Chapter 7.5, Problem 14E

a.

To determine

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

Answer to Problem 14E

The sum is $A+B=\left[\begin{array}{cc}-2& 0\\ 6& 3\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]$

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,

  $\begin{array}{l}A+B=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]+\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]\\ A+B=\left[\begin{array}{cc}1-3& 2-2\\ 2+4& 1+2\end{array}\right]\\ A+B=\left[\begin{array}{cc}-2& 0\\ 6& 3\end{array}\right]\end{array}$

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 14E

The difference is $A-B=\left[\begin{array}{cc}4& 4\\ -2& -1\end{array}\right]$

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]$

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,

  $\begin{array}{l}A-B=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]-\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]\\ A-B=\left[\begin{array}{cc}1+3& 2+2\\ 2-4& 1-2\end{array}\right]\\ A-B=\left[\begin{array}{cc}4& 4\\ -2& -1\end{array}\right]\end{array}$

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 14E

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

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]$

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,

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

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 14E

The differenceis $3A-2B=\left[\begin{array}{cc}9& 10\\ -2& -1\end{array}\right]$ .

Explanation of Solution

Given information:

The matrices are,

  $A=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\text{and}B=\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]$

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,

  $\begin{array}{l}3A-2B=3\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]-2\left[\begin{array}{cc}-3& -2\\ 4& 2\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}1\left(3\right)& 2\left(3\right)\\ 2\left(3\right)& 1\left(3\right)\end{array}\right]-\left[\begin{array}{cc}-3\left(2\right)& -2\left(2\right)\\ 4\left(2\right)& 2\left(2\right)\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}3& 6\\ 6& 3\end{array}\right]-\left[\begin{array}{cc}-6& -4\\ 8& 4\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}3-\left(-6\right)& 6-\left(-4\right)\\ 6-8& 3-4\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}3+6& 6+4\\ -2& -1\end{array}\right]\\ 3A-2B=\left[\begin{array}{cc}9& 10\\ -2& -1\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.