Chapter 7.5, Problem 52E

To determine

To evaluate:the matrix expression. Use the matrix capabilities of a graphing utility to verify your answer.

Answer to Problem 52E

The result given matrix expression is

  $\left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}18& -20\\ -4& 5\end{array}\right]$ .

Explanation of Solution

Given information:

The matrix expression is

  $\left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]$

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 $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}+\cdots +a_{in}{b}_{nj}$ .

Calculations:

Let us consider the given expression simplify as shown:

  $\begin{array}{c}\left(\left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\right)\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]=\left(\left[\begin{array}{cc}(6\times 0)+(5\times -1)+(-1\times 4)& (6\times 3)+(5\times -3)+(-1\times 1)\\ (1\times 0)+(-2\times -1)+(0\times 4)& (1\times 3)+(-2\times -3)+(0\times 1)\end{array}\right]\right)\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}0-5-4& 18-15-1\\ 0+2+0& 3+6+0\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}-9& 2\\ 2& 9\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]\\ =\left[\begin{array}{cc}-9\times \left(-2\right)+2\times 0& -9\times 2+2\times \left(-1\right)\\ 2\times \left(-2\right)+9\times 0& 2\times 2+9\times \left(-1\right)\end{array}\right]\\ =\left[\begin{array}{cc}18& -20\\ -4& 5\end{array}\right]\end{array}$

Calculation:

Here, Since A is a $2\times 3$ andB is a $3\times 2$ matrix, therefore, there product will be of dimension $2\times 2$ and hence it will be possible.

Again, AB is a $2\times 2$ matrix and Cis $2\times 2$ , therefore, there product will be of dimension $2\times 2$ and hence it will be possible.

Let us consider the following given matrices.

  $\left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]$

Use the matrix capabilities of a graphing utility to evaluate the product and for that we feed the matrices and get the result as

  $\begin{array}{c}>A:=matrix\left(\left[6,5,-1\right],\left[1,-2,0\right]\right)\\ \left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\\ B:=matrix\left(\left[0,3\right],\left[-1,-3\right],\left[4,1\right]\right)\\ \left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\\ C:=matrix\left(\left[-2,2\right],\left[0,-1\right]\right)\\ \left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]\\ D:=multiply\left(A,B,C\right)\\ \left[\begin{array}{cc}18& -20\\ -4& 5\end{array}\right]\end{array}$

Conclusion:

Performing the same matrix expression in Symbolab graphing utility, the result is

  $\left[\begin{array}{ccc}6& 5& -1\\ 1& -2& 0\end{array}\right]\left[\begin{array}{cc}0& 3\\ -1& -3\\ 4& 1\end{array}\right]\left[\begin{array}{cc}-2& 2\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}18& -20\\ -4& 5\end{array}\right]$

Thus, we got the same results. Hence, verified.