Chapter 16.3, Problem 5WE

To determine

To find: product of the matrices.

Answer to Problem 5WE

The product of given matrices is:

  $\left[\begin{array}{cc}-3& 24\\ -1& 14\end{array}\right]$

Explanation of Solution

Given:

The matrices:

  $\left[\begin{array}{cc}3& 0\\ 1& 2\end{array}\right]\left[\begin{array}{cc}-1& 8\\ 0& 3\end{array}\right]$

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

Calculation:

First check if this product is possible.

Note that here number of columns of first matrix is 2 and number of rows of 2^nd matricesis 2 which are same. Thus this product is possible.

Now find this product $AB=\left[c_{ij}\right]$ using formula $c_{ij}=a_{i1}{b}_{1j}+a_{i2}{b}_{2j}+a_{i3}{b}_{3j}\mathrm{...}+a_{in}{b}_{nj}$ .

  $\begin{array}{c}\left[\begin{array}{cc}3& 0\\ 1& 2\end{array}\right]\left[\begin{array}{cc}-1& 8\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}\left(3\right)\left(-1\right)+\left(0\right)\left(0\right)& \left(3\right)\left(8\right)+\left(0\right)\left(3\right)\\ \left(1\right)\left(-1\right)+\left(2\right)\left(0\right)& \left(1\right)\left(8\right)+\left(2\right)\left(3\right)\end{array}\right]\\ =\left[\begin{array}{cc}-3& 24+0\\ -1+0& 8+6\end{array}\right]\\ =\left[\begin{array}{cc}-3& 24\\ -1& 14\end{array}\right]\end{array}$

Thus, product of given matrices is:

  $\left[\begin{array}{cc}-3& 24\\ -1& 14\end{array}\right]$