Chapter 8.4, Problem 100E

a.

To determine

To explain: why each equation is an example of the given property of determinants. Using a graphing utility verify the results.

Explanation of Solution

Given:

(A and B are square matrices).

If B is obtained from A by adding a multiple of a row of A to another row of A or by adding a multiple of a column of A to another column of A, then $\left|B\right|=\left|A\right|$

  $\left|\begin{array}{cc}1& -3\\ 5& 2\end{array}\right|=\left|\begin{array}{cc}1& -33\\ 0& 17\end{array}\right|$

Concept used:

Steps to use a graphing calculator for finding the determinant of a square matrix:

1. In the graphing device press “ $2^{\text{nd}}{x}^{-1}$” for entering the matrix
2. Choose “ EDIT” option for changing the order of the matrix
3. Put the entries of the matrix.
4. Quit the screen
5. Press “ $2^{\text{nd}}{x}^{-1}$ ” button again and select “MATH” option, then select “ $\det \left(\left|A\right|\right)$” and enter the matrix A for finding the determinant

Calculation:

Let $A=\left[\begin{array}{cc}1& -3\\ 5& 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1& -33\\ 0& 17\end{array}\right]$

Two matrices A and B are given, to prove: $\left|B\right|=\left|A\right|$

For the determinant of $A=\left[\begin{array}{cc}1& -3\\ 5& 2\end{array}\right]$ do step 1 and 2

  

Quit the screen and do step 4 and 5

  

Now for the determinant of $B=\left[\begin{array}{cc}1& -33\\ 0& 17\end{array}\right]$ do step 1 and 2

  

Quit the screen and do step 4 and 5

  

Hence the determinant of B is 17

Here it can be conclude that on adding a multiple of a row to another row of a matrix gives the determinant which has the same value as the determinant of the initial matrix.

Manually $\left|A\right|=\left|\begin{array}{cc}1& -3\\ 5& 2\end{array}\right|=2-\left(-3\times 5\right)=2+15=17$

Matrix B is obtained by adding row 1 multiplied by -5 to row 2 in matrix A.

  $\left|B\right|=\left|\begin{array}{cc}1& -33\\ 0& 17\end{array}\right|=17\times 1-0=17$

Here also it can be conclude that on adding a multiple of a row to another row of a matrix gives the determinant which has the same value as the determinant of the initial matrix.

b.

To determine

To explain: why each equation is an example of the given property of determinants. Use a graphing utility to verify the results.

Explanation of Solution

Given:

(A and B are square matrices).

If B is obtained from A by adding a multiple of a row of A to another row of A or by adding a multiple of a column of A to another column of A, then $\left|B\right|=\left|A\right|$

  $\left|\begin{array}{ccc}5& 4& 2\\ 2& -3& 4\\ 7& 6& 3\end{array}\right|=\left|\begin{array}{ccc}1& 10& -6\\ 2& -3& 4\\ 7& 6& 3\end{array}\right|$

Concept used:

Steps to use a graphing calculator for finding the determinant of a square matrix:

1. In the graphing device press “ $2^{\text{nd}}{x}^{-1}$” for entering the matrix
2. Choose “ EDIT” option for changing the order of the matrix
3. Put the entries of the matrix.
4. Quit the screen
5. Press “ $2^{\text{nd}}{x}^{-1}$ ” button again and select “MATH” option, then select “ $\det \left(\left|A\right|\right)$” and enter the matrix A for finding the determinant

Calculation:

Let $A=\left[\begin{array}{ccc}5& 4& 2\\ 2& -3& 4\\ 7& 6& 3\end{array}\right]$ and $B=\left[\begin{array}{ccc}1& 10& -6\\ 2& -3& 4\\ 7& 6& 3\end{array}\right]$

First find $\left|A\right|=\left|\begin{array}{ccc}5& 4& 2\\ 2& -3& 4\\ 7& 6& 3\end{array}\right|$

Do step1 and step 2

  

Quit the screen and do step 4 and step 5

  

Hence $\left|A\right|=-11$

For determinant of $B=\left[\begin{array}{ccc}1& 10& -6\\ 2& -3& 4\\ 7& 6& 3\end{array}\right]$

Do step 1 and step 2

  

Quit the screen and do step 4 and step 5

  

Hence $\left|B\right|=-11$

Here it can be conclude that on adding a multiple of a row to another row of a matrix gives the determinant value same as the determinant of the initial matrix.

On manual calculation

Expanding along $R_1$

  $\begin{array}{l}\left|A\right|=5\left|\begin{array}{cc}-3& 4\\ 6& 3\end{array}\right|-4\left|\begin{array}{cc}2& 4\\ 7& 3\end{array}\right|+2\left|\begin{array}{cc}2& -3\\ 7& 6\end{array}\right|\\ \left|A\right|=5\left(-9-24\right)-4\left(6-28\right)+2\left(12+21\right)\\ \left|A\right|=-11\end{array}$

Here we notice that matrix B is obtained by adding $R_1$ multiplied by -5 to row 2 in matrix A.

Now we find $\left|B\right|=\left|\begin{array}{ccc}1& 10& -6\\ 2& -3& 4\\ 7& 6& 3\end{array}\right|$

Expanding along $R_1$

  $\begin{array}{l}\left|B\right|=1\left|\begin{array}{cc}-3& 4\\ 6& 3\end{array}\right|-10\left|\begin{array}{cc}2& 4\\ 7& 3\end{array}\right|-6\left|\begin{array}{cc}2& -3\\ 7& 6\end{array}\right|\\ \left|B\right|=1\left(-9-24\right)-10\left(6-28\right)-6\left(12+21\right)\\ \left|B\right|=-11\end{array}$

Here also it can be conclude that on adding a multiple of a row to another row of a matrix gives the determinant value same as the determinant of the initial matrix.