Chapter 7.2, Problem 28E

Solution

a.

To calculate: The matrix multiplication $AB$ by using the definition of the matrix multiplication.

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

The multiplication is $\left[\begin{array}{cccc}-3& 2& 1& 3\\ 2& 1& 0& -1\\ -1& 2& 3& -4\\ 4& 0& 2& -1\end{array}\right]$ .

Given Information:

The matrix are,

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

Formula Used:

The matrix multiplication is possible if and only if order of the first matrix column equal to the order of the second matrix row.

Calculation:

Consider the given matrix,

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

It can be observed that the number of column of matrix $A$ is 4 and the number of row of the matrix $B$ is 4. So the multiplication of the matrix $AB$ exists.

Now, find the value of $AB$ .

  $\begin{array}{c}AB=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]\cdot \left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]\\ =\left[\begin{array}{cccc}0+0-3+0& 0+0+2+0& 0+0+1+0& 0+0+3+0\\ 0+2+0+0& 0+1+0+0& 0+0+0+0& 0-1+0+0\\ -1+0+0+0& 2+0+0+0& 3+0+0+0& -4+0+0+0\\ 0+0+0+4& 0+0+0+0& 0+0+0+2& 0+0+0-1\end{array}\right]\\ =\left[\begin{array}{cccc}-3& 2& 1& 3\\ 2& 1& 0& -1\\ -1& 2& 3& -4\\ 4& 0& 2& -1\end{array}\right]\end{array}$

Now, use the graphing calculator to find the product of the matrix.

  

Hence, the obtained matrix after multiplication is $\left[\begin{array}{cccc}-3& 2& 1& 3\\ 2& 1& 0& -1\\ -1& 2& 3& -4\\ 4& 0& 2& -1\end{array}\right]$ .

b.

To calculate: The matrix multiplication $BA$ by using the definition of the matrix multiplication.

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

The multiplication is $\left[\begin{array}{cccc}3& 2& -1& -4\\ 0& 1& 2& -1\\ 1& 2& -3& 3\\ 2& 0& 4& -1\end{array}\right]$ .

Given Information:

The matrix are,

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

Formula Used:

The matrix multiplication is possible if and only if order of the first matrix column equal to the order of the second matrix row.

Calculation:

Consider the given matrix,

  $A=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]$ , $B=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]$

It can be observed that the number of column of matrix $B$ is 4 and the number of row of the matrix $A$ is 4. So the multiplication of the matrix $BA$ exists.

Now, find the value of $BA$ .

  $\begin{array}{c}BA=\left[\begin{array}{cccc}-1& 2& 3& -4\\ 2& 1& 0& -1\\ -3& 2& 1& 3\\ 4& 0& 2& -1\end{array}\right]\cdot \left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{cccc}0+0+3+0& 0+2+0+0& -1+0+0+0& 0+0+0-4\\ 0+0+0+0& 0+1+0-0& 2+0+0+0& 0+0+0-1\\ 0+0+1+0& 0+2+0+0& -3+0+0+0& 0+0+0+3\\ 0+0+2-0& 0+0+0+0& 4+0+0+0& 0+0+0-1\end{array}\right]\\ =\left[\begin{array}{cccc}3& 2& -1& -4\\ 0& 1& 2& -1\\ 1& 2& -3& 3\\ 2& 0& 4& -1\end{array}\right]\end{array}$

Now, use the graphing calculator to find the product of the matrix.

  

Hence, the multiplication of matrix is $\left[\begin{array}{cccc}3& 2& -1& -4\\ 0& 1& 2& -1\\ 1& 2& -3& 3\\ 2& 0& 4& -1\end{array}\right]$ .