3_01_3A_Basic_Techniques

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Cofactors and 2 x 2 Determinants

##### Outcomes

1. Evaluate the determinant of a square matrix using either Laplace Expansion or row operations.

2. Demonstrate the effects that row operations have on determinants.

3. Verify the following:

1. The determinant of a product of matrices is the product of the determinants.

2. The determinant of a matrix is equal to the determinant of its transpose.

Let \\A\\ be an \\n\times n\\ matrix. That is, let \\A\\ be a square matrix. The determinant of \\A\\, denoted by \\\det \left( A\right)\\ is a very important number which we will explore throughout this section.

If \\A\\ is a 2\\\times 2\\ matrix, the determinant is given by the following formula.

##### Definition \\\PageIndex{1}\\: Determinant of a Two By Two Matrix

Let \\A=\left$$ \begin{array}{rr} a & b \\ c & d \end{array} \right$$ .\\ Then \\\det \left( A\right) = ad-cb\nonumber \\

The determinant is also often denoted by enclosing the matrix with two vertical lines. Thus \\\det \left$$ \begin{array}{rr} a & b \\ c & d \end{array} \right$$ =\left\| \begin{array}{rr} a & b \\ c & d \end{array} \right\| =ad - bc\nonumber \\

The following is an example of finding the determinant of a \\2 \times 2\\ matrix.

##### Example \\\PageIndex{1}\\: A Two by Two Determinant

Find \\\det\left(A\right)\\ for the matrix \\A = \left$$ \begin{array}{rr} 2 & 4 \\ -1 & 6 \end{array} \right$$ .\\

###### Solution

From Definition $\PageIndex{1}$, \\\det \left( A\right) = \left( 2\right) \left( 6\right) -\left( -1\right) \left( 4\right) = 12 + 4 = 16\nonumber \\

The \\2 \times 2\\ determinant can be used to find the determinant of larger matrices. We will now explore how to find the determinant of a \\3 \times 3\\ matrix, using several tools including the \\2 \times 2\\ determinant.

We begin with the following definition.

##### Definition \\\PageIndex{2}\\: The \\ij^{th}\\ Minor of a Matrix

Let \\A\\ be a \\3\times 3\\ matrix. The \\ij^{th}\\ minor of \\A\\, denoted as \\minor\left( A\right) \_{ij},\\ is the determinant of the \\2\times 2\\ matrix which results from deleting the \\i^{th}\\ row and the \\j^{th}\\ column of \\A\\.

In general, if \\A\\ is an \\n\times n\\ matrix, then the \\ij^{th}\\ minor of \\A\\ is the determinant of the \\n-1 \times n-1\\ matrix which results from deleting the \\i^{th}\\ row and the \\j^{th}\\ column of \\A\\.

Hence, there is a minor associated with each entry of \\A\\. Consider the following example which demonstrates this definition.

##### Example \\\PageIndex{2}\\: Finding Minors of a Matrix

Let \\A = \left$$ \begin{array}{rrr} 1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 2 & 1 \end{array} \right$$\nonumber \\ Find \\minor\left( A\right) \_{12}\\ and \\minor\left( A\right) \_{23}\\.

###### Solution

First we will find \\minor\left( A\right) \_{12}\\. By Definition $\PageIndex{2}$, this is the determinant of the \\2\times 2\\ matrix which results when you delete the first row and the second column. This minor is given by \\minor \left(A\right)\_{12} = \det \left$$ \begin{array}{rr} 4 & 2 \\ 3 & 1 \end{array} \right$$\nonumber \\ Using Definition $\PageIndex{1}$, we see that \\\det \left$$ \begin{array}{rr} 4 & 2 \\ 3 & 1 \end{array} \right$$ = \left(4\right)\left(1\right) - \left(3\right)\left(2\right) = 4 - 6 = -2\nonumber\\

Therefore \\minor \left(A\right)\_{12} = -2\\.

Similarly, \\minor\left(A\right)\_{23}\\ is the determinant of the \\2\times 2\\ matrix which results when you delete the second row and the third column. This minor is therefore \\minor \left(A\right)\_{23} = \det \left$$ \begin{array}{rr} 1 & 2 \\ 3 & 2 \end{array} \right$$ = -4\nonumber \\ Finding the other minors of \\A\\ is left as an exercise.

The \\ij^{th}\\ minor of a matrix \\A\\ is used in another important definition, given next.

##### Definition \\\PageIndex{3}\\: The \\ij^{th}\\ Cofactor of a Matrix

Suppose \\A\\ is an \\n\times n\\ matrix. The \\ij^{th}\\ cofactor, denoted by \\\mathrm{cof}\left( A\right) \_{ij}\\ is defined to be \\\mathrm{cof}\left( A\right) \_{ij} = \left( -1\right) ^{i+j} minor\left(A\right)\_{ij}\nonumber \\

It is also convenient to refer to the cofactor of an entry of a matrix as follows. If \\a\_{ij}\\ is the \\ij^{th}\\ entry of the matrix, then its cofactor is just \\\mathrm{cof}\left( A\right) \_{ij}.\\

##### Example \\\PageIndex{3}\\: Finding Cofactors of a Matrix

Consider the matrix \\A=\left$$ \begin{array}{rrr} 1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 2 & 1 \end{array} \right$$\nonumber \\ Find \\\mathrm{cof}\left( A\right) \_{12}\\ and \\\mathrm{cof}\left( A\right) \_{23}\\.

###### Solution

We will use Definition $\PageIndex{3}$ to compute these cofactors.

First, we will compute \\\mathrm{cof}\left( A\right) \_{12}\\. Therefore, we need to find \\minor\left(A\right)\_{12}\\. This is the determinant of the \\2\times 2\\ matrix which results when you delete the first row and the second column. Thus \\minor\left(A\right)\_{12}\\ is given by \\\det \left$$ \begin{array}{rr} 4 & 2 \\ 3 & 1 \end{array} \right$$ = -2\nonumber \\ Then, \\\mathrm{cof}\left( A\right) \_{12}=\left( -1\right) ^{1+2} minor\left(A\right)\_{12} =\left( -1\right) ^{1+2}\left( -2\right) =2\nonumber \\ Hence, \\\mathrm{cof}\left( A\right) \_{12}=2\\.

Similarly, we can find \\\mathrm{cof}\left( A\right) \_{23}\\. First, find \\minor\left(A\right)\_{23}\\, which is the determinant of the \\2\times 2\\ matrix which results when you delete the second row and the third column. This minor is therefore \\\det \left$$ \begin{array}{rr} 1 & 2 \\ 3 & 2 \end{array} \right$$ = -4\nonumber \\ Hence, \\\mathrm{cof}\left( A\right) \_{23}=\left( -1\right) ^{2+3} minor\left(A\right)\_{23} =\left( -1\right) ^{2+3}\left( -4\right) =4\nonumber \\

You may wish to find the remaining cofactors for the above matrix. Remember that there is a cofactor for every entry in the matrix.

We have now established the tools we need to find the determinant of a \\3 \times3\\ matrix.

##### Definition \\\PageIndex{4}\\: The Determinant of a Three By Three Matrix

Let \\A\\ be a \\3\times 3\\ matrix. Then, \\\det \left(A\right)\\ is calculated by picking a row (or column) and taking the product of each entry in that row (column) with its cofactor and adding these products together.

This process when applied to the \\i^{th}\\ row (column) is known as expanding along the \\i^{th}\\ row (column) as is given by \\\det \left(A\right) = a\_{i1}\mathrm{cof}(A)\_{i1} + a\_{i2}\mathrm{cof}(A)\_{i2} + a\_{i3}\mathrm{cof}(A)\_{i3}\nonumber \\

When calculating the determinant, you can choose to expand any row or any column. Regardless of your choice, you will always get the same number which is the determinant of the matrix \\A.\\ This method of evaluating a determinant by expanding along a row or a column is called Laplace Expansion or Cofactor Expansion.

Consider the following example.

##### Example \\\PageIndex{4}\\: Finding the Determinant of a Three by Three Matrix

Let \\A=\left$$ \begin{array}{rrr} 1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 2 & 1 \end{array} \right$$\nonumber \\ Find \\\det\left(A\right)\\ using the method of Laplace Expansion.

###### Solution

First, we will calculate \\\det \left(A\right)\\ by expanding along the first column. Using Definition $\PageIndex{4}$, we take the \\1\\ in the first column and multiply it by its cofactor, \\1 \left( -1\right) ^{1+1}\left\| \begin{array}{rr} 3 & 2 \\ 2 & 1 \end{array} \right\| = (1)(1)(-1) = -1\nonumber \\ Similarly, we take the \\4\\ in the first column and multiply it by its cofactor, as well as with the \\3\\ in the first column. Finally, we add these numbers together, as given in the following equation. \\\det \left(A\right) = 1 \overset{ \mathrm{cof}\left( A\right) \_{11}}{\overbrace{\left( -1\right) ^{1+1}\left\| \begin{array}{rr} 3 & 2 \\ 2 & 1 \end{array} \right\| }}+4 \overset{\mathrm{cof}\left( A\right) \_{21}}{\overbrace{\left( -1\right) ^{2+1}\left\| \begin{array}{rr} 2 & 3 \\ 2 & 1 \end{array} \right\| }}+3 \overset{\mathrm{cof}\left( A\right) \_{31}}{\overbrace{\left( -1\right) ^{3+1}\left\| \begin{array}{rr} 2 & 3 \\ 3 & 2 \end{array} \right\| }}\nonumber \\ Calculating each of these, we obtain \\\det \left(A\right) = 1 \left(1\right)\left(-1\right) + 4 \left(-1\right)\left(-4\right) + 3 \left(1\right)\left(-5\right) = -1 + 16 + -15 = 0\nonumber \\ Hence, \\\det\left(A\right) = 0\\.

As mentioned in Definition $\PageIndex{4}$, we can choose to expand along any row or column. Let’s try now by expanding along the second row. Here, we take the \\4\\ in the second row and multiply it to its cofactor, then add this to the \\3\\ in the second row multiplied by its cofactor, and the \\2\\ in the second row multiplied by its cofactor. The calculation is as follows. \\\det \left(A\right) = 4 \overset{\mathrm{cof}\left( A\right) \_{21}}{\overbrace{\left( -1\right) ^{2+1}\left\| \begin{array}{rr} 2 & 3 \\ 2 & 1 \end{array} \right\| }}+3 \overset{\mathrm{cof}\left( A\right) \_{22}}{\overbrace{\left( -1\right) ^{2+2}\left\| \begin{array}{rr} 1 & 3 \\ 3 & 1 \end{array} \right\| }}+2 \overset{\mathrm{cof}\left( A\right) \_{23}}{\overbrace{\left( -1\right) ^{2+3}\left\| \begin{array}{rr} 1 & 2 \\ 3 & 2 \end{array} \right\| }}\nonumber \\

Calculating each of these products, we obtain \\\det \left(A\right) = 4\left(-1\right)\left(-2\right) + 3\left(1\right)\left(-8\right) + 2 \left(-1\right)\left(-4\right) = 0\nonumber \\

You can see that for both methods, we obtained \\\det \left(A\right) = 0\\.

As mentioned above, we will always come up with the same value for \\\det \left(A\right)\\ regardless of the row or column we choose to expand along. You should try to compute the above determinant by expanding along other rows and columns. This is a good way to check your work, because you should come up with the same number each time!

We present this idea formally in the following theorem.

##### Theorem \\\PageIndex{1}\\: The Determinant is Well Defined

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

We have now looked at the determinant of \\2 \times 2\\ and \\3 \times 3\\ matrices. It turns out that the method used to calculate the determinant of a \\3 \times 3\\ matrix can be used to calculate the determinant of any sized matrix. Notice that Definition $\PageIndex{2}$, Definition $\PageIndex{3}$ and Definition $\PageIndex{4}$ can all be applied to a matrix of any size.

For example, the \\ij^{th}\\ minor of a \\4 \times 4\\ matrix is the determinant of the \\3 \times 3\\ matrix you obtain when you delete the \\i^{th}\\ row and the \\j^{th}\\ column. Just as with the \\3 \times 3\\ determinant, we can compute the determinant of a \\4 \times 4\\ matrix by Laplace Expansion, along any row or column

Consider the following example.

##### Example \\\PageIndex{5}\\: Determinant of a Four by Four Matrix

Find \\\det \left( A\right)\\ where \\A=\left$$ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 5 & 4 & 2 & 3 \\ 1 & 3 & 4 & 5 \\ 3 & 4 & 3 & 2 \end{array} \right$$\nonumber \\

###### Solution

As in the case of a \\3\times 3\\ matrix, you can expand this along any row or column. Lets pick the third column. Then, using Laplace Expansion, \\\det \left( A\right) = 3\left( -1\right) ^{1+3}\left\vert \begin{array}{rrr} 5 & 4 & 3 \\ 1 & 3 & 5 \\ 3 & 4 & 2 \end{array} \right\vert +2\left( -1\right) ^{2+3}\left\vert \begin{array}{rrr} 1 & 2 & 4 \\ 1 & 3 & 5 \\ 3 & 4 & 2 \end{array} \right\vert +\nonumber \\ \\4\left( -1\right) ^{3+3}\left\vert \begin{array}{rrr} 1 & 2 & 4 \\ 5 & 4 & 3 \\ 3 & 4 & 2 \end{array} \right\vert +3\left( -1\right) ^{4+3}\left\vert \begin{array}{rrr} 1 & 2 & 4 \\ 5 & 4 & 3 \\ 1 & 3 & 5 \end{array} \right\vert\nonumber \\

Now, you can calculate each \\3 \times 3\\ determinant using Laplace Expansion, as we did above. You should complete these as an exercise and verify that \\\det \left( A \right)= -12\\.

The following provides a formal definition for the determinant of an \\n \times n\\ matrix. You may wish to take a moment and consider the above definitions for \\2 \times 2\\ and \\3 \times 3\\ determinants in context of this definition.

##### Definition \\\PageIndex{5}\\: The Determinant of an \\n\times n\\ Matrix

Let \\A\\ be an \\n\times n\\ matrix where \\n\geq 2\\ and suppose the determinant of an \\\left( n-1\right) \times \left( n-1\right)\\ has been defined. Then \\\det \left( A\right) =\sum\_{j=1}^{n}a\_{ij}\mathrm{cof}\left( A\right) \_{ij}=\sum\_{i=1}^{n}a\_{ij}\mathrm{cof}\left( A\right) \_{ij}\nonumber \\ The first formula consists of expanding the determinant along the \\i^{th}\\ row and the second expands the determinant along the \\j^{th}\\ column.

In the following sections, we will explore some important properties and characteristics of the determinant.

The Determinant of a Triangular Matrix

There is a certain type of matrix for which finding the determinant is a very simple procedure. Consider the following definition.

##### Definition \\\PageIndex{6}\\: Triangular Matrices

A matrix \\A\\ is upper triangular if \\a\_{ij}=0\\ whenever \\i\>j\\. Thus the entries of such a matrix below the main diagonal equal \\0\\, as shown. Here, \\\ast\\ refers to any nonzero number. \\ \left$$ \begin{array}{cccc} \ast & \ast & \cdots & \ast \\ 0 & \ast & \cdots & \vdots \\ \vdots & \vdots & \ddots & \ast \\ 0 & \cdots & 0 & \ast \end{array} \right$$\nonumber \\ A lower triangular matrix is defined similarly as a matrix for which all entries above the main diagonal are equal to zero.

The following theorem provides a useful way to calculate the determinant of a triangular matrix.

##### Theorem \\\PageIndex{2}\\: Determinant of a Triangular Matrix

Let \\A\\ be an upper or lower triangular matrix. Then \\\det \left( A\right)\\ is obtained by taking the product of the entries on the main diagonal.

The verification of this Theorem can be done by computing the determinant using Laplace Expansion along the first row or column.

Consider the following example.

##### Example \\\PageIndex{6}\\: Determinant of a Triangular Matrix

Let \\A=\left$$ \begin{array}{rrrr} 1 & 2 & 3 & 77 \\ 0 & 2 & 6 & 7 \\ 0 & 0 & 3 & 33.7 \\ 0 & 0 & 0 & -1 \end{array} \right$$\nonumber \\ Find \\\det \left( A\right) .\\

###### Solution

From Theorem $\PageIndex{2}$, it suffices to take the product of the elements on the main diagonal. Thus \\\det \left( A\right) =1\times 2\times 3\times \left( -1\right) =-6.\\

Without using Theorem $\PageIndex{2}$, you could use Laplace Expansion. We will expand along the first column. This gives \\\begin{aligned} \det \left(A\right) = &1\left\| \begin{array}{rrr} 2 & 6 & 7 \\ 0 & 3 & 33.7 \\ 0 & 0 & -1 \end{array} \right\| +0\left( -1\right) ^{2+1}\left\| \begin{array}{rrr} 2 & 3 & 77 \\ 0 & 3 & 33.7 \\ 0 & 0 & -1 \end{array} \right\| + \\ &0\left( -1\right) ^{3+1}\left\| \begin{array}{rrr} 2 & 3 & 77 \\ 2 & 6 & 7 \\ 0 & 0 & -1 \end{array} \right\| +0\left( -1\right) ^{4+1}\left\| \begin{array}{rrr} 2 & 3 & 77 \\ 2 & 6 & 7 \\ 0 & 3 & 33.7 \end{array} \right\|\end{aligned}\\ and the only nonzero term in the expansion is \\1\left\| \begin{array}{rrr} 2 & 6 & 7 \\ 0 & 3 & 33.7 \\ 0 & 0 & -1 \end{array} \right\|\nonumber \\ Now find the determinant of this \\3 \times 3\\ matrix, by expanding along the first column to obtain \\\det \left(A\right) = 1\times \left( 2\times \left\| \begin{array}{rr} 3 & 33.7 \\ 0 & -1 \end{array} \right\| +0\left( -1\right) ^{2+1}\left\| \begin{array}{rr} 6 & 7 \\ 0 & -1 \end{array} \right\| +0\left( -1\right) ^{3+1}\left\| \begin{array}{rr} 6 & 7 \\ 3 & 33.7 \end{array} \right\| \right)\nonumber \\ \\=1\times 2\times \left\| \begin{array}{rr} 3 & 33.7 \\ 0 & -1 \end{array} \right\|\nonumber \\ Next use Definition $\PageIndex{1}$ to find the determinant of this \\2 \times 2\\ matrix, which is just \\3 \times -1 - 0 \times 33.7 = -3\\. Putting all these steps together, we have \\\det \left(A\right) = 1\times 2\times 3\times \left( -1\right) =-6\nonumber \\ which is just the product of the entries down the main diagonal of the original matrix!

You can see that while both methods result in the same answer, Theorem $\PageIndex{2}$ provides a much quicker method.

In the next section, we explore some important properties of determinants.

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3_02_3A_Properties_of_Determinants

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Properties of Determinants I: Examples

There are many important properties of determinants. Since many of these properties involve the row operations discussed in Chapter 1, we recall that definition now.

##### Definition \\\PageIndex{1}\\: Row Operations

The row operations consist of the following

1. Switch two rows.

2. Multiply a row by a nonzero number.

3. Replace a row by a multiple of another row added to itself.

We will now consider the effect of row operations on the determinant of a matrix. In future sections, we will see that using the following properties can greatly assist in finding determinants. This section will use the theorems as motivation to provide various examples of the usefulness of the properties.

The first theorem explains the affect on the determinant of a matrix when two rows are switched.

##### Theorem \\\PageIndex{1}\\: Switching Rows

Let \\A\\ be an \\n\times n\\ matrix and let \\B\\ be a matrix which results from switching two rows of \\A.\\ Then \\\det \left( B\right) = - \det \left( A\right) .\\

When we switch two rows of a matrix, the determinant is multiplied by \\-1\\. Consider the following example.

##### Example \\\PageIndex{1}\\: Switching Two Rows

Let \\A=\left$$ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right$$\\ and let \\B=\left$$ \begin{array}{rr} 3 & 4 \\ 1 & 2 \end{array} \right$$\\. Knowing that \\\det \left( A \right) =-2\\, find \\\det \left( B \right)\\.

###### Solution

By Definition 3.1.1, \\\det \left(A\right) = 1 \times 4 - 3 \times 2 = -2\\. Notice that the rows of \\B\\ are the rows of \\A\\ but switched. By Theorem $\PageIndex{1}$ since two rows of \\A\\ have been switched, \\\det \left(B\right) = - \det \left(A\right) = - \left(-2\right) = 2\\. You can verify this using Definition 3.1.1.

The next theorem demonstrates the effect on the determinant of a matrix when we multiply a row by a scalar.

##### Theorem \\\PageIndex{2}\\: Multiplying a Row by a Scalar

Let \\A\\ be an \\n\times n\\ matrix and let \\B\\ be a matrix which results from multiplying some row of \\A\\ by a scalar \\k\\. Then \\\det \left( B\right) = k \det \left( A\right)\\.

Notice that this theorem is true when we multiply *one* row of the matrix by \\k\\. If we were to multiply *two* rows of \\A\\ by \\k\\ to obtain \\B\\, we would have \\\det \left(B\right) = k^2 \det \left(A\right)\\. Suppose we were to multiply all \\n\\ rows of \\A\\ by \\k\\ to obtain the matrix \\B\\, so that \\B = kA\\. Then, \\\det \left(B\right) = k^n \det \left(A\right)\\. This gives the next theorem.

##### Theorem \\\PageIndex{3}\\: Scalar Multiplication

Let \\A\\ and \\B\\ be \\n \times n\\ matrices and \\k\\ a scalar, such that \\B = kA\\. Then \\\det(B) = k^n \det(A)\\.

Consider the following example.

##### Example \\\PageIndex{2}\\: Multiplying a Row by 5

Let \\A=\left$$ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right$$ ,\\ B=\left$$ \begin{array}{rr} 5 & 10 \\ 3 & 4 \end{array} \right$$ .\\ Knowing that \\\det \left( A \right) =-2\\, find \\\det \left( B \right)\\.

###### Solution

By Definition 3.1.1, \\\det \left( A\right) =-2.\\ We can also compute \\\det \left(B\right)\\ using Definition 3.1.1, and we see that \\\det \left(B\right) = -10\\.

Now, let’s compute \\\det \left(B\right)\\ using Theorem $\PageIndex{2}$ and see if we obtain the same answer. Notice that the first row of \\B\\ is \\5\\ times the first row of \\A\\, while the second row of \\B\\ is equal to the second row of \\A\\. By Theorem $\PageIndex{2}$, \\\det \left( B \right) = 5 \times \det \left( A \right) = 5 \times -2 = -10.\\

You can see that this matches our answer above.

Finally, consider the next theorem for the last row operation, that of adding a multiple of a row to another row.

##### Theorem \\\PageIndex{4}\\: Adding a Multiple of a Row to Another Row

Let \\A\\ be an \\n\times n\\ matrix and let \\B\\ be a matrix which results from adding a multiple of a row to another row. Then \\\det \left( A\right) =\det \left( B \right)\\.

Therefore, when we add a multiple of a row to another row, the determinant of the matrix is unchanged. Note that if a matrix \\A\\ contains a row which is a multiple of another row, \\\det \left(A\right)\\ will equal \\0\\. To see this, suppose the first row of \\A\\ is equal to \\-1\\ times the second row. By Theorem $\PageIndex{4}$, we can add the first row to the second row, and the determinant will be unchanged. However, this row operation will result in a row of zeros. Using Laplace Expansion along the row of zeros, we find that the determinant is \\0\\.

Consider the following example.

##### Example \\\PageIndex{3}\\: Adding a Row to Another Row

Let \\A=\left$$ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right$$\\ and let \\B=\left$$ \begin{array}{rr} 1 & 2 \\ 5 & 8 \end{array} \right$$ .\\ Find \\\det \left(B\right)\\.

###### Solution

By Definition 3.1.1, \\\det \left(A\right) = -2\\. Notice that the second row of \\B\\ is two times the first row of \\A\\ added to the second row. By Theorem $\PageIndex{1}$, \\\det \left( B\right) = \det \left( A \right) =-2\\. As usual, you can verify this answer using Definition 3.1.1.

##### Example \\\PageIndex{4}\\: Multiple of a Row

Let \\A = \left$$ \begin{array}{rr} 1 & 2 \\ 2 & 4 \end{array} \right$$\\. Show that \\\det \left( A \right) = 0\\.

###### Solution

Using Definition 3.1.1, the determinant is given by \\\det \left( A \right) = 1 \times 4 - 2 \times 2 = 0\nonumber \\

However notice that the second row is equal to \\2\\ times the first row. Then by the discussion above following Theorem $\PageIndex{4}$ the determinant will equal \\0\\.

Until now, our focus has primarily been on row operations. However, we can carry out the same operations with columns, rather than rows. The three operations outlined in Definition $\PageIndex{1}$ can be done with columns instead of rows. In this case, in Theorems $\PageIndex{1}$, $\PageIndex{2}$, and $\PageIndex{4}$ you can replace the word, "row" with the word "column".

There are several other major properties of determinants which do not involve row (or column) operations. The first is the determinant of a product of matrices.

##### Theorem \\\PageIndex{5}\\: Determinant of a Product

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

In order to find the determinant of a product of matrices, we can simply take the product of the determinants.

Consider the following example.

##### Example \\\PageIndex{5}\\: The Determinant of a Product

Compare \\\det \left( AB\right)\\ and \\\det \left( A\right) \det \left( B\right)\\ for \\A=\left$$ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right$$ ,B=\left$$ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right$$\nonumber \\

###### Solution

First compute \\AB\\, which is given by \\AB=\left$$ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right$$ \left$$ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right$$ = \left$$ \begin{array}{rr} 11 & 4 \\ -1 & -4 \end{array} \right$$\nonumber \\ and so by Definition 3.1.1 \\\det \left( AB\right) =\det \left$$ \begin{array}{rr} 11 & 4 \\ -1 & -4 \end{array} \right$$ = -40\nonumber \\

Now \\\det \left( A\right) =\det \left$$ \begin{array}{rr} 1 & 2 \\ -3 & 2 \end{array} \right$$ = 8\nonumber \\ and \\\det \left( B\right) =\det \left$$ \begin{array}{rr} 3 & 2 \\ 4 & 1 \end{array} \right$$ = -5\nonumber \\

Computing \\\det \left(A\right) \times \det \left(B\right)\\ we have \\8 \times -5 = -40\\. This is the same answer as above and you can see that \\\det \left( A\right) \det \left( B\right) =8\times \left( -5\right) =-40 = \det \left(AB\right)\\.

Consider the next important property.

##### Theorem \\\PageIndex{6}\\: Determinant of the Transpose

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

This theorem is illustrated in the following example.

##### Example \\\PageIndex{6}\\: Determinant of the Transpose

Let \\A = \left$$ \begin{array}{rr} 2 & 5 \\ 4 & 3 \end{array} \right$$\nonumber \\ Find \\\det \left(A^T\right)\\.

###### Solution

First, note that \\A^{T} = \left$$ \begin{array}{rr} 2 & 4 \\ 5 & 3 \end{array} \right$$\nonumber \\

Using Definition 3.1.1, we can compute \\\det \left(A\right)\\ and \\\det \left(A^T\right)\\. It follows that \\\det \left(A\right) = 2 \times 3 - 4 \times 5 = -14\\ and \\\det \left(A^T\right) = 2 \times 3 - 5 \times 4 = -14\\. Hence, \\\det \left(A\right) = \det \left(A^T\right)\\.

The following provides an essential property of the determinant, as well as a useful way to determine if a matrix is invertible.

##### Theorem \\\PageIndex{7}\\: Determinant of the Inverse

Let \\A\\ be an \\n \times n\\ matrix. Then \\A\\ is invertible if and only if \\\det(A) \neq 0\\. If this is true, it follows that \\\det(A^{-1}) = \frac{1}{\det(A)}\nonumber \\

Consider the following example.

##### Example \\\PageIndex{7}\\: Determinant of an Invertible Matrix

Let \\A = \left$$ \begin{array}{rr} 3 & 6 \\ 2 & 4 \end{array} \right$$, B = \left$$ \begin{array}{rr} 2 & 3 \\ 5 & 1 \end{array} \right$$\\. For each matrix, determine if it is invertible. If so, find the determinant of the inverse.

###### Solution

Consider the matrix \\A\\ first. Using Definition 3.1.1 we can find the determinant as follows: \\\det \left( A \right) = 3 \times 4 - 2 \times 6 = 12 - 12 = 0\nonumber \\ By Theorem $\PageIndex{7}$ \\A\\ is not invertible.

Now consider the matrix \\B\\. Again by Definition 3.1.1 we have \\\det \left( B \right) = 2 \times 1 - 5 \times 3 = 2 - 15 = -13\nonumber \\ By Theorem $\PageIndex{7}$ \\B\\ is invertible and the determinant of the inverse is given by \\\begin{aligned} \det \left( A^{-1} \right) &= \frac{1}{\det(A)} \\ &= \frac{1}{-13} \\ &= -\frac{1}{13}\end{aligned}\\

Properties of Determinants II: Some Important Proofs

This section includes some important proofs on determinants and cofactors.

First we recall the definition of a determinant. If \\A=\left$$ a\_{ij} \right$$\\ is an \\n\times n\\ matrix, then \\\det A\\ is defined by computing the expansion along the first row: \\\label{E1} \det A=\sum\_{i=1}^n a\_{1,i} \mathrm{cof}(A)\_{1,i}.\\ If \\n=1\\ then \\\det A=a\_{1,1}\\.

The following example is straightforward and strongly recommended as a means for getting used to definitions.

##### Example \\\PageIndex{8}\\:

$1$ Let \\E\_{ij}\\ be the elementary matrix obtained by interchanging \\i\\th and \\j\\th rows of \\I\\. Then \\\det E\_{ij}=-1\\.

$2$ Let \\E\_{ik}\\ be the elementary matrix obtained by multiplying the \\i\\th row of \\I\\ by \\k\\. Then \\\det E\_{ik}=k\\.

$3$ Let \\E\_{ijk}\\ be the elementary matrix obtained by multiplying \\i\\th row of \\I\\ by \\k\\ and adding it to its \\j\\th row. Then \\\det E\_{ijk}=1\\.

$4$ If \\C\\ and \\B\\ are such that \\CB\\ is defined and the \\i\\th row of \\C\\ consists of zeros, then the \\i\\th row of \\CB\\ consists of zeros.

$5$ If \\E\\ is an elementary matrix, then \\\det E=\det E^T\\.

Many of the proofs in section use the Principle of Mathematical Induction. This concept is discussed in Appendix A.2 and is reviewed here for convenience. First we check that the assertion is true for \\n=2\\ (the case \\n=1\\ is either completely trivial or meaningless).

Next, we assume that the assertion is true for \\n-1\\ (where \\n\geq 3\$ and prove it for \\n\\. Once this is accomplished, by the Principle of Mathematical Induction we can conclude that the statement is true for all \\n\times n\\ matrices for every \\n\geq 2\\.

If \\A\\ is an \\n\times n\\ matrix and \\1\leq j \leq n\\, then the matrix obtained by removing \\1\\st column and \\j\\th row from \\A\\ is an \\n-1\times n-1\\ matrix (we shall denote this matrix by \\A(j)\\ below). Since these matrices are used in computation of cofactors \\\mathrm{cof}(A)\_{1,i}\\, for \\1\leq i\neq n\\, the inductive assumption applies to these matrices.

Consider the following lemma.

##### Lemma \\\PageIndex{1}\\:

If \\A\\ is an \\n\times n\\ matrix such that one of its rows consists of zeros, then \\\det A=0\\.

Proof

We will prove this lemma using Mathematical Induction.

If \\n=2\\ this is easy (check!).

Let \\n\geq 3\\ be such that every matrix of size \\n-1\times n-1\\ with a row consisting of zeros has determinant equal to zero. Let \\i\\ be such that the \\i\\th row of \\A\\ consists of zeros. Then we have \\a\_{ij}=0\\ for \\1\leq j\leq n\\.

Fix \\j\in \\1,2, \dots ,n\\\\ such that \\j\neq i\\. Then matrix \\A(j)\\ used in computation of \\\mathrm{cof}(A)\_{1,j}\\ has a row consisting of zeros, and by our inductive assumption \\\mathrm{cof}(A)\_{1,j}=0\\.

On the other hand, if \\j=i\\ then \\a\_{1,j}=0\\. Therefore \\a\_{1,j}\mathrm{cof}(A)\_{1,j}=0\\ for all \\j\\ and by \\\eqref{E1}\\ we have \\\det A=\sum\_{j=1}^n a\_{1,j} \mathrm{cof}(A)\_{1,j}=0\nonumber \\ as each of the summands is equal to 0.

##### Lemma \\\PageIndex{2}\\:

Assume \\A\\, \\B\\ and \\C\\ are \\n\times n\\ matrices that for some \\1\leq i\leq n\\ satisfy the following.

1. \\j\\th rows of all three matrices are identical, for \\j\neq i\\.

2. Each entry in the \\j\\th row of \\A\\ is the sum of the corresponding entries in \\j\\th rows of \\B\\ and \\C\\.

Then \\\det A=\det B+\det C\\.

Proof

This is not difficult to check for \\n=2\\ (do check it!).

Now assume that the statement of Lemma is true for \\n-1\times n-1\\ matrices and fix \\A,B\\ and \\C\\ as in the statement. The assumptions state that we have \\a\_{l,j}=b\_{l,j}=c\_{l,j}\\ for \\j\neq i\\ and for \\1\leq l\leq n\\ and \\a\_{l,i}=b\_{l,i}+c\_{l,i}\\ for all \\1\leq l\leq n\\. Therefore \\A(i)=B(i)=C(i)\\, and \\A(j)\\ has the property that its \\i\\th row is the sum of \\i\\th rows of \\B(j)\\ and \\C(j)\\ for \\j\neq i\\ while the other rows of all three matrices are identical. Therefore by our inductive assumption we have \\\mathrm{cof}(A)\_{1j}=\mathrm{cof}(B)\_{1j}+\mathrm{cof}(C)\_{1j}\\ for \\j\neq i\\.

By \\\eqref{E1}\\ we have (using all equalities established above) \\\begin{aligned} \det A&=\sum\_{l=1}^n a\_{1,l} \mathrm{cof}(A)\_{1,l}\\ &=\sum\_{l\neq i} a\_{1,l}(\mathrm{cof}(B)\_{1,l}+\mathrm{cof}(C)\_{1,l})+ (b\_{1,i}+c\_{1,i})\mathrm{cof}(A)\_{1,i}\\ &= \det B+\det C\end{aligned}\\ This proves that the assertion is true for all \\n\\ and completes the proof.

##### Theorem \\\PageIndex{8}\\:

Let \\A\\ and \\B\\ be \\n\times n\\ matrices.

1. If \\A\\ is obtained by interchanging \\i\\th and \\j\\th rows of \\B\\ (with \\i\neq j\$, then \\\det A=-\det B\\.

2. If \\A\\ is obtained by multiplying \\i\\th row of \\B\\ by \\k\\ then \\\det A=k\det B\\.

3. If two rows of \\A\\ are identical then \\\det A=0\\.

4. If \\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\\.

Proof

We prove all statements by induction. The case \\n=2\\ is easily checked directly (and it is strongly suggested that you do check it).

We assume \\n\geq 3\\ and (1)–(4) are true for all matrices of size \\n-1\times n-1\\.

$1$ We prove the case when \\j=i+1\\, i.e., we are interchanging two consecutive rows.

Let \\l\in \\1, \dots, n\\\setminus \\i,j\\\\. Then \\A(l)\\ is obtained from \\B(l)\\ by interchanging two of its rows (draw a picture) and by our assumption \\\label{E2} \mathrm{cof}(A)\_{1,l}=-\mathrm{cof}(B)\_{1,l}.\\

Now consider \\a\_{1,i} \mathrm{cof}(A)\_{1,l}\\. We have that \\a\_{1,i}=b\_{1,j}\\ and also that \\A(i)=B(j)\\. Since \\j=i+1\\, we have \$-1)^{1+j}=(-1)^{1+i+1}=-(-1)^{1+i}\nonumber \\ and therefore \\a\_{1i}\mathrm{cof}(A)\_{1i}=-b\_{1j} \mathrm{cof}(B)\_{1j}\\ and \\a\_{1j}\mathrm{cof}(A)\_{1j}=-b\_{1i} \mathrm{cof}(B)\_{1i}\\. Putting this together with \\\eqref{E2}\\ into \\\eqref{E1}\\ we see that if in the formula for \\\det A\\ we change the sign of each of the summands we obtain the formula for \\\det B\\. \\\det A=\sum\_{l=1}^n a\_{1l}\mathrm{cof}(A)\_{1l} =-\sum\_{l=1}^n b\_{1l} B\_{1l} =\det B.\nonumber \\

We have therefore proved the case of (1) when \\j=i+1\\. In order to prove the general case, one needs the following fact. If \\i\

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 \\\eqref{E1}\\, 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 $\PageIndex{2}$, we have that \\\det A=\det B+\det C\nonumber \\ 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\\.

##### Theorem \\\PageIndex{9}\\:

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

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 $\PageIndex{8}$ and Theorem $\PageIndex{8}$.

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 $\PageIndex{8}$ the last row of \\CB\\ consists of zeros. By Lemma $\PageIndex{1}$ 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\nonumber \\ 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) 0 =0\nonumber \\ hence \\\det AB=0=\det A \det B\\.

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

##### Theorem \\\PageIndex{10}\\:

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

Proof

Note first that the conclusion is true if \\A\\ is elementary by (5) of Example $\PageIndex{8}$.

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 $\PageIndex{9}$ we have \\\det (A^T)=\det (C^T)\cdot \det (E_m^T)\cdot \dots \cdot \det (E_2^T)\cdot \det(E_1).\nonumber \\ By (5) of Example $\PageIndex{8}$ 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.

##### Theorem \\\PageIndex{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}.\nonumber \\ 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 $\PageIndex{8}$ we have \\\det A=-\det B.\nonumber\\ Now we have \\\det B=\sum\_{i=1}^n b\_{1,i} \mathrm{cof}(B)\_{1,i}.\nonumber \\ 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}\nonumber \\ 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 cofactor along any column yields the same result.

Finally, since \\\det A=\det A^T\\ by Theorem $\PageIndex{10}$, 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.

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3_03_3A_Finding_Determinants_using_Row_Operations

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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.

Example \\\PageIndex{1}\\: Finding a Determinant

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

###### 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$$ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -9 & -13 & -17 \\ 0 & -3 & -8 & -13 \\ 0 & -2 & -10 & -3 \end{array} \right$$\nonumber \\ 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$$ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & 0 & 11 & 22 \\ 0 & -3 & -8 & -13 \\ 0 & 6 & 30 & 9 \end{array} \right$$\nonumber \\ 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$$ \begin{array}{rrrr} 1 & 2 & 3 & 4 \\ 0 & -3 & -8 & -13 \\ 0 & 0 & 11 & 22 \\ 0 & 0 & 14 & -17 \end{array} \right$$\nonumber \\

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\nonumber \\ 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 $\PageIndex{1}$, 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.

Example \\\PageIndex{2}\\: Find the Determinant

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

###### 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$$ \begin{array}{rrrr} 1 & 2 & 3 & 2 \\ 0 & -5 & -1 & -1 \\ 0 & -3 & -4 & 1 \\ 0 & -10 & -8 & -4 \end{array} \right$$\nonumber \\ 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$$ \begin{array}{rrrr} 1 & -8 & 3 & 2 \\ 0 & 0 & -1 & -1 \\ 0 & -8 & -4 & 1 \\ 0 & 10 & -8 & -4 \end{array} \right$$\nonumber \\ 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$$ \begin{array}{rrrr} 1 & 0 & 7 & 1 \\ 0 & 0 & -1 & -1 \\ 0 & -8 & -4 & 1 \\ 0 & 10 & -8 & -4 \end{array} \right$$\nonumber \\ 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\nonumber \\ 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\nonumber \\

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.

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3_04_3A_Applications_of_the_Determinant

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##### 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.

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^{th}\\ entry is the \\ij^{th}\\ cofactor of \\A\\. The formal definition is as follows.

##### Definition \\\PageIndex{1}\\: 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^{th}\\ cofactor of \\A\\.

Note that \\\mathrm{cof}\left(A\right)\_{ij}\\ denotes the \\ij^{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$$ \begin{array}{rr} a & b \\ c & d \end{array} \right$$\nonumber \\ then \\{adj}\left(A\right)\\ is given by \\{adj}\left(A\right) = \left$$ \begin{array}{rr} d & -b \\ -c & a \end{array} \right$$\nonumber \\

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\\.

##### Theorem \\\PageIndex{1}\\: The Inverse and the Determinant

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

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)\nonumber \\

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.

##### Example \\\PageIndex{1}\\: Find Inverse Using the Determinant

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

###### Solution

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

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$$ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -6 & -8 \\ 0 & 0 & -2 \end{array} \right$$\nonumber \\ 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$$ \begin{array}{rrr} -2 & -2 & 6 \\ 4 & -2 & 0 \\ 2 & 8 & -6 \end{array} \right$$\nonumber \\ Here, the \\ij^{th}\\ entry is the \\ij^{th}\\ cofactor of the original matrix \\A\\ which you can verify. Therefore, from Theorem $\PageIndex{1}$, 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$$ \begin{array}{rrr} -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{2}{3} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{array} \right$$\nonumber \\

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\nonumber \\ 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}\\.

##### Example \\\PageIndex{2}\\: Find the Inverse From a Formula

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

###### 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)\nonumber \\

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}\nonumber \\

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$$ \begin{array}{rrr} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{array} \right$$\nonumber \\

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$$ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right$$\nonumber \\ 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 $\PageIndex{1}$.

##### Theorem \\\PageIndex{1}\\: The Inverse and the Determinant

Proof

(of Theorem $\PageIndex{1}$) 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}\nonumber \\ 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 his case. Thus we obtain: \\A \\ {adj}\left(A\right) = {\det \left(A\right)} I\nonumber \\ Similarly we can verify that: \\{adj}\left(A\right)A = {\det \left(A\right)} I\nonumber \\ 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)\nonumber \\ 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)\nonumber \\

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.

##### Example \\\PageIndex{3}\\: Inverse for Non-Constant Matrix

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

###### 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$$ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right$$\nonumber \\ 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$$ \begin{array}{ccc} e^{-t} & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array} \right$$\nonumber \\

Cramer’s Rule

Another context in which the formula given in Theorem $\PageIndex{1}$ 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\nonumber \\ Let \\x_i\\ be the \\i^{th}\\ entry of \\X\\ and \\b_j\\ be the \\j^{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}\nonumber \\ where \\{adj}\left(A\right)\_{ij}\\ is the \\ij^{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$$ \begin{array}{ccccc} \ast & \cdots & b\_{1} & \cdots & \ast \\ \vdots & & \vdots & & \vdots \\ \ast & \cdots & b\_{n} & \cdots & \ast \end{array} \right$$\nonumber \\ where here the \\i^{th}\\ column of \\A\\ is replaced with the column vector \\\left$$ b\_{1}\cdots \cdot ,b\_{n}\right$$ ^{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.

##### Procedure \\\PageIndex{1}\\: 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)}\nonumber \\ where \\A\_{i}\\ is the matrix obtained by replacing the \\i^{th}\\ column of \\A\\ with the column matrix \\B = \left$$ \begin{array}{c} b_1 \\ \vdots \\ b_n \end{array} \right$$\nonumber \\

We illustrate this procedure in the following example.

##### Example \\\PageIndex{4}\\: 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$$ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right$$\nonumber \\

###### Solution

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

In order to find \\x\\, we calculate \\x = \frac{\det \left(A\_{1}\right)}{\det \left(A\right)}\nonumber \\ 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$$ \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 3 & -3 & 2 \end{array} \right$$\nonumber \\

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}\nonumber \\

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$$ \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 3 & 2 \end{array} \right$$\nonumber \\

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}\nonumber \\

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

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}\nonumber \\

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.

Using Cramer’s Rule

##### Example \\\PageIndex{5}\\: 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$$ \begin{array}{c} 1 \\ t \\ {0.05in}t^{2} \end{array} \right$$\nonumber \\

###### Solution

We are asked to find the value of \\z\\ in the solution. We will solve using Cramer’s rule. Thus \\z={.05in} \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}\nonumber \\

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.

##### Example \\\PageIndex{6}\\: 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}\\.

###### Solution

We want to find a polynomial given by \\p(x) = r_0 + r_1x_1 + r_2x_2^2\nonumber \\ 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$$ \begin{array}{rrr\|r} 1 & 1 & 1 & 4 \\ 1 & 2 & 4 & 9 \\ 1 & 3 & 9 & 12 \end{array} \right$$\nonumber\\

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

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\nonumber \\

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.

##### Procedure \\\PageIndex{2}\\: 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}\nonumber \\

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}\nonumber \\

3. Set up the augmented matrix of this system of equations \\\left$$ \begin{array}{rrrrr\|r} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & y_1 \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & y_2 \\ \vdots & \vdots & \vdots & &\vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & y_n \\ \end{array} \right$$\nonumber \\

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\\.

This procedure motivates the following theorem.

##### Theorem \\\PageIndex{2}\\: 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.

##### Example \\\PageIndex{7}\\: 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)\\.

###### Solution

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

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$$ \begin{array}{rrrr\|r} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 2 \\ 1 & 3 & 9 & 27 & 22 \\ 1 & 5 & 25 & 125 & 66 \end{array} \right$$\nonumber\\

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

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\\.

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3_E_3A_nExercises

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> 原页: https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/03%3A_Determinants/3.E%3A_nExercises

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Exercise \\\PageIndex{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$$\begin{array}{cc}4&3\\6&2\end{array}\right$$\\

Exercise \\\PageIndex{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}\\

Exercise \\\PageIndex{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$$\begin{array}{cccc}1&2&3&2\\1&3&2&3\\4&1&5&0\\1&2&1&2\end{array}\right$$\\

Answer

1. The answer is \\31\\.

2. The answer is \\375\\.

3. The answer is \\-2\\.

Exercise \\\PageIndex{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\|\nonumber\\

Answer

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

Exercise \\\PageIndex{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\|\nonumber\\

Answer

\\\left\|\begin{array}{ccc}1&2&1\\1&0&1\\2&1&1\end{array}\right\|=2\nonumber\\

Exercise \\\PageIndex{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\|\nonumber\\

Answer

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

Exercise \\\PageIndex{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\|\nonumber\\

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\nonumber\\

Exercise \\\PageIndex{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$$\begin{array}{cccc}2&3&15&0\\0&4&1&7\\0&0&-3&5\\0&0&0&1\end{array}\right$$\\

Exercise \\\PageIndex{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$$\begin{array}{cc}a&c\\b&d\end{array}\right$$\nonumber\\

Answer

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

Exercise \\\PageIndex{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$$\begin{array}{cc}c&d\\a&b\end{array}\right$$\nonumber\\

Answer

In this case two rows were switched and so the resulting determinant is \\−1\\ times the first

Exercise \\\PageIndex{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$$\begin{array}{cc}a&b\\a+c&b+d\end{array}\right$$\nonumber\\

Answer

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

Exercise \\\PageIndex{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$$\begin{array}{cc}a&b\\2c&2d\end{array}\right$$\nonumber\\

Answer

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

Exercise \\\PageIndex{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$$\begin{array}{cc}b&a\\d&c\end{array}\right$$\nonumber\\

Answer

In this case the two columns were switched so the determinant of the second is \\−1\\ times the determinant of the first.

Exercise \\\PageIndex{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.

Exercise \\\PageIndex{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\\.

Exercise \\\PageIndex{16}\\

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

Answer

\\\begin{array}{c}\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 \\ \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\end{array}\nonumber\\

Exercise \\\PageIndex{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$$\begin{array}{cc}-1&0\\0&-1\end{array}\right$$\\.

Exercise \\\PageIndex{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\\.

Exercise \\\PageIndex{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) = 1\\, or \\-1\\.

Exercise \\\PageIndex{20}\\

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

Answer

\\\det(A) = \det(S^{−1}BS) = \det(S^{−1})\det(B)\det(S) = \det(B)\det(S^{−1}S) = \det(B)\\.

Exercise \\\PageIndex{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) ≥ 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\\.

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.

Exercise \\\PageIndex{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\|\nonumber\\

Answer

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

Exercise \\\PageIndex{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\|\nonumber\\

Answer

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

Exercise \\\PageIndex{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\|\nonumber\\

Answer

One can row reduce this using only row operation 3 to \\\left$$\begin{array}{cccc}1&2&1&2\\0&-5&-5&-3 \\ 0&0&2&\frac{9}{5} \\ 0&0&0&-\frac{63}{10}\end{array}\right$$\nonumber\\ 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\nonumber\\

Exercise \\\PageIndex{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\|\nonumber\\

Answer

One can row reduce this using only row operation 3 to \\\left$$\begin{array}{cccc}1&4&1&2\\0&-10&-5&-3 \\ 0&0&2&\frac{19}{5} \\ 0&0&0&-\frac{211}{20}\end{array}\right$$\nonumber\\ 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\nonumber\\

Exercise \\\PageIndex{26}\\

Let \\A=\left$$\begin{array}{ccc}1&2&3\\0&2&1\\3&1&0\end{array}\right$$\nonumber\\ 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 \\\begin{aligned}\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$$\begin{array}{ccc}\frac{1}{13}&-\frac{3}{13}&\frac{4}{13} \\ -\frac{3}{13}&\frac{9}{13}&\frac{1}{13} \\ \frac{6}{13}&-\frac{5}{13}&-\frac{2}{13}\end{array}\right$$\end{aligned}\\

Exercise \\\PageIndex{27}\\

Let \\A=\left$$\begin{array}{ccc}1&2&0\\0&2&1\\3&1&1\end{array}\right$$\nonumber\\ 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$$\begin{array}{ccc}\frac{1}{7}&-\frac{2}{7}&\frac{2}{7} \\ \frac{3}{7}&\frac{1}{7}&-\frac{1}{7} \\ -\frac{6}{7}&\frac{5}{7}&\frac{2}{7}\end{array}\right$$\\

Exercise \\\PageIndex{28}\\

Let \\A=\left$$\begin{array}{ccc}1&3&3\\2&4&1\\0&1&1\end{array}\right$$\nonumber\\ 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\nonumber\\ so it has an inverse which is \\\left$$\begin{array}{ccc}1&0&-3 \\ -\frac{2}{3}&\frac{1}{3}&\frac{5}{3} \\ \frac{2}{3}&-\frac{1}{3}&-\frac{2}{3}\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{29}\\

Let \\A=\left$$\begin{array}{ccc}1&2&3\\0&2&1\\2&6&7\end{array}\right$$\nonumber\\ 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.

Exercise \\\PageIndex{30}\\

Let \\A=\left$$\begin{array}{ccc}1&0&3\\1&0&1\\3&1&0\end{array}\right$$\nonumber\\ 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\nonumber\\ and so it has an inverse. The inverse turns out to equal \\\left$$\begin{array}{ccc}-\frac{1}{2}&\frac{3}{2}&0 \\ \frac{3}{2}&-\frac{9}{2}&1 \\ \frac{1}{2}&-\frac{1}{2}&0\end{array}\right$$\nonumber\\

Exercise \\\PageIndex{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$$\begin{array}{ccc}1&2&1\\2&3&0\\0&1&2\end{array}\right$$\\

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\\

Exercise \\\PageIndex{32}\\

Consider the matrix \\A=\left$$\begin{array}{ccc}1&0&0\\0&\cos t&-\sin t \\ 0&\sin t&\cos t\end{array}\right$$\nonumber\\ 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\\

Exercise \\\PageIndex{33}\\

Consider the matrix \\A=\left$$\begin{array}{ccc}1&t&t^2 \\ 0&1&2t \\ t&0&2\end{array}\right$$\nonumber\\ 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\nonumber\\ and so it has no inverse when \\t=-\sqrt$$3$${2}\\

Exercise \\\PageIndex{34}\\

Consider the matrix \\A=\left$$\begin{array}{ccc}e^t &\cosh t&\sinh t \\ e^t&\sinh t&\cosh t \\ e^t&\cosh t&\sinh t\end{array}\right$$\nonumber\\ 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\nonumber\\ and so this matrix fails to have a nonzero determinant at any value of \\t\\.

Exercise \\\PageIndex{35}\\

Consider the matrix \\A=\left$$\begin{array}{ccc}e^t &e^{-t}\cos t&e^{-t}\sin t \\ e^t&-e^{-t}\cos t-e^{-t}\sin t &-e^{-t}\sin t+e^{-t}\cos t \\ e^t&2e^{-t}\sin t&-2e^{-t}\cos t\end{array}\right$$\nonumber\\ 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\nonumber\\ and so this matrix is always invertible.

Exercise \\\PageIndex{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\\.

Exercise \\\PageIndex{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 $\PageIndex{36}$.

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\nonumber\\ and so it follows from the above problem that \$BA−I)X = 0\\. Since \\X\\ is arbitrary, it follows that \\BA = I\\.

Exercise \\\PageIndex{38}\\

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

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}.\nonumber\\ Hence the inverse is \\\begin{array}{l}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$$\begin{array}{ccc}e^{-t}&0&0 \\ 0&e^{-t}(\cos t+\sin t)&-(\sin t)e^{-t} \\ 0&-e^{-t}(\cos t-\sin t)&(\cos t)e^{-t}\end{array}\right$$ \end{array}\nonumber\\

Exercise \\\PageIndex{39}\\

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

Answer

\\\begin{array}{l} \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$$\begin{array}{ccc}\frac{1}{2}e^{-t}&0&\frac{1}{2}e^{-t} \\ \frac{1}{2}\cos t+\frac{1}{2}\sin t&-\sin t&\frac{1}{2}\sin t-\frac{1}{2}\cos t \\ \frac{1}{2}\sin t-\frac{1}{2}\cos t&\cos t&-\frac{1}{2}\cos t-\frac{1}{2}\sin t\end{array}\right$$\end{array}\nonumber\\

Exercise \\\PageIndex{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

Exercise \\\PageIndex{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.

Exercise \\\PageIndex{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.

Exercise \\\PageIndex{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$$\\

Exercise \\\PageIndex{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}\nonumber\\

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\nonumber\\

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