Chapter 11.4, Problem 30AYU

To determine

To calculate: The product of the matrices $\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$ .

Answer to Problem 30AYU

The product of the matrices $\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$ is $\left[\begin{array}{cc}6& 32\\ 1& 3\\ -2& -4\end{array}\right]$ .

Explanation of Solution

Given information:

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

Formula used:

To find the product AB of two matrices A and B, the number of column in matrix A must be equal to number of rows in matrix B.

Calculation:

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

Recall that to find the product AB of two matrices A and B, the number of column in matrix A must be equal to number of rows in matrix B.

  $\begin{array}{c}\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}8-2+0& 24+2+6\\ 0+1+0& 0-1+4\\ -2+0+0& -6+0+2\end{array}\right]\\ =\left[\begin{array}{cc}6& 32\\ 1& 3\\ -2& -4\end{array}\right]\end{array}$

Thus, the product of the matrices $\left[\begin{array}{ccc}4& -2& 3\\ 0& 1& 2\\ -1& 0& 1\end{array}\right]\left[\begin{array}{cc}2& 6\\ 1& -1\\ 0& 2\end{array}\right]$ is $\left[\begin{array}{cc}6& 32\\ 1& 3\\ -2& -4\end{array}\right]$ .