Chapter 16.3, Problem 7WE

To determine

To find: product of the matrices.

Answer to Problem 7WE

The product of given matrices is:

  $\left[\begin{array}{ccc}-3& 0& 6\\ 15& 12& 2\\ 3& 0& -6\end{array}\right]$

Explanation of Solution

Given:

The matrices:

  $\left[\begin{array}{ccc}0& 2& -1\\ 4& 1& 0\\ 0& -1& 2\end{array}\right]\left[\begin{array}{ccc}4& 3& 0\\ -1& 0& 2\\ 1& 0& -2\end{array}\right]$

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 3 and number of rows of 2^nd matricesis 3 which are same. Thus this product is possible.

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

  $\left[\begin{array}{ccc}0& 2& -1\\ 4& 1& 0\\ 0& -1& 2\end{array}\right]\left[\begin{array}{ccc}4& 3& 0\\ -1& 0& 2\\ 1& 0& -2\end{array}\right]=\left[\begin{array}{ccc}\left(0\right)\left(4\right)+\left(2\right)\left(-1\right)+\left(-1\right)\left(1\right)& \left(0\right)\left(3\right)+\left(2\right)\left(0\right)+\left(-1\right)\left(0\right)& \left(0\right)\left(0\right)+\left(2\right)\left(2\right)+\left(-1\right)\left(-2\right)\\ \left(4\right)\left(4\right)+\left(1\right)\left(-1\right)+\left(0\right)\left(1\right)& \left(4\right)\left(3\right)+\left(1\right)\left(0\right)+\left(0\right)\left(0\right)& \left(4\right)\left(0\right)+\left(1\right)\left(2\right)+\left(0\right)\left(-2\right)\\ \left(0\right)\left(4\right)+\left(-1\right)\left(-1\right)+\left(2\right)\left(1\right)& \left(0\right)\left(3\right)+\left(-1\right)\left(0\right)+\left(2\right)\left(0\right)& \left(0\right)\left(0\right)+\left(-1\right)\left(2\right)+\left(2\right)\left(-2\right)\end{array}\right]$

  $=\left[\begin{array}{ccc}0-2-1& 0+0+0& 0+4+2\\ 16-1+0& 12+0+0& 0+2+0\\ 0+1+2& 0+0+0& 0-2-4\end{array}\right]$

  $=\left[\begin{array}{ccc}-3& 0& 6\\ 15& 12& 2\\ 3& 0& -6\end{array}\right]$

Thus, product of given matrices is:

  $\left[\begin{array}{ccc}-3& 0& 6\\ 15& 12& 2\\ 3& 0& -6\end{array}\right]$