Chapter 8.1, Problem 144E

(a).

To determine

To calculate: the difference of two matrices.

Answer to Problem 144E

The difference of two matrix $A$ and $B$ that is $\underset{\_}{A-B=\left[\begin{array}{cc}10& 19\\ -12& -5\end{array}\right]}$ .

Explanation of Solution

Given information:

Two matrices for which difference that is $A-B$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..

Calculation:

Two matrices for which difference that is $A-B$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..

Calculate $A-B$

  $A-B=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]-\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$

  $=\left[\begin{array}{cc}10-0& 7-\left(-12\right)\\ \left(-4\right)-8& 6-11\end{array}\right]$

  $=\left[\begin{array}{cc}10& 19\\ -12& -5\end{array}\right]$

Hence $\underset{\_}{A-B=\left[\begin{array}{cc}10& 19\\ -12& -5\end{array}\right]}$ .

(b).

To determine

To calculate: the expression involving two matrices.

Answer to Problem 144E

  $\underset{\_}{2B-3A=\left[\begin{array}{cc}-30& -45\\ 28& 4\end{array}\right]}$ .

Explanation of Solution

Given information:

Two matrices for which expression that is $2B-3A$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..

Calculation:

Two matrices for which expression that is $2B-3A$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..

Calculate $2B-3A$

  $2B-3A=2\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]-3\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$

  $=\left[\begin{array}{cc}2\times 0& 2\times \left(-12\right)\\ 2\times 8& 2\times 11\end{array}\right]-\left[\begin{array}{cc}3\times 10& 3\times 7\\ 3\times \left(-4\right)& 3\times 6\end{array}\right]$

  $=\left[\begin{array}{cc}0& -24\\ 16& 22\end{array}\right]-\left[\begin{array}{cc}30& 21\\ -12& 18\end{array}\right]$

  $=\left[\begin{array}{cc}0-30& -24-21\\ 16-\left(-12\right)& 22-18\end{array}\right]$

  $=\left[\begin{array}{cc}-30& -45\\ 28& 4\end{array}\right]$

Hence $\underset{\_}{2B-3A=\left[\begin{array}{cc}-30& -45\\ 28& 4\end{array}\right]}$ .

(c).

To determine

To calculate: the product of two matrices.

Answer to Problem 144E

The product of two matrix $A$ and $B$ that is $\underset{\_}{AB=\left[\begin{array}{cc}56& -43\\ 48& 114\end{array}\right]}$ .

Explanation of Solution

Given information:

Two matrix for which product that is $AB$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ .

Formula used:

To calculate product of two matrix , prerequisite is number of columns in first matrix must be equal to number of rows in second matrix.

Calculation:

Two matrix for which product $AB$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..........

As number of columns of matrix $A$ is equal to number of number of rows of matrix $B$ that is 2 , so $AB$ can be calculated.

Calculate product

  $AB=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$

  $=\left[\begin{array}{cc}10\times 0+7\times 8& 10\times (-12)+7\times 11\\ (-4)\times 0+6\times 8& (-4)\times (-12)+6\times 11\end{array}\right]$

  $=\left[\begin{array}{cc}0+56& -120+77\\ 0+48& 48+66\end{array}\right]$

  $=\left[\begin{array}{cc}56& -43\\ 48& 114\end{array}\right]$

Hence the product of two matrix $A$ and $B$ that is $\underset{\_}{AB=\left[\begin{array}{cc}56& -43\\ 48& 114\end{array}\right]}$ .

(d).

To determine

To calculate: the product of two matrices.

Answer to Problem 144E

The product of two matrix $B$ and $A$ that is $\underset{\_}{BA=\left[\begin{array}{cc}48& -72\\ 36& 122\end{array}\right]}$ .

Explanation of Solution

Given information:

Two matrix for which product that is $BA$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ .

Formula used:

To calculate product of two matrix , prerequisite is number of columns in first matrix must be equal to number of rows in second matrix.

Calculation:

Two matrix for which product $BA$ needs to be calculated are as follows.

  $A=\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]$ ..........

As number of columns of matrix $B$ is equal to number of number of rows of matrix $A$ that is 2 , so $BA$ can be calculated.

Calculate product

  $BA=\left[\begin{array}{cc}0& -12\\ 8& 11\end{array}\right]\left[\begin{array}{cc}10& 7\\ -4& 6\end{array}\right]$

  $=\left[\begin{array}{cc}0\times 10+(-12)\times (-4)& 0\times 7+(-12)\times 6\\ 8\times 10+11\times (-4)& 8\times 7+11\times 6\end{array}\right]$

  $=\left[\begin{array}{cc}0+48& 0-72\\ 80-44& 56+66\end{array}\right]$

  $=\left[\begin{array}{cc}48& -72\\ 36& 122\end{array}\right]$

Hence the product of two matrix $B$ and $A$ that is $\underset{\_}{BA=\left[\begin{array}{cc}48& -72\\ 36& 122\end{array}\right]}$ .