Chapter 7.5, Problem 47E

(a).

To determine

To calculate: the product of two matrices.

Answer to Problem 47E

The product of two matrix $A$ and $B$ that is $\underset{\_}{AB=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]}$ .

Explanation of Solution

Given information:

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

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

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.

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

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

Calculate product

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

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

  $=\left[\begin{array}{cc}3-3& -9-1\\ 1+9& -3+3\end{array}\right]$

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

Hence the product of two matrix $A$ and $B$ that is $\underset{\_}{AB=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]}$ .

(b).

To determine

To calculate: the product of two matrices.

Answer to Problem 47E

The product of two matrix $B$ and $A$ that is $\underset{\_}{BA=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]}$ .

Explanation of Solution

Given information:

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

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

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.

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

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

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

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

  $=\left[\begin{array}{cc}3-3& -1-9\\ 9+1& -3+3\end{array}\right]$

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

Hence the product of two matrix $B$ and $A$ that is $\underset{\_}{BA=\left[\begin{array}{cc}0& -10\\ 10& 0\end{array}\right]}$ .

(c).

To determine

To calculate: the square of a matrix.

Answer to Problem 47E

The square of matrix $A$ that is $\underset{\_}{A^2=\left[\begin{array}{cc}8& -6\\ 6& 8\end{array}\right]}$

Explanation of Solution

Given information:

The matrix for which the square that is $A^2=A.A$ needs to be calculated is as follows.

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

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:

The matrix for which square that is $A^2=A.A$ needs to be calculated is as follows.

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

Calculate the square

  $A^2=A.A$

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

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

  $=\left[\begin{array}{cc}9-1& -3-3\\ 3+3& -1+9\end{array}\right]$

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

Hence the square of matrix $A$ that is $\underset{\_}{A^2=\left[\begin{array}{cc}8& -6\\ 6& 8\end{array}\right]}$ .