Chapter 7.5, Problem 13E

a.

To determine

To find $A+B$ and graphing check your result by utility.

Answer to Problem 13E

The sum is $A+B=\left[\begin{array}{cc}8& -1\\ 1& 7\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The sum of both matrices is,

  $\begin{array}{l}A+B=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]+\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]\\ A+B=\left[\begin{array}{cc}5+3& -2+1\\ 3-2& 1+6\end{array}\right]\\ A+B=\left[\begin{array}{cc}8& -1\\ 1& 7\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.

b.

To determine

To find $A-B$ and graphing check your result by utility.

Answer to Problem 13E

The difference is $A-B=\left[\begin{array}{cc}2& -3\\ 5& -5\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

  $\begin{array}{l}A-B=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]\\ A-B=\left[\begin{array}{cc}5-3& -2-1\\ 3+2& 1-6\end{array}\right]\\ A-B=\left[\begin{array}{cc}2& -3\\ 5& -5\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.

c.

To determine

To find $3A$ and graphing check your result by utility.

Answer to Problem 13E

  $3A=\left[\begin{array}{cc}15& -6\\ 9& 3\end{array}\right]$ .

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

  $\begin{array}{l}3A=3\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]\\ 3A=\left[\begin{array}{cc}5\left(3\right)& -2\left(3\right)\\ 3\left(3\right)& 1\left(3\right)\end{array}\right]\\ 3A=\left[\begin{array}{cc}15& -6\\ 9& 3\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.

d.

To determine

To find $3A-2B$ and graphing check your result by utility.

Answer to Problem 13E

The differenceis $3A-2B=\left[\begin{array}{cc}9& -8\\ 13& -9\end{array}\right]$ .

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]$

Concept used:

Operations like addition and subtraction of matrices are only possible when they are of same dimension.

Dimension of matrix with a rows and b columns is given by $a\times b$ .

If a matrix is multiplied by a scalar then each element of the matrix is multiplied by the same scalar.

Calculation:

The difference of both matrices is,

  $\begin{array}{l}3A-2B=3\left[\begin{array}{cc}5& -2\\ 3& 1\end{array}\right]-2\left[\begin{array}{cc}3& 1\\ -2& 6\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}5\left(3\right)& -2\left(3\right)\\ 3\left(3\right)& 1\left(3\right)\end{array}\right]-\left[\begin{array}{cc}3\left(2\right)& 1\left(2\right)\\ -2\left(2\right)& 6\left(2\right)\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}15& -6\\ 9& 3\end{array}\right]-\left[\begin{array}{cc}6& 2\\ -4& 12\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}15-6& -6-2\\ 9-\left(-4\right)& 3-12\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}9& -8\\ 9+4& -9\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}9& -8\\ 13& -9\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.