$. Then, $\det\left(A^T\right) = \det \left( A \right) $.

Proof: Note first that the conclusion is true if $A$ is elementary by (5) of Example 8.

Let $C$ be the reduced row-echelon form of $A$. Then we can write $A= E_1\cdot E_2\cdot \dots\cdot E_m C$. Then $A^T=C^T\cdot E_m^T\cdot \dots \cdot E_2^T\cdot E_1$. By Theorem 9 we have $\det (A^T)=\det (C^T)\cdot \det (E_m^T)\cdot \dots \cdot \det (E_2^T)\cdot \det(E_1). $ By (5) of Example 8 we have that $\det E_j=\det E_j^T$ for all $j$. Also, $\det C$ is either 0 or 1 (depending on whether $C=I$ or not) and in either case $\det C=\det C^T$. Therefore $\det A=\det A^T$.

The above discussions allow us to now prove Theorem 3.1.1. It is restated below.

定理 11 Theorem 11

Expanding an $n\times n$ matrix along any row or column always gives the same result, which is the determinant.

Proof: We first show that the determinant can be computed along any row. The case $n=1$ does not apply and thus let $n \geq 2$.

Let $A$ be an $n\times n$ matrix and fix $j>1$. We need to prove that $\det A=\sum_{i=1}^n a_{j,i} \mathrm{cof}(A)_{j,i}. $ Let us prove the case when $j=2$.

Let $B$ be the matrix obtained from $A$ by interchanging its $1$st and $2$nd rows. Then by Theorem 8 we have $\det A=-\det B.$ Now we have $\det B=\sum_{i=1}^n b_{1,i} \mathrm{cof}(B)_{1,i}. $ Since $B$ is obtained by interchanging the $1$st and $2$nd rows of $A$ we have that $b_{1,i}=a_{2,i}$ for all $i$ and one can see that $minor(B)_{1,i}=minor(A)_{2,i}$.

Further, $\mathrm{cof}(B)_{1,i}=(-1)^{1+i} minor B_{1,i}=- (-1)^{2+i} minor (A)_{2,i} = - \mathrm{cof}(A)_{2,i} $ hence $\det B=-\sum_{i=1}^n a_{2,i} \mathrm{cof}(A)_{2,i}$, and therefore $\det A=-\det B= \sum_{i=1}^n a_{2,i} \mathrm{cof}(A)_{2,i}$ as desired.

The case when $j>2$ is very similar; we still have $minor(B)_{1,i}=minor (A)_{j,i}$ but checking that $\det B=-\sum_{i=1}^n a_{j,i} \mathrm{cof}(A)_{j,i}$ is slightly more involved.

Now the cofactor expansion along column $j$ of $A$ is equal to the cofactor expansion along row $j$ of $A^T$, which is by the above result just proved equal to the cofactor expansion along row 1 of $A^T$, which is equal to the cofactor expansion along column $1$ of $A$. Thus the cofactor expansion along any column yields the same result.

Finally, since $\det A=\det A^T$ by Theorem 10, we conclude that the cofactor expansion along row $1$ of $A$ is equal to the cofactor expansion along row $1$ of $A^T$, which is equal to the cofactor expansion along column $1$ of $A$. Thus the proof is complete.

Finding Determinants using Row Operations 用行运算求行列式

Theorems 3.2.1, 3.2.2 and 3.2.4 illustrate how row operations affect the determinant of a matrix. In this section, we look at two examples where row operations are used to find the determinant of a large matrix. Recall that when working with large matrices, Laplace Expansion is effective but timely, as there are many steps involved. This section provides useful tools for an alternative method. By first applying row operations, we can obtain a simpler matrix to which we apply Laplace Expansion.

While working through questions such as these, it is useful to record your row operations as you go along. Keep this in mind as you read through the next example.

示例 1：求一个行列式 Finding a Determinant

Find the determinant of the matrix $A=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 1 & 2 & 3 \\ 4 & 5 & 4 & 3 \\ 2 & 2 & -4 & 5 \end{array} \right] $.

解 Solution

We will use the properties of determinants outlined above to find $\det \left(A\right)$. First, add $-5$ times the first row to the second row. Then add $-4$ times the first row to the third row, and $-2$ times the first row to the fourth row. This yields the matrix $B=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -9 & -13 & -17 \\ 0 & -3 & -8 & -13 \\ 0 & -2 & -10 & -3 \end{array} \right] $ Notice that the only row operation we have done so far is adding a multiple of a row to another row. Therefore, by Theorem 3.2.4, $\det \left(B\right) = \det \left(A\right).$

At this stage, you could use Laplace Expansion to find $\det \left(B\right)$. However, we will continue with row operations to find an even simpler matrix to work with.

Add $-3$ times the third row to the second row. By Theorem 3.2.4 this does not change the value of the determinant. Then, multiply the fourth row by $-3$. This results in the matrix $C=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & 0 & 11 & 22 \\ 0 & -3 & -8 & -13 \\ 0 & 6 & 30 & 9 \end{array} \right] $ Here, $\det \left(C\right) = -3 \det \left(B\right)$, which means that $\det \left( B\right) =\left(-\frac{1}{3}\right) \det \left( C\right) $.

Since $\det \left(A\right) = \det \left(B\right)$, we now have that $\det \left(A\right) = \left(-\frac{1}{3}\right) \det \left( C\right)$. Again, you could use Laplace Expansion here to find $\det \left(C\right)$. However, we will continue with row operations.

Now replace the add $2$ times the third row to the fourth row. This does not change the value of the determinant by Theorem 3.2.4. Finally switch the third and second rows. This causes the determinant to be multiplied by $-1.$ Thus $\det \left( C\right) = -\det \left( D\right)$ where $D=\left[ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -3 & -8 & -13 \\ 0 & 0 & 11 & 22 \\ 0 & 0 & 14 & -17 \end{array} \right] $.

Hence, $\det \left(A\right) = \left(-\frac{1}{3}\right) \det \left( C\right) = \left(\frac{1}{3}\right) \det \left( D\right)$.

You could do more row operations or you could note that this can be easily expanded along the first column. Then, expand the resulting $3 \times 3$ matrix also along the first column. This results in $\det \left( D\right) =1\left( -3\right) \left\vert \begin{array}{cc} 11 & 22 \\ 14 & -17 \end{array} \right\vert = 1485 $ and so $\det \left( A\right) =\left(\frac{1}{3}\right) \left( 1485\right) =495.$

You can see that by using row operations, we can simplify a matrix to the point where Laplace Expansion involves only a few steps. In Example 1, we also could have continued until the matrix was in upper triangular form, and taken the product of the entries on the main diagonal. Whenever computing the determinant, it is useful to consider all the possible methods and tools.

Consider the next example.

示例 2：求行列式 Find the Determinant

Find the determinant of the matrix $A = \left[ \begin{array}{rrrr} 1 & 2 & 3 & 2 \\ 1 & -3 & 2 & 1 \\ 2 & 1 & 2 & 5 \\ 3 & -4 & 1 & 2 \end{array} \right] $.

解 Solution

Once again, we will simplify the matrix through row operations. Add $-1$ times the first row to the second row. Next add $-2$ times the first row to the third and finally take $-3$ times the first row and add to the fourth row. This yields $B = \left[ \begin{array}{rrrr} 1 & 2 & 3 & 2 \\ 0 & -5 & -1 & -1 \\ 0 & -3 & -4 & 1 \\ 0 & -10 & -8 & -4 \end{array} \right] $ By Theorem 3.2.4, $\det \left(A\right) = \det \left(B\right).$

Remember you can work with the columns also. Take $-5$ times the fourth column and add to the second column. This yields $C = \left[ \begin{array}{rrrr} 1 & -8 & 3 & 2 \\ 0 & 0 & -1 & -1 \\ 0 & -8 & -4 & 1 \\ 0 & 10 & -8 & -4 \end{array} \right] $ By Theorem 3.2.4 $\det \left(A\right) = \det \left(C\right).$

Now take $-1$ times the third row and add to the top row. This gives. $D = \left[ \begin{array}{rrrr} 1 & 0 & 7 & 1 \\ 0 & 0 & -1 & -1 \\ 0 & -8 & -4 & 1 \\ 0 & 10 & -8 & -4 \end{array} \right] $ which by Theorem 3.2.4 has the same determinant as $A.$

Now, we can find $\det \left(D\right)$ by expanding along the first column as follows. You can see that there will be only one non zero term. $\det \left(D\right) = 1 \det \left[ \begin{array}{rrr} 0 & -1 & -1 \\ -8 & -4 & 1 \\ 10 & -8 & -4 \end{array} \right] + 0 + 0 + 0 $ Expanding again along the first column, we have $\det \left(D\right) = 1 \left ( 0 + 8\det \left[ \begin{array}{rr} -1 & -1 \\ -8 & -4 \end{array} \right] +10\det \left[ \begin{array}{rr} -1 & -1 \\ -4 & 1 \end{array} \right] \right) = -82 $.

Now since $\det \left(A\right) = \det \left(D\right)$, it follows that $\det \left(A\right) = -82$.

Remember that you can verify these answers by using Laplace Expansion on $A$. Similarly, if you first compute the determinant using Laplace Expansion, you can use the row operation method to verify.

Applications of the Determinant 行列式的应用

学习目标 Outcomes

1. Use determinants to determine whether a matrix has an inverse, and evaluate the inverse using cofactors.
2. Apply Cramer's Rule to solve a $2\times 2$ or a $3\times 3$ linear system.
3. Given data points, find an appropriate interpolating polynomial and use it to estimate points.

1. 用行列式判断一个矩阵是否有逆，并用代数余子式求逆矩阵。
2. 应用克莱默法则（Cramer's Rule）求解 $2\times 2$ 或 $3\times 3$ 线性方程组。
3. 给定数据点，求合适的插值多项式，并用来估计点值。

逆的公式 A Formula for the Inverse

The determinant of a matrix also provides a way to find the inverse of a matrix. Recall the definition of the inverse of a matrix in Definition 2.6.1. We say that $A^{-1}$, an $n \times n$ matrix, is the inverse of $A$, also $n \times n$, if $AA^{-1} = I$ and $A^{-1}A=I$.

We now define a new matrix called the cofactor matrix of $A$. The cofactor matrix of $A$ is the matrix whose $ij^{\mathrm{th}}$ entry is the $ij^{\mathrm{th}}$ cofactor of $A$. The formal definition is as follows.

定义 1：代数余子式矩阵 The Cofactor Matrix

Let $A=\left[ a_{ij}\right]$ be an $n\times n$ matrix. Then the cofactor matrix of $A$, denoted $\mathrm{cof}\left( A\right)$, is defined by $\mathrm{cof}\left( A\right) =\left[ \mathrm{cof}\left(A\right)_{ij}\right]$ where $\mathrm{cof}\left(A\right)_{ij}$ is the $ij^{\mathrm{th}}$ cofactor of $A$.

Note that $\mathrm{cof}\left(A\right)_{ij}$ denotes the $ij^{\mathrm{th}}$ entry of the cofactor matrix.

We will use the cofactor matrix to create a formula for the inverse of $A$. First, we define the adjugate of $A$ to be the transpose of the cofactor matrix. We can also call this matrix the classical adjoint of $A$, and we denote it by $adj \left(A\right)$.

In the specific case where $A$ is a $2 \times 2$ matrix given by $A = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right] $ then ${adj}\left(A\right)$ is given by ${adj}\left(A\right) = \left[ \begin{array}{rr} d & -b \\ -c & a \end{array} \right] $.

In general, ${adj}\left(A\right)$ can always be found by taking the transpose of the cofactor matrix of $A$. The following theorem provides a formula for $A^{-1}$ using the determinant and adjugate of $A$.

定理 1：逆与行列式 The Inverse and the Determinant

Let $A$ be an $n\times n$ matrix. Then $A \cdot {adj}\left(A\right) = {adj}\left(A\right)A = {\det \left(A\right)} I $.

Moreover $A$ is invertible if and only if $\det \left(A\right) \neq 0$. In this case we have: $A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right) $.

Notice that the first formula holds for any $n \times n$ matrix $A$, and in the case $A$ is invertible we actually have a formula for $A^{-1}$.

Consider the following example.

示例 1：用行列式求逆 Find Inverse Using the Determinant

Find the inverse of the matrix $A=\left[ \begin{array}{rrr} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 1 \end{array} \right] $ using the formula in Theorem 1.

解 Solution

According to Theorem 1, $A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right) $.

First we will find the determinant of this matrix. Using Theorems 3.2.1, 3.2.2, and 3.2.4, we can first simplify the matrix through row operations. First, add $-3$ times the first row to the second row. Then add $-1$ times the first row to the third row to obtain $B = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -6 & -8 \\ 0 & 0 & -2 \end{array} \right] $ By Theorem 3.2.4, $\det \left(A\right) = \det \left(B\right)$. By Theorem 3.1.2, $\det \left(B\right) = 1 \times -6 \times -2 = 12$. Hence, $\det \left(A\right) = 12$.

Now, we need to find ${adj} \left(A\right)$. To do so, first we will find the cofactor matrix of $A$. This is given by $\mathrm{cof}\left( A\right) = \left[ \begin{array}{rrr} -2 & -2 & 6 \\ 4 & -2 & 0 \\ 2 & 8 & -6 \end{array} \right] $ Here, the $ij^{\mathrm{th}}$ entry is the $ij^{\mathrm{th}}$ cofactor of the original matrix $A$ which you can verify. Therefore, from Theorem 1, the inverse of $A$ is given by $A^{-1} = \frac{1}{12}\left[ \begin{array}{rrr} -2 & -2 & 6 \\ 4 & -2 & 0 \\ 2 & 8 & -6 \end{array} \right] ^{T}= \left[ \begin{array}{rrr} -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{2}{3} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{array} \right] $.

Remember that we can always verify our answer for $A^{-1}$. Compute the product $AA^{-1}$ and $A^{-1}A$ and make sure each product is equal to $I$.

Compute $A^{-1}A$ as follows. $A^{-1}A = \left[ \begin{array}{rrr} -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{2}{3} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{array} \right] \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 1 \end{array} \right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I $ You can verify that $AA^{-1} = I$ and hence our answer is correct.

We will look at another example of how to use this formula to find $A^{-1}$.

示例 2：用公式求逆 Find the Inverse From a Formula

Find the inverse of the matrix $A=\left[ \begin{array}{rrr} \frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{6} & \frac{1}{3} & - \frac{1}{2} \\ -\frac{5}{6} & \frac{2}{3} & - \frac{1}{2} \end{array} \right] $ using the formula given in Theorem 1.

解 Solution

First we need to find $\det \left(A\right)$. This step is left as an exercise and you should verify that $\det \left(A\right) = \frac{1}{6}.$ The inverse is therefore equal to $A^{-1} = \frac{1}{(1/6)} {adj} \left(A\right) = 6 {adj} \left(A\right) $.

We continue to calculate as follows. Here we show the $2 \times 2$ determinants needed to find the cofactors. $A^{-1} = 6\left[ \begin{array}{rrr} \left\| \begin{array}{rr} \frac{1}{3} & -\frac{1}{2} \\ \frac{2}{3} & -\frac{1}{2} \end{array} \right\| & -\left\| \begin{array}{rr} -\frac{1}{6} & -\frac{1}{2} \\ -\frac{5}{6} & -\frac{1}{2} \end{array} \right\| & \left\| \begin{array}{rr} -\frac{1}{6} & \frac{1}{3} \\ -\frac{5}{6} & \frac{2}{3} \end{array} \right\| \\ -\left\| \begin{array}{rr} 0 & \frac{1}{2} \\ \frac{2}{3} & -\frac{1}{2} \end{array} \right\| & \left\| \begin{array}{rr} \frac{1}{2} & \frac{1}{2} \\ -\frac{5}{6} & -\frac{1}{2} \end{array} \right\| & -\left\| \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{5}{6} & \frac{2}{3} \end{array} \right\| \\ \left\| \begin{array}{rr} 0 & \frac{1}{2} \\ \frac{1}{3} & -\frac{1}{2} \end{array} \right\| & -\left\| \begin{array}{rr} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{6} & -\frac{1}{2} \end{array} \right\| & \left\| \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{1}{6} & \frac{1}{3} \end{array} \right\| \end{array} \right] ^{T} $.

Expanding all the $2\times 2$ determinants, this yields $A^{-1} = 6\left[ \begin{array}{rrr} \frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ \frac{1}{3} & \frac{1}{6} & -\frac{1}{3} \\ -\frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} \right] ^{T}= \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{array} \right] $.

Again, you can always check your work by multiplying $A^{-1}A$ and $AA^{-1}$ and ensuring these products equal $I$. $A^{-1}A = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{array} \right] \left[ \begin{array}{rrr} \frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{6} & \frac{1}{3} & - \frac{1}{2} \\ -\frac{5}{6} & \frac{2}{3} & - \frac{1}{2} \end{array} \right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] $ This tells us that our calculation for $A^{-1}$ is correct. It is left to the reader to verify that $AA^{-1} = I$.

The verification step is very important, as it is a simple way to check your work! If you multiply $A^{-1}A$ and $AA^{-1}$ and these products are not both equal to $I$, be sure to go back and double check each step. One common error is to forget to take the transpose of the cofactor matrix, so be sure to complete this step.

We will now prove Theorem 1.

定理 1 的证明 Proof of Theorem 1

Proof: Recall that the $(i,j)$-entry of ${adj}(A)$ is equal to $\mathrm{cof}(A)_{ji}$. Thus the $(i,j)$-entry of $B=A\cdot {adj}(A)$ is : $B_{ij}=\sum_{k=1}^n a_{ik} {adj} (A)_{kj}= \sum_{k=1}^n a_{ik} \mathrm{cof} (A)_{jk} $ By the cofactor expansion theorem, we see that this expression for $B_{ij}$ is equal to the determinant of the matrix obtained from $A$ by replacing its $j$th row by $a_{i1}, a_{i2}, \dots a_{in}$ — i.e., its $i$th row.

If $i=j$ then this matrix is $A$ itself and therefore $B_{ii}=\det A$. If on the other hand $i\neq j$, then this matrix has its $i$th row equal to its $j$th row, and therefore $B_{ij}=0$ in this case. Thus we obtain: $A \cdot {adj}\left(A\right) = {\det \left(A\right)} I $. Similarly we can verify that: ${adj}\left(A\right)A = {\det \left(A\right)} I $. And this proves the first part of the theorem.

Further if $A$ is invertible, then by Theorem 3.2.5 we have: $1 = \det \left( I \right) = \det \left( A A^{-1} \right) = \det \left( A \right) \det \left( A^{-1} \right) $ and thus $\det \left( A \right) \neq 0$. Equivalently, if $\det \left( A \right) = 0$, then $A$ is not invertible.

Finally if $\det \left( A \right) \neq 0$, then the above formula shows that $A$ is invertible and that: $A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right) $.

This completes the proof.

This method for finding the inverse of $A$ is useful in many contexts. In particular, it is useful with complicated matrices where the entries are functions, rather than numbers.

Consider the following example.

示例 3：非常数矩阵的逆 Inverse for Non-Constant Matrix

Suppose $A\left( t\right) =\left[ \begin{array}{ccc} e^{t} & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & -\sin t & \cos t \end{array} \right] $ Show that $A\left( t\right) ^{-1}$ exists and then find it.

解 Solution

First note $\det \left( A\left( t\right) \right) = e^{t}(\cos^2 t + \sin^2 t) = e^{t}\neq 0$ so $A\left( t\right) ^{-1}$ exists.

The cofactor matrix is $C\left( t\right) =\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right] $ and so the inverse is $\frac{1}{e^{t}}\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right] ^{T}= \left[ \begin{array}{ccc} e^{-t} & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array} \right] $.

克莱默法则 Cramer's Rule

Another context in which the formula given in Theorem 1 is important is Cramer's Rule. Recall that we can represent a system of linear equations in the form $AX=B$, where the solutions to this system are given by $X$. Cramer's Rule gives a formula for the solutions $X$ in the special case that $A$ is a square invertible matrix. Note this rule does not apply if you have a system of equations in which there is a different number of equations than variables (in other words, when $A$ is not square), or when $A$ is not invertible.

Suppose we have a system of equations given by $AX=B$, and we want to find solutions $X$ which satisfy this system. Then recall that if $A^{-1}$ exists, $\begin{aligned} AX&=B \\ A^{-1}\left(AX\right)&=A^{-1}B \\ \left(A^{-1}A\right)X&=A^{-1}B \\ IX&=A^{-1}B \\ X &= A^{-1}B \end{aligned}$ Hence, the solutions $X$ to the system are given by $X=A^{-1}B$. Since we assume that $A^{-1}$ exists, we can use the formula for $A^{-1}$ given above. Substituting this formula into the equation for $X$, we have $X=A^{-1}B=\frac{1}{\det \left( A\right) }{adj}\left( A\right)B $. Let $x_i$ be the $i^{\mathrm{th}}$ entry of $X$ and $b_j$ be the $j^{\mathrm{th}}$ entry of $B$. Then this equation becomes $x_i = \sum_{j=1}^{n}\left[ a_{ij}\right]^{-1}b_{j}=\sum_{j=1}^{n}\frac{1} {\det \left( A\right) } {adj}\left( A\right) _{ij}b_{j} $ where ${adj}\left(A\right)_{ij}$ is the $ij^{\mathrm{th}}$ entry of ${adj}\left(A\right)$.

By the formula for the expansion of a determinant along a column, $x_{i}=\frac{1}{\det \left( A\right) }\det \left[ \begin{array}{ccccc} \ast & \cdots & b_{1} & \cdots & \ast \\ \vdots & & \vdots & & \vdots \\ \ast & \cdots & b_{n} & \cdots & \ast \end{array} \right] $ where here the $i^{\mathrm{th}}$ column of $A$ is replaced with the column vector $\left[ b_{1}\cdots ,b_{n}\right] ^{T}$. The determinant of this modified matrix is taken and divided by $\det \left( A\right)$. This formula is known as Cramer's rule.

We formally define this method now.

过程 1：使用克莱默法则 Using Cramer's Rule

Suppose $A$ is an $n\times n$ invertible matrix and we wish to solve the system $AX=B$ for $X =\left[ x_{1},\cdots ,x_{n}\right] ^{T}.$ Then Cramer's rule says $x_{i}= \frac{\det \left(A_{i}\right)}{\det \left(A\right)} $ where $A_{i}$ is the matrix obtained by replacing the $i^{\mathrm{th}}$ column of $A$ with the column matrix $B = \left[ \begin{array}{c} b_1 \\ \vdots \\ b_n \end{array} \right] $.

We illustrate this procedure in the following example.

示例 4：使用克莱默法则 Using Cramer's Rule

Find $x,y,z$ if $\left[ \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] $.

解 Solution

We will use method outlined in Procedure 1 to find the values for $x,y,z$ which give the solution to this system. Let $B = \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] $.

In order to find $x$, we calculate $x = \frac{\det \left(A_{1}\right)}{\det \left(A\right)} $ where $A_1$ is the matrix obtained from replacing the first column of $A$ with $B$.

Hence, $A_1$ is given by $A_1 = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 3 & -3 & 2 \end{array} \right] $.

Therefore, $x= \frac{\det \left(A_{1}\right)}{\det \left(A\right)} = \frac{\left\| \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 3 & -3 & 2 \end{array} \right\| }{\left\| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right\| }=\frac{1}{2} $.

Similarly, to find $y$ we construct $A_2$ by replacing the second column of $A$ with $B$. Hence, $A_2$ is given by $A_2 = \left[ \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 3 & 2 \end{array} \right] $.

Therefore, $y=\frac{\det \left(A_{2}\right)}{\det \left(A\right)} = \frac{\left\| \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 3 & 2 \end{array} \right\| }{\left\| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right\| }=-\frac{1}{7} $.

Similarly, $A_3$ is constructed by replacing the third column of $A$ with $B$. Then, $A_3$ is given by $A_3 = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 2 \\ 2 & -3 & 3 \end{array} \right] $.

Therefore, $z$ is calculated as follows: $z= \frac{\det \left(A_{3}\right)}{\det \left(A\right)} = \frac{\left\| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 2 \\ 2 & -3 & 3 \end{array} \right\| }{\left\| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right\| }=\frac{11}{14} $.

Cramer's Rule gives you another tool to consider when solving a system of linear equations.

We can also use Cramer's Rule for systems of non linear equations. Consider the following system where the matrix $A$ has functions rather than numbers for entries.

示例 5：对非常数矩阵用克莱默法则 Use Cramer's Rule for Non-Constant Matrix

Solve for $z$ if $\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{c} 1 \\ t \\ t^{2} \end{array} \right] $.

解 Solution

We are asked to find the value of $z$ in the solution. We will solve using Cramer's rule. Thus $z= \frac{\left\| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & e^{t}\cos t & t \\ 0 & -e^{t}\sin t & t^{2} \end{array} \right\| }{\left\| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right\| }= t\left( \left( \cos t\right) t+\sin t\right) e^{-t} $.

多项式插值 Polynomial Interpolation

In studying a set of data that relates variables $x$ and $y$, it may be the case that we can use a polynomial to "fit" to the data. If such a polynomial can be established, it can be used to estimate values of $x$ and $y$ which have not been provided.

Consider the following example.

示例 6：多项式插值 Polynomial Interpolation

Given data points $(1,4), (2,9), (3,12)$, find an interpolating polynomial $p(x)$ of degree at most $2$ and then estimate the value corresponding to $x = \frac{1}{2}$.

解 Solution

We want to find a polynomial given by $p(x) = r_0 + r_1x_1 + r_2x_2^2 $ such that $p(1)=4, p(2)=9$ and $p(3)=12$. To find this polynomial, substitute the known values in for $x$ and solve for $r_0, r_1$, and $r_2$. $\begin{aligned} p(1) &= r_0 + r_1 + r_2 = 4 \\ p(2) &= r_0 + 2r_1 + 4r_2 = 9 \\ p(3) &= r_0 + 3r_1 + 9r_2 = 12 \end{aligned}$.

Writing the augmented matrix, we have $\left[ \begin{array}{rrr|r} 1 & 1 & 1 & 4 \\ 1 & 2 & 4 & 9 \\ 1 & 3 & 9 & 12 \end{array} \right] $.

After row operations, the resulting matrix is $\left[ \begin{array}{rrr|r} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & -1 \end{array} \right] $.

Therefore the solution to the system is $r_0 = -3, r_1 = 8, r_2 = -1$ and the required interpolating polynomial is $p(x) = -3 + 8x - x^2 $.

To estimate the value for $x = \frac{1}{2}$, we calculate $p(\frac{1}{2})$: $\begin{aligned} p(\frac{1}{2}) &= -3 + 8(\frac{1}{2}) - (\frac{1}{2})^2 \\ &= -3 + 4 - \frac{1}{4} \\ &= \frac{3}{4} \end{aligned}$.

This procedure can be used for any number of data points, and any degree of polynomial. The steps are outlined below.

过程 2：求插值多项式 Finding an Interpolation Polynomial

Suppose that values of $x$ and corresponding values of $y$ are given, such that the actual relationship between $x$ and $y$ is unknown. Then, values of $y$ can be estimated using an interpolating polynomial $p(x)$. If given $x_1, ..., x_n$ and the corresponding $y_1, ..., y_n$, the procedure to find $p(x)$ is as follows:

1. The desired polynomial $p(x)$ is given by $p(x) = r_0 + r_1 x + r_2 x^2 + ... + r_{n-1}x^{n-1} $.
2. $p(x_i) = y_i$ for all $i = 1, 2, ...,n$ so that $\begin{array}{c} r_0 + r_1x_1 + r_2 x_1^2 + ... + r_{n-1}x_1^{n-1} = y_1 \\ r_0 + r_1x_2 + r_2 x_2^2 + ... + r_{n-1}x_2^{n-1} = y_2 \\ \vdots \\ r_0 + r_1x_n + r_2 x_n^2 + ... + r_{n-1}x_n^{n-1} = y_n \end{array} $.
3. Set up the augmented matrix of this system of equations $\left[ \begin{array}{rrrrr|r} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & y_1 \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & y_2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & y_n \end{array} \right] $.
4. Solving this system will result in a unique solution $r_0, r_1, \cdots, r_{n-1}$. Use these values to construct $p(x)$, and estimate the value of $p(a)$ for any $x=a$.

1. 所求多项式 $p(x)$ 为 $p(x) = r_0 + r_1 x + r_2 x^2 + ... + r_{n-1}x^{n-1} $。
2. 对所有 $i = 1, 2, ...,n$ 有 $p(x_i) = y_i$，即 $\begin{array}{c} r_0 + r_1x_1 + r_2 x_1^2 + ... + r_{n-1}x_1^{n-1} = y_1 \\ r_0 + r_1x_2 + r_2 x_2^2 + ... + r_{n-1}x_2^{n-1} = y_2 \\ \vdots \\ r_0 + r_1x_n + r_2 x_n^2 + ... + r_{n-1}x_n^{n-1} = y_n \end{array} $。
3. 列出这个方程组的增广矩阵 $\left[ \begin{array}{rrrrr|r} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & y_1 \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & y_2 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & y_n \end{array} \right] $。
4. 求解这个方程组会得到一个唯一解 $r_0, r_1, \cdots, r_{n-1}$。用这些值构造 $p(x)$，并对任意 $x=a$ 估计 $p(a)$ 的值。

This procedure motivates the following theorem.

定理 2：多项式插值 Polynomial Interpolation

Given $n$ data points $(x_1, y_1), (x_2, y_2), \cdots, (x_n, y_n)$ with the $x_i$ distinct, there is a unique polynomial $p(x) = r_0 + r_1x + r_2x^2 + \cdots + r_{n-1}x^{n-1}$ such that $p(x_i) = y_i$ for $i=1,2,\cdots, n$. The resulting polynomial $p(x)$ is called the interpolating polynomial for the data points.

We conclude this section with another example.

示例 7：多项式插值 Polynomial Interpolation

Consider the data points $(0,1), (1,2), (3,22), (5,66)$. Find an interpolating polynomial $p(x)$ of degree at most three, and estimate the value of $p(2)$.

解 Solution

The desired polynomial $p(x)$ is given by: $p(x) = r_0 + r_1 x + r_2x^2 + r_3x^3 $.

Using the given points, the system of equations is $\begin{aligned} p(0) &= r_0 = 1 \\ p(1) &= r_0 + r_1 + r_2 + r_3 = 2 \\ p(3) &= r_0 + 3r_1 + 9r_2 + 27r_3 = 22 \\ p(5) &= r_0 + 5r_1 + 25r_2 + 125r_3 = 66 \end{aligned}$.

The augmented matrix is given by: $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 2 \\ 1 & 3 & 9 & 27 & 22 \\ 1 & 5 & 25 & 125 & 66 \end{array} \right] $.

The resulting matrix is $\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right] $.

Therefore, $r_0 = 1, r_1 = -2, r_2 = 3, r_3 = 0$ and $p(x) = 1 -2x + 3x^2$. To estimate the value of $p(2)$, we compute $p(2) = 1 -2(2) + 3(2^2) = 1 - 4 + 12 = 9$.

习题 Exercises

习题 1 Exercise 1

Find the determinants of the following matrices.

1. $\left[\begin{array}{cc}1&3\\0&2\end{array}\right]$
2. $\left[\begin{array}{cc}0&3\\0&2\end{array}\right]$
3. $\left[\begin{array}{cc}4&3\\6&2\end{array}\right]$

1. $\left[\begin{array}{cc}1&3\\0&2\end{array}\right]$
2. $\left[\begin{array}{cc}0&3\\0&2\end{array}\right]$
3. $\left[\begin{array}{cc}4&3\\6&2\end{array}\right]$

习题 2 Exercise 2

Let $A=\left[\begin{array}{ccc}1&2&4\\0&1&3\\-2&5&1\end{array}\right]$. Find the following.

1. $minor(A)_{11}$
2. $minor(A)_{21}$
3. $minor(A)_{32}$
4. $cof(A)_{11}$
5. $cof(A)_{21}$
6. $cof(A)_{32}$

1. $minor(A)_{11}$
2. $minor(A)_{21}$
3. $minor(A)_{32}$
4. $cof(A)_{11}$
5. $cof(A)_{21}$
6. $cof(A)_{32}$

习题 3 Exercise 3

Find the determinants of the following matrices.

1. $\left[\begin{array}{ccc}1&2&3\\3&2&2\\0&9&8\end{array}\right]$
2. $\left[\begin{array}{ccc}4&3&2\\1&7&8\\3&-9&3\end{array}\right]$
3. $\left[\begin{array}{cccc}1&2&3&2\\1&3&2&3\\4&1&5&0\\1&2&1&2\end{array}\right]$

1. $\left[\begin{array}{ccc}1&2&3\\3&2&2\\0&9&8\end{array}\right]$
2. $\left[\begin{array}{ccc}4&3&2\\1&7&8\\3&-9&3\end{array}\right]$
3. $\left[\begin{array}{cccc}1&2&3&2\\1&3&2&3\\4&1&5&0\\1&2&1&2\end{array}\right]$

Answer: The answers are $31$, $375$, and $-2$ respectively.

习题 4 Exercise 4

Find the following determinant by expanding along the first row and second column. $\left\|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right\|$

Answer: $\left\|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right\|=6$

习题 5 Exercise 5

Find the following determinant by expanding along the first column and third row. $\left\|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right\|$

Answer: $\left\|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right\|=2$

习题 6 Exercise 6

Find the following determinant by expanding along the second row and first column. $\left\|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right\|$

Answer: $\left\|\begin{array}{ccc}1&2&1\\2&1&3\\2&1&1\end{array}\right\|=6$

习题 7 Exercise 7

Compute the determinant by cofactor expansion. Pick the easiest row or column to use. $\left\|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right\|$

Answer: $\left\|\begin{array}{cccc}1&0&0&1\\2&1&1&0\\0&0&0&2\\2&1&3&1\end{array}\right\|=-4$

习题 8 Exercise 8

Find the determinant of the following matrices.

1. $A=\left[\begin{array}{cc}1&-34\\0&2\end{array}\right]$
2. $A=\left[\begin{array}{ccc}4&3&14\\0&-2&0\\0&0&5\end{array}\right]$
3. $A=\left[\begin{array}{cccc}2&3&15&0\\0&4&1&7\\0&0&-3&5\\0&0&0&1\end{array}\right]$

1. $A=\left[\begin{array}{cc}1&-34\\0&2\end{array}\right]$
2. $A=\left[\begin{array}{ccc}4&3&14\\0&-2&0\\0&0&5\end{array}\right]$
3. $A=\left[\begin{array}{cccc}2&3&15&0\\0&4&1&7\\0&0&-3&5\\0&0&0&1\end{array}\right]$

习题 9 Exercise 9

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&c\\b&d\end{array}\right]$

Answer: It does not change the determinant. This was just taking the transpose.

习题 10 Exercise 10

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}c&d\\a&b\end{array}\right]$

Answer: In this case two rows were switched and so the resulting determinant is $-1$ times the first.

习题 11 Exercise 11

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&b\\a+c&b+d\end{array}\right]$

Answer: The determinant is unchanged. It was just the first row added to the second.

习题 12 Exercise 12

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}a&b\\2c&2d\end{array}\right]$

Answer: The second row was multiplied by $2$ so the determinant of the result is $2$ times the original determinant.

习题 13 Exercise 13

An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant. $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\to\cdots\to\left[\begin{array}{cc}b&a\\d&c\end{array}\right]$

Answer: In this case the two columns were switched so the determinant of the second is $-1$ times the determinant of the first.

习题 14 Exercise 14

Let $A$ be an $r\times r$ matrix and suppose there are $r-1$ rows (columns) such that all rows (columns) are linear combinations of these $r-1$ rows (columns). Show $\det(A) = 0$.

Answer: If the determinant is nonzero, then it will remain nonzero with row operations applied to the matrix. However, by assumption, you can obtain a row of zeros by doing row operations. Thus the determinant must have been zero after all.

习题 15 Exercise 15

Show $\det(aA) = a^n \det(A)$ for an $n\times n$ matrix $A$ and scalar $a$.

Answer: $\det(aA) = \det(aIA) = \det(aI)\det(A) = a^n \det(A)$. The matrix which has $a$ down the main diagonal has determinant equal to $a^n$.

习题 16 Exercise 16

Construct $2\times 2$ matrices $A$ and $B$ to show that the $\det A\det B = \det(AB)$.

Answer: $\det\left(\left[\begin{array}{cc}1&2\\3&4\end{array}\right]\left[\begin{array}{cc}-1&2\\-5&6\end{array}\right]\right)=-8$ and $\det\left[\begin{array}{cc}1&2\\3&4\end{array}\right]\det\left[\begin{array}{cc}-1&2\\-5&6\end{array}\right]=-2\times 4=-8$.

习题 17 Exercise 17

Is it true that $\det(A+B) = \det(A)+\det(B)$? If this is so, explain why. If it is not so, give a counter example.

Answer: This is not true at all. Consider $A=\left[\begin{array}{cc}1&0\\0&1\end{array}\right], B=\left[\begin{array}{cc}-1&0\\0&-1\end{array}\right]$.

习题 18 Exercise 18

An $n\times n$ matrix is called nilpotent if for some positive integer, $k$ it follows $A^k = 0$. If $A$ is a nilpotent matrix and $k$ is the smallest possible integer such that $A^k = 0$, what are the possible values of $\det(A)$?

Answer: It must be $0$ because $0 = \det(0) = \det (A^k) = (\det(A))^k$.

习题 19 Exercise 19

A matrix is said to be orthogonal if $A^TA = I$. Thus the inverse of an orthogonal matrix is just its transpose. What are the possible values of $\det(A)$ if $A$ is an orthogonal matrix?

Answer: You would need $\det (AA^T) = \det(A)\det (A^T) = \det(A)^2 = 1$ and so $\det(A)$ is $1$ or $-1$.

习题 20 Exercise 20

Let $A$ and $B$ be two $n\times n$ matrices. $A \sim B$ ($A$ is similar to $B$) means there exists an invertible matrix $P$ such that $A = P^{-1}BP$. Show that if $A \sim B$, then $\det(A) = \det(B)$.

Answer: $\det(A) = \det(P^{-1}BP) = \det(P^{-1})\det(B)\det(P) = \det(B)\det(P^{-1}P) = \det(B)$.

习题 21 Exercise 21

Tell whether each statement is true or false. If true, provide a proof. If false, provide a counter example.

1. If $A$ is a $3\times 3$ matrix with a zero determinant, then one column must be a multiple of some other column.
2. If any two columns of a square matrix are equal, then the determinant of the matrix equals zero.
3. For two $n\times n$ matrices $A$ and $B$, $\det(A+B) = \det(A) +\det(B)$.
4. For an $n\times n$ matrix $A$, $\det(3A) = 3 \det(A)$.
5. If $A^{-1}$ exists then $\det(A^{-1}) = \det(A)^{-1}$.
6. If $B$ is obtained by multiplying a single row of $A$ by $4$ then $\det(B) = 4 \det(A)$.
7. For $A$ an $n\times n$ matrix, $\det(-A) = (-1)^n \det(A)$.
8. If $A$ is a real $n\times n$ matrix, then $\det (A^TA) \geq 0$.
9. If $A^k = 0$ for some positive integer $k$, then $\det(A) = 0$.
10. If $AX = 0$ for some $X\neq 0$, then $\det(A) = 0$.

1. 若 $A$ 是 $3\times 3$ 矩阵且行列式为零，则必有一列是另一列的倍数。
2. 若方阵的任意两列相等，则该矩阵的行列式等于零。
3. 对两个 $n\times n$ 矩阵 $A$、$B$，$\det(A+B) = \det(A) +\det(B)$。
4. 对 $n\times n$ 矩阵 $A$，$\det(3A) = 3 \det(A)$。
5. 若 $A^{-1}$ 存在，则 $\det(A^{-1}) = \det(A)^{-1}$。
6. 若 $B$ 是把 $A$ 的某一行乘以 $4$ 所得，则 $\det(B) = 4 \det(A)$。
7. 对 $n\times n$ 矩阵 $A$，$\det(-A) = (-1)^n \det(A)$。
8. 若 $A$ 是实 $n\times n$ 矩阵，则 $\det (A^TA) \geq 0$。
9. 若对某个正整数 $k$ 有 $A^k = 0$，则 $\det(A) = 0$。
10. 若对某个非零的 $X$ 有 $AX = 0$，则 $\det(A) = 0$。

Answer: (1) False, consider $\left[\begin{array}{ccc}1&1&2\\-1&5&4\\0&3&3\end{array}\right]$. (2) True. (3) False. (4) False. (5) True. (6) False. (7) True. (8) True. (9) True. (10) True.

习题 22 Exercise 22

Find the determinant using row operations to first simplify. $\left\|\begin{array}{ccc}1&2&1\\2&3&2\\-4&1&2\end{array}\right\|$

Answer: $\left\|\begin{array}{ccc}1&2&1\\2&3&2\\-4&1&2\end{array}\right\|=-6$

习题 23 Exercise 23

Find the determinant using row operations to first simplify. $\left\|\begin{array}{ccc}2&1&3\\2&4&2\\1&4&-5\end{array}\right\|$

Answer: $\left\|\begin{array}{ccc}2&1&3\\2&4&2\\1&4&-5\end{array}\right\|=-32$

习题 24 Exercise 24

Find the determinant using row operations to first simplify. $\left\|\begin{array}{cccc}1&2&1&2\\3&1&-2&3\\-1&0&3&1\\2&3&2&-2\end{array}\right\|$

Answer: One can row reduce this using only row operation 3 to $\left[\begin{array}{cccc}1&2&1&2\\0&-5&-5&-3\\0&0&2&\frac{9}{5}\\0&0&0&-\frac{63}{10}\end{array}\right]$ and therefore, the determinant is $-63$. $\left\|\begin{array}{cccc}1&2&1&2\\3&1&-2&3\\-1&0&3&1\\2&3&2&-2\end{array}\right\|=63$

习题 25 Exercise 25

Find the determinant using row operations to first simplify. $\left\|\begin{array}{cccc}1&4&1&2\\3&2&-2&3\\-1&0&3&3\\2&1&2&-2\end{array}\right\|$

Answer: One can row reduce this using only row operation 3 to $\left[\begin{array}{cccc}1&4&1&2\\0&-10&-5&-3\\0&0&2&\frac{19}{5}\\0&0&0&-\frac{211}{20}\end{array}\right]$ Thus the determinant is given by $\left\|\begin{array}{cccc}1&4&1&2\\3&2&-2&3\\-1&0&3&3\\2&1&2&-2\end{array}\right\|=211$

习题 26 Exercise 26

Let $A=\left[\begin{array}{ccc}1&2&3\\0&2&1\\3&1&0\end{array}\right]$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse which involves the cofactor matrix.

Answer: $\det\left[\begin{array}{ccc}1&2&3\\0&2&1\\3&1&0\end{array}\right]=-13$, and so it has an inverse. This inverse is $\frac{1}{-13}\left[\begin{array}{rrr}\left\|\begin{array}{cc}2&1\\1&0\end{array}\right\|&-\left\|\begin{array}{cc}0&1\\3&0\end{array}\right\|&\left\|\begin{array}{cc}0&2\\3&1\end{array}\right\|\\-\left\|\begin{array}{cc}2&3\\1&0\end{array}\right\|&\left\|\begin{array}{cc}1&3\\3&0\end{array}\right\|&-\left\|\begin{array}{cc}1&2\\3&1\end{array}\right\|\\\left\|\begin{array}{cc}2&3\\2&1\end{array}\right\|&-\left\|\begin{array}{cc}1&3\\0&1\end{array}\right\|&\left\|\begin{array}{cc}1&2\\0&2\end{array}\right\|\end{array}\right]^T =\frac{1}{-13}\left[\begin{array}{ccc}-1&3&-6\\3&-9&5\\-4&-1&2\end{array}\right]^T =\left[\begin{array}{ccc}\frac{1}{13}&-\frac{3}{13}&\frac{4}{13}\\-\frac{3}{13}&\frac{9}{13}&\frac{1}{13}\\\frac{6}{13}&-\frac{5}{13}&-\frac{2}{13}\end{array}\right]$.

习题 27 Exercise 27

Let $A=\left[\begin{array}{ccc}1&2&0\\0&2&1\\3&1&1\end{array}\right]$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Answer: $\det\left[\begin{array}{ccc}1&2&0\\0&2&1\\3&1&1\end{array}\right]=7$, so it has an inverse. This inverse is $\frac{1}{7}\left[\begin{array}{ccc}1&3&-6\\-2&1&5\\2&-1&2\end{array}\right]^T=\left[\begin{array}{ccc}\frac{1}{7}&-\frac{2}{7}&\frac{2}{7}\\\frac{3}{7}&\frac{1}{7}&-\frac{1}{7}\\-\frac{6}{7}&\frac{5}{7}&\frac{2}{7}\end{array}\right]$.

习题 28 Exercise 28

Let $A=\left[\begin{array}{ccc}1&3&3\\2&4&1\\0&1&1\end{array}\right]$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Answer: $\det\left[\begin{array}{ccc}1&3&3\\2&4&1\\0&1&1\end{array}\right]=3$, so it has an inverse which is $\left[\begin{array}{ccc}1&0&-3\\-\frac{2}{3}&\frac{1}{3}&\frac{5}{3}\\\frac{2}{3}&-\frac{1}{3}&-\frac{2}{3}\end{array}\right]$.

习题 29 Exercise 29

Let $A=\left[\begin{array}{ccc}1&2&3\\0&2&1\\2&6&7\end{array}\right]$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

习题 30 Exercise 30

Let $A=\left[\begin{array}{ccc}1&0&3\\1&0&1\\3&1&0\end{array}\right]$ Determine whether the matrix $A$ has an inverse by finding whether the determinant is non zero. If the determinant is nonzero, find the inverse using the formula for the inverse.

Answer: $\det\left[\begin{array}{ccc}1&0&3\\1&0&1\\3&1&0\end{array}\right]=2$, and so it has an inverse. The inverse turns out to equal $\left[\begin{array}{ccc}-\frac{1}{2}&\frac{3}{2}&0\\\frac{3}{2}&-\frac{9}{2}&1\\\frac{1}{2}&-\frac{1}{2}&0\end{array}\right]$.

习题 31 Exercise 31

For the following matrices, determine if they are invertible. If so, use the formula for the inverse in terms of the cofactor matrix to find each inverse. If the inverse does not exist, explain why.

1. $\left[\begin{array}{cc}1&1\\1&2\end{array}\right]$
2. $\left[\begin{array}{ccc}1&2&3\\0&2&1\\4&1&1\end{array}\right]$
3. $\left[\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right]$

1. $\left[\begin{array}{cc}1&1\\1&2\end{array}\right]$
2. $\left[\begin{array}{ccc}1&2&3\\0&2&1\\4&1&1\end{array}\right]$
3. $\left[\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right]$

Answer: (1) $\left\|\begin{array}{cc}1&1\\1&2\end{array}\right\|=1$. (2) $\left\|\begin{array}{ccc}1&2&3\\0&2&1\\4&1&1\end{array}\right\|=-15$. (3) $\left\|\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right\|=0$.

习题 32 Exercise 32

Consider the matrix $A=\left[\begin{array}{ccc}1&0&0\\0&\cos t&-\sin t\\0&\sin t&\cos t\end{array}\right]$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Answer: No. It has a nonzero determinant for all $t$.

习题 33 Exercise 33

Consider the matrix $A=\left[\begin{array}{ccc}1&t&t^2\\0&1&2t\\t&0&2\end{array}\right]$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Answer: $\det\left[\begin{array}{ccc}1&t&t^2\\0&1&2t\\t&0&2\end{array}\right]=t^3+2$, and so it has no inverse when $t=\sqrt[3]{-2}$.

习题 34 Exercise 34

Consider the matrix $A=\left[\begin{array}{ccc}e^t &\cosh t&\sinh t\\e^t&\sinh t&\cosh t\\e^t&\cosh t&\sinh t\end{array}\right]$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Answer: $\det\left[\begin{array}{ccc}e^t&\cosh t&\sinh t\\e^t&\sinh t&\cosh t\\e^t&\cosh t&\sinh t\end{array}\right]=0$ and so this matrix fails to have a nonzero determinant at any value of $t$.

习题 35 Exercise 35

Consider the matrix $A=\left[\begin{array}{ccc}e^t &e^{-t}\cos t&e^{-t}\sin t\\e^t&-e^{-t}\cos t-e^{-t}\sin t &-e^{-t}\sin t+e^{-t}\cos t\\e^t&2e^{-t}\sin t&-2e^{-t}\cos t\end{array}\right]$ Does there exist a value of $t$ for which this matrix fails to have an inverse? Explain.

Answer: $\det\left[\begin{array}{ccc}e^t&e^{-t}\cos t&e^{-t}\sin t\\e^t&-e^{-t}\cos t-e^{-t}\sin t&-e^{-t}\sin t+e^{-t}\cos t\\e^t&2e^{-t}\sin t&-2e^{-t}\cos t\end{array}\right]=5e^{-t}\neq 0$ and so this matrix is always invertible.

习题 36 Exercise 36

Show that if $\det(A)\neq 0$ for $A$ an $n\times n$ matrix, it follows that if $AX = 0$, then $X = 0$.

Answer: If $\det(A) \neq 0$, then $A^{-1}$ exists and so you could multiply on both sides on the left by $A^{-1}$ and obtain that $X = 0$.

习题 37 Exercise 37

Suppose $A,B$ are $n\times n$ matrices and that $AB = I$. Show that then $BA = I$. Hint: First explain why $\det(A)$, $\det(B)$ are both nonzero. Then $(AB)A = A$ and then show $BA(BA-I) = 0$. From this use what is given to conclude $A(BA-I) = 0$. Then use Exercise 36.

Answer: You have $1 = \det(A)\det(B)$. Hence both $A$ and $B$ have inverses. Letting $X$ be given, $A(BA-I)X = (AB)AX - AX = AX - AX = 0$ and so it follows from the above problem that $(BA-I)X = 0$. Since $X$ is arbitrary, it follows that $BA = I$.

习题 38 Exercise 38

Use the formula for the inverse in terms of the cofactor matrix to find the inverse of the matrix $A=\left[\begin{array}{ccc}e^t&0&0\\0&e^t\cos t&e^t\sin t\\0&e^t\cos t-e^t\sin t&e^t\cos t+e^t\sin t\end{array}\right]$.

Answer: $\det\left[\begin{array}{ccc}e^t&0&0\\0&e^t\cos t&e^t\sin t\\0&e^t\cos t-e^t\sin t&e^t\cos t+e^t\sin t\end{array}\right]=e^{3t}$. Hence the inverse is $e^{-3t}\left[\begin{array}{ccc}e^{2t}&0&0\\0&e^{2t}\cos t+e^{2t}\sin t&-(e^{2t}\cos t-e^{2t}\sin t)\\0&-e^{2t}\sin t&e^{2t}\cos t\end{array}\right]^T =\left[\begin{array}{ccc}e^{-t}&0&0\\0&e^{-t}(\cos t+\sin t)&-\sin t \cdot e^{-t}\\0&-e^{-t}(\cos t-\sin t)&\cos t \cdot e^{-t}\end{array}\right]$.

习题 39 Exercise 39

Find the inverse, if it exists, of the matrix $A=\left[\begin{array}{ccc}e^t&\cos t&\sin t\\e^t&-\sin t&\cos t\\e^t&-\cos t&-\sin t\end{array}\right]$.

Answer: $\left[\begin{array}{ccc}e^t&\cos t&\sin t\\e^t&-\sin t&\cos t\\e^t&-\cos t&-\sin t\end{array}\right]^{-1} =\left[\begin{array}{ccc}\frac{1}{2}e^{-t}&0&\frac{1}{2}e^{-t}\\\frac{1}{2}\cos t+\frac{1}{2}\sin t&-\sin t&\frac{1}{2}\sin t-\frac{1}{2}\cos t\\\frac{1}{2}\sin t-\frac{1}{2}\cos t&\cos t&-\frac{1}{2}\cos t-\frac{1}{2}\sin t\end{array}\right]$.

习题 40 Exercise 40

Suppose $A$ is an upper triangular matrix. Show that $A^{-1}$ exists if and only if all elements of the main diagonal are non zero. Is it true that $A^{-1}$ will also be upper triangular? Explain. Could the same be concluded for lower triangular matrices?

Answer: The given condition is what it takes for the determinant to be non zero. Recall that the determinant of an upper triangular matrix is just the product of the entries on the main diagonal.

习题 41 Exercise 41

If $A$, $B$, and $C$ are each $n\times n$ matrices and $ABC$ is invertible, show why each of $A$, $B$, and $C$ are invertible.

Answer: This follows because $\det(ABC) = \det(A)\det(B)\det(C)$ and if this product is nonzero, then each determinant in the product is nonzero and so each of these matrices is invertible.

习题 42 Exercise 42

Decide if this statement is true or false: Cramer's rule is useful for finding solutions to systems of linear equations in which there is an infinite set of solutions.

Answer: False.

习题 43 Exercise 43

Use Cramer's rule to find the solution to $\begin{aligned}x+2y&=1\\ 2x-y&=2\end{aligned}$.

Answer: Solution is: $x = 1, y = 0$.

习题 44 Exercise 44

Use Cramer's rule to find the solution to $\begin{array}{c}x+2y+z=1\\ 2x-y-z=2\\ x+z=1\end{array}$.

Answer: Solution is: $x = 1, y = 0, z = 0$. For example, $y=\frac{\left\|\begin{array}{ccc}1&1&1\\2&2&-1\\1&1&1\end{array}\right\|}{\left\|\begin{array}{ccc}1&2&1\\2&-1&-1\\1&0&1\end{array}\right\|}=0$.