Chapter 7.5, Problem 37E

To determine

To find: The product of two given matrices.

Answer to Problem 37E

The required product of the given matrices is $AB=\left[\begin{array}{cc}-2& 51\\ -8& 33\\ 0& 27\end{array}\right]$ .

Explanation of Solution

Given: The two matrices are $A=\left[\begin{array}{cc}-1& 6\\ -4& 5\\ 0& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 3\\ 0& 9\end{array}\right]$ .

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 $A=\left[\begin{array}{cc}-1& 6\\ -4& 5\\ 0& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}2& 3\\ 0& 9\end{array}\right]$ .

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-

  $AB$

  $=\left[\begin{array}{cc}-1& 6\\ -4& 5\\ 0& 3\end{array}\right]\left[\begin{array}{cc}2& 3\\ 0& 9\end{array}\right]$

  $=\left[\begin{array}{cc}-1\times 2+6\times 0& -1\times 3+6\times 9\\ -4\times 2+5\times 0& -4\times 3+5\times 9\\ 0\times 2+3\times 0& 0\times 3+3\times 9\end{array}\right]$

  $=\left[\begin{array}{cc}-2+0& -3+54\\ -8+0& -12+45\\ 0& 0+27\end{array}\right]$

  $=\left[\begin{array}{cc}-2& 51\\ -8& 33\\ 0& 27\end{array}\right]$

Thus, the required product is $AB=\left[\begin{array}{cc}-2& 51\\ -8& 33\\ 0& 27\end{array}\right]$