Chapter 7.5, Problem 37E

To determine

To find product of below matrices AB, if possible.

  $A=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right],B=\left[\begin{array}{ccc}\frac{1}{3}& 0& 0\\ 0& -\frac{1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$

Answer to Problem 37E

The product of given matrices is:

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

Explanation of Solution

Given:

The matrices:

  $A=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right],B=\left[\begin{array}{ccc}\frac{1}{3}& 0& 0\\ 0& -\frac{1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$

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:

In order to find product of below matrices AB

  $A=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right],B=\left[\begin{array}{ccc}\frac{1}{3}& 0& 0\\ 0& -\frac{1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]$

First check if this product is possible.

Note that here number of columns of A and number of rows of B are same, that is 3. Thus this product is possible.

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

  $\begin{array}{c}AB=\left[\begin{array}{ccc}6& 0& 0\\ 0& 4& 0\\ 0& 0& -2\end{array}\right]\left[\begin{array}{ccc}\frac{1}{3}& 0& 0\\ 0& -\frac{1}{4}& 0\\ 0& 0& \frac{1}{6}\end{array}\right]\\ =\left[\begin{array}{ccc}6\times \frac{1}{3}+0\times 0+0\times 0& 6\times 0+0\times \frac{-1}{4}+0\times 0& 6\times 0+0\times 0+0\times \frac{1}{6}\\ 0\times \frac{1}{3}+4\times 0+0\times 0& 0\times 0+4\times \frac{-1}{4}+0\times 0& 0\times 0+4\times 0+0\times \frac{1}{6}\\ 0\times \frac{1}{3}+0\times 0+\left(-2\right)\times 0& 0\times 0+0\times \frac{-1}{4}+\left(-2\right)\times 0& 0\times 0+0\times 0+\left(-2\right)\times \frac{1}{6}\end{array}\right]\\ =\left[\begin{array}{ccc}2& 0& 0\\ 0& -1& 0\\ 0& 0& -\frac{1}{3}\end{array}\right]\end{array}$

Note that the number of rows and number of columns in matrix AB 3 and 3 respectively.

Thus, product of given matrices is:

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