Chapter 8, Problem 4CT

a.

To determine

To calculate: $A-B$ , where A and B are the matrices.

Answer to Problem 4CT

  $A-B=\left[\begin{array}{cc}1& 5\\ 0& -4\end{array}\right]$

Explanation of Solution

Given information:

Given matrices are $A=\left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]$

Consider, the matrices A and B as, $A=\left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]$

Now,

  $\begin{array}{c}A-B=\left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]-\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]\\ =\left[\begin{array}{cc}6-5& 5-0\\ -5-\left(-5\right)& -5-\left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}1& 5\\ -5+5& -5+1\end{array}\right]\\ =\left[\begin{array}{cc}1& 5\\ 0& -4\end{array}\right]\end{array}$

Therefore, $A-B=\left[\begin{array}{cc}1& 5\\ 0& -4\end{array}\right]$

b.

To determine

To calculate: $3C$ , where C is a matrix.

Answer to Problem 4CT

  $3C=\left[\begin{array}{ccc}6& -3& 12\\ 0& 18& -9\end{array}\right]$

Explanation of Solution

Given information:

Matrix C is given as $C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Consider the given matrix as $C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Now,

  $\begin{array}{c}3C=3\times \left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]\\ =\left[\begin{array}{ccc}3\times 2& 3\times \left(-1\right)& 3\times 4\\ 3\times 0& 3\times 6& 3\times \left(-3\right)\end{array}\right]\\ =\left[\begin{array}{ccc}6& -3& 12\\ 0& 18& -9\end{array}\right]\end{array}$

Therefore, $3C=\left[\begin{array}{ccc}6& -3& 12\\ 0& 18& -9\end{array}\right]$

c.

To determine

To calculate: $3A-2B$ , where A and B are the matrices.

Answer to Problem 4CT

  $3A-2B=\left[\begin{array}{cc}8& 15\\ -5& -13\end{array}\right]$

Explanation of Solution

Given information:

Given matrices are $A=\left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]$

Consider the given matrices as $A=\left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]$

Now,

  $3A=3\times \left[\begin{array}{cc}6& 5\\ -5& -5\end{array}\right]=\left[\begin{array}{cc}3\times 6& 3\times 5\\ 3\times \left(-5\right)& 3\times \left(-5\right)\end{array}\right]=\left[\begin{array}{cc}18& 15\\ -15& -15\end{array}\right]$

And

  $2B=2\times \left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]=\left[\begin{array}{cc}2\times 5& 2\times 0\\ 2\times \left(-5\right)& 2\times \left(-1\right)\end{array}\right]=\left[\begin{array}{cc}10& 0\\ -10& -2\end{array}\right]$

Then

  $\begin{array}{c}3A-2B=\left[\begin{array}{cc}18& 15\\ -15& -15\end{array}\right]-\left[\begin{array}{cc}10& 0\\ -10& -2\end{array}\right]\\ =\left[\begin{array}{cc}18-10& 15-0\\ -15-\left(-10\right)& -15-\left(-2\right)\end{array}\right]\\ =\left[\begin{array}{cc}8& 15\\ -15+10& -15+2\end{array}\right]\\ =\left[\begin{array}{cc}8& 15\\ -5& -13\end{array}\right]\end{array}$

Therefore, $3A-2B=\left[\begin{array}{cc}8& 15\\ -5& -13\end{array}\right]$

d.

To determine

To calculate: $BC$ , where B and C are the matrices.

Answer to Problem 4CT

  $BC=\left[\begin{array}{ccc}10& -5& 20\\ -10& -1& -17\end{array}\right]$

Explanation of Solution

Given information:

Given matrices are $B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]\text{ and }C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Formula used:

If A and B are two matrices, then these matrices are confirmable for product that is $A\times B$ only if number of columns of matrix A is equal to number of rows of matrix B.

Consider the given matrices as $B=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]\text{ and }C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Now,

  $\begin{array}{c}BC=\left[\begin{array}{cc}5& 0\\ -5& -1\end{array}\right]\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]\\ =\left[\begin{array}{ccc}5\times 2+0\times 0& 5\times \left(-1\right)+0\times 6& 5\times 4+0\times \left(-3\right)\\ -5\times 2+\left(-1\right)\times 0& -5\times \left(-1\right)+\left(-1\right)\times 6& -5\times 4+\left(-1\right)\times \left(-3\right)\end{array}\right]\\ =\left[\begin{array}{ccc}10& -5& 20\\ -10& 5-6& -20+3\end{array}\right]\\ =\left[\begin{array}{ccc}10& -5& 20\\ -10& -1& -17\end{array}\right]\end{array}$

Therefore, $BC=\left[\begin{array}{ccc}10& -5& 20\\ -10& -1& -17\end{array}\right]$

e.

To determine

To calculate: $C^2$ , where C is the matrix.

Answer to Problem 4CT

  $C^2\text{ does not exist}\text{.}$

Explanation of Solution

Given information:

Matrix C is given as $C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Consider the given matrix as $C=\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]$

Now,

  $\begin{array}{c}{C}^2=C\times C\\ ={\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]}_{2\times 3}{\left[\begin{array}{ccc}2& -1& 4\\ 0& 6& -3\end{array}\right]}_{2\times 3}\end{array}$

This multiplication is not possible. Because number of columns of first matrix is not equal to number of rows of second matrix.

Therefore, $C^2$ does not exist.