Chapter 7.5, Problem 38E

To determine

To find: The product of two given matrices.

Answer to Problem 38E

The required product of the given matrices is $AB=\left[\begin{array}{cc}11& 1\\ -11& 5\\ 54& -6\end{array}\right]$ .

Explanation of Solution

Given: The two matrices are $A=\left[\begin{array}{cc}3& -1\\ 4& -5\\ 2& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}6& 0\\ 7& -1\end{array}\right]$ .

Concept Used: Matrix multiplication is possible if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. Thus, two matrices $A={\left[a_{ij}\right]}_{m\times n}$ of order $m\times n$ and $B={\left[b_{ij}\right]}_{n\times p}$ of order $n\times p$ can be multiplied which results in a matrix of order $m\times p$ given by $AB={\left[a_{ij}{b}_{ji}\right]}_{m\times p}$ , i.e.- the respective terms in the rows of the first matrix is multiplied to the respective terms in the columns of the second matrix, or vice versa.

Calculations: The given matrices are $A=\left[\begin{array}{cc}3& -1\\ 4& -5\\ 2& 6\end{array}\right]$ and $B=\left[\begin{array}{cc}6& 0\\ 7& -1\end{array}\right]$ .

Since number of columns in A is equal to the number of rows in B , hence AB exists which will be a matrix of order $3\times 2$ .

Thus the required product is-

  $AB$

  $=\left[\begin{array}{cc}3& -1\\ 4& -5\\ 2& 6\end{array}\right]\left[\begin{array}{cc}6& 0\\ 7& -1\end{array}\right]$

  $=\left[\begin{array}{cc}3\times 6-1\times 7& 3\times 0-1\times (-1)\\ 4\times 6-5\times 7& 4\times 0-5\times (-1)\\ 2\times 6+6\times 7& 2\times 0+6\times (-1)\end{array}\right]$

  $=\left[\begin{array}{cc}18-7& 0+1\\ 24-35& 0+5\\ 12+42& 0-6\end{array}\right]$

  $=\left[\begin{array}{cc}11& 1\\ -11& 5\\ 54& -6\end{array}\right]$

Thus, the required product is $AB=\left[\begin{array}{cc}11& 1\\ -11& 5\\ 54& -6\end{array}\right]$