Chapter 7.5, Problem 43E

(a).

To determine

To find $AB$ and check result by matrix calculator.

Answer to Problem 43E

  $AB=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]$

Explanation of Solution

Given:

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

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Here, Since $A\text{ and }B$ are matrices of dimension $2\times 2$ , therefore, there product will be of dimension $2\times 2$ and hence it will be possible.

  $\begin{array}{c}AB=\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]\left[\begin{array}{cc}1& -3\\ 3& 1\end{array}\right]\\ =\left[\begin{array}{cc}3\times 1+\left(-1\right)\times 3& 3\times \left(-3\right)+\left(-1\right)\times 1\\ 1\times 1+\left(3\right)\times 3& 1\times \left(-3\right)+\left(3\right)\times 1\end{array}\right]\\ =\left[\begin{array}{cc}3-3& -9-1\\ 1+9& -3+3\end{array}\right]\\ =\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.

(b).

To determine

To find $BA$ and check result by matrix calculator.

Answer to Problem 43E

  $BA=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]$

Explanation of Solution

Given:

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

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Here, Since $B\text{and}A$ are matrices of dimension $2\times 2$ , therefore, there product will be of dimension $2\times 2$ and hence it will be possible.

  $\begin{array}{c}BA=\left[\begin{array}{cc}1& -3\\ 3& 1\end{array}\right]\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]\\ =\left[\begin{array}{cc}3\times 1+\left(-3\right)\times 1& 3\times \left(-3\right)+\left(-1\right)\times 1\\ 1\times 1+\left(3\right)\times 3& 3\times \left(-1\right)+\left(3\right)\times 1\end{array}\right]\\ =\left[\begin{array}{cc}3-3& -9-1\\ 1+9& -3+3\end{array}\right]\\ =\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.

(c).

To determine

To find $A^2$ and check result by matrix calculator.

Answer to Problem 43E

  $A^2=\left[\begin{array}{cc}8& -6\\ 6& 8\end{array}\right]$

Explanation of Solution

Given:

Consider $A=\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]$ .

Concept used:

If $A=\left[a_{ij}\right]$ is an $m\times n$ matrix and $B=\left[b_{ij}\right]$ is an $n\times p$ matrix, then the product $AB$ is an $m\times p$ matrix given by $AB=\left[c_{ij}\right]$ , where $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculation:

Here, Sincematrix A is of dimension $2\times 2$ , therefore, product $A^2$ will be of dimension $2\times 2$ and hence it will be possible.

  $\begin{array}{c}{A}^2=A\times A=\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]\left[\begin{array}{cc}3& -1\\ 1& 3\end{array}\right]\\ =\left[\begin{array}{cc}3\times 3+\left(-1\right)\times 1& 3\times \left(-1\right)+\left(-1\right)\times 3\\ 1\times 3+\left(3\right)\times 1& 1\times \left(-1\right)+\left(3\right)\times 3\end{array}\right]\\ =\left[\begin{array}{cc}9-1& -3-3\\ 3+3& -1+9\end{array}\right]\\ =\left[\begin{array}{cc}8& -6\\ 6& 8\end{array}\right]\end{array}$

Now, by graphing utility of matrix capability,

  

Hence, the result is verified.