Q13. - Algebra 2

Question

Determine whether each pair of matrices are inverses.13.

$\mathbf{X}\mathbf{=}\left[\begin{array}{cc}\frac{1}{3}& \mathbf{-}\frac{2}{3}\\ \frac{2}{3}& \mathbf{-}\frac{1}{3}\end{array}\right]$

$\mathbf{Y}\mathbf{=}\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$

Step-by-Step Solution

Answer

The given matrices are not inverse of each other.

Step 1 ­- Definition of inverse of matrix.

A square matrixB is said to be an inverse of the square matrixA if$\mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{=}\mathbf{I}$ whereL is an identity matrix of the same order as that of matrixA orB .

Step 2 ­- Find the product of the given matrices.

The product of the given matrices is:

$\begin{array}{c}X\cdot Y=\left[\begin{array}{cc}\frac{1}{3}& -\frac{2}{3}\\ \frac{2}{3}& -\frac{1}{3}\end{array}\right]\cdot \left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]\\ =\left[\begin{array}{cc}\left(\frac{1}{3}\right)\left(1\right)+(-\frac{2}{3})\left(2\right)& \left(\frac{1}{3}\right)\left(2\right)+(-\frac{2}{3})\left(1\right)\\ \left(\frac{2}{3}\right)\left(1\right)+(-\frac{1}{3})\left(2\right)& \left(\frac{2}{3}\right)\left(2\right)+(-\frac{1}{3})\left(1\right)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{3}+\left(\frac{-4}{3}\right)& \frac{2}{3}+(-\frac{2}{3})\\ \frac{2}{3}+(-\frac{2}{3})& \frac{4}{3}+\left(\frac{-1}{3}\right)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1}{3}-\frac{4}{3}& \frac{2}{3}-\frac{2}{3}\\ \frac{2}{3}-\frac{2}{3}& \frac{4}{3}-\frac{1}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{1-4}{3}& 0\\ 0& \frac{4-1}{3}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{-3}{3}& 0\\ 0& \frac{3}{3}\end{array}\right]\\ =\left[\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right]\end{array}$

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.