Chapter 11.4, Problem 26AYU

In Problems 27-34, determine whether the product is defined. If it is defined, find the product; if it is not, say "not defined."

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

To determine

The product of the matrix is defined and if so, find the product.

Answer to Problem 26AYU

Solution:

The product is DEFINED and is $=\left[\begin{array}{ccc}-22& 29& 8\\ -10& 17& 6\end{array}\begin{array}{c}-1\\ -1\end{array}\right]$.

Explanation of Solution

Given:

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

Calculation:

Matrix $A$ is a $2\times 2$ matrix and matrix $B$ is a $2\times 4$ matrix. Since the number of columns in $A$ is equal to the number of rows in $B$, the product $AB$ is DEFINED.

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

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

$=\left[\begin{array}{ccc}-24+2& 24+5& 4+4\\ -12+2& 12+5& 2+4\end{array}\begin{array}{c}0-1\\ 0-1\end{array}\right]$

$=\left[\begin{array}{ccc}-22& 29& 8\\ -10& 17& 6\end{array}\begin{array}{c}-1\\ -1\end{array}\right]$