Chapter 3.8, Problem 50SR

To determine

To find: The product of the matrices $\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]$ and $\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]$ .

Answer to Problem 50SR

The product of the matrices $\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]$ and $\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]$ is $\left[\begin{array}{cc}34& 10\\ -21& -5\end{array}\right]$ .

Explanation of Solution

Given:

The matrices are $\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]$ and $\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]$ .

Calculation:

The first matrix $\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]$ is of the order $2\times 2$ and the second matrix $\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]$ is of the order $2\times 2$ .

The product of two matrices is defined as the number of columns of the first matrix is equal to the number of rows of the second matrix.

Multiply the given matrices as follows:

  $\begin{array}{c}\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]=\left[\begin{array}{cc}4\left(6\right)+2\cdot \left(5\right)& 4\cdot 2+2\cdot \left(1\right)\\ -1\left(6\right)+\left(-3\right)\left(5\right)& \left(-1\right)\cdot 2+\left(-3\right)\left(1\right)\end{array}\right]\\ =\left[\begin{array}{cc}24+10& 8+2\\ -6-15& -2-3\end{array}\right]\\ =\left[\begin{array}{cc}34& 10\\ -21& -5\end{array}\right]\end{array}$

Thus, the product of the matrices $\left[\begin{array}{cc}4& 2\\ -1& -3\end{array}\right]$ and $\left[\begin{array}{cc}6& 2\\ 5& 1\end{array}\right]$ is $\left[\begin{array}{cc}34& 10\\ -21& -5\end{array}\right]$ .