Q4. - Algebra 2

Question

Determine whether each pair of matrices are inverses.

4. $\mathbf{A}\mathbf{=}\left[\begin{array}{cc}2& -1\\ 1& -3\end{array}\right]$,width="124" height="90" style="max-width: none; vertical-align: -40px;" $\mathbf{B}\mathbf{=}\left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& \mathbf{-}\frac{1}{3}\end{array}\right]$

 

Step-by-Step Solution

Answer

The given matrices are not inverse of each other.

Step 1 ­- Definition of inverse of matrix.

A square matrix $\mathbf{B}$ is said to be an inverse of the square matrix $\mathbf{A}$ if $\mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{=}\mathbf{I}$where$\mathbf{I}$ is an identity matrix of the same order as that of matrix $\mathbf{A}$ or$\mathbf{B}$

Step 2 ­- Find the product of the given matrices.

The product of the given matrices is:

$\begin{array}{c}A\cdot B=\left[\begin{array}{cc}2& -1\\ 1& -3\end{array}\right]\cdot \left[\begin{array}{cc}\frac{1}{2}& 0\\ 0& -\frac{1}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\left(2\right)\left(\frac{1}{2}\right)+\left(-1\right)\left(0\right)& \left(2\right)\left(0\right)+\left(-1\right)\left(-\frac{1}{3}\right)\\ \left(1\right)\left(\frac{1}{2}\right)+\left(-3\right)\left(0\right)& \left(1\right)\left(0\right)+\left(-3\right)\left(\frac{-1}{3}\right)\end{array}\right]\\ =\left[\begin{array}{cc}1+0& 0+\frac{1}{3}\\ \frac{1}{2}+0& 0+1\end{array}\right]\\ =\left[\begin{array}{cc}1& \frac{1}{3}\\ \frac{1}{2}& 1\end{array}\right]\end{array}$

As the product of the given matrices is not equal to the identity matrix, therefore the given matrices are not inverse of each other.

Step 3 ­- Write the conclusion.

The given matrices are not inverse of each other.