Q14. - Algebra 2

Question

Determine whether each pair of matrices are inverses.14.

$\mathbf{C}\mathbf{=}\left[\begin{array}{cc}1& 5\\ 1& -2\end{array}\right]$

$\mathbf{D}\mathbf{=}\left[\begin{array}{cc}\frac{2}{7}& \frac{5}{7}\\ \frac{1}{7}& \mathbf{-}\frac{1}{7}\end{array}\right]$

Step-by-Step Solution

Answer

The given matrices are inverse of each other.

Step 1 ­- Definition of inverse of matrix.

A square matrixB is said to be an inverse of the square matrix Aif$\mathbf{AB}\mathbf{=}\mathbf{BA}\mathbf{=}\mathbf{I}$ 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:

$\begin{array}{c}C\cdot D=\left[\begin{array}{cc}1& 5\\ 1& -2\end{array}\right]\cdot \left[\begin{array}{cc}\frac{2}{7}& \frac{5}{7}\\ \frac{1}{7}& -\frac{1}{7}\end{array}\right]\\ =\left[\begin{array}{cc}\left(1\right)\left(\frac{2}{7}\right)+\left(5\right)\left(\frac{1}{7}\right)& \left(1\right)\left(\frac{5}{7}\right)+\left(5\right)(-\frac{1}{7})\\ \left(1\right)\left(\frac{2}{7}\right)+(-2)\left(\frac{1}{7}\right)& \left(1\right)\left(\frac{5}{7}\right)+(-2)(-\frac{1}{7})\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{7}+\left(\frac{5}{7}\right)& \frac{5}{7}+(-\frac{5}{7})\\ \frac{2}{7}+(-\frac{2}{7})& \frac{5}{7}+\left(\frac{2}{7}\right)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{7}+\frac{5}{7}& \frac{5}{7}-\frac{5}{7}\\ \frac{2}{7}-\frac{2}{7}& \frac{5}{7}+\frac{2}{7}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2+5}{7}& 0\\ 0& \frac{5+2}{7}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{7}{7}& 0\\ 0& \frac{7}{7}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

$\begin{array}{c}D.C=\left[\begin{array}{cc}\frac{2}{7}& \frac{5}{7}\\ \frac{1}{7}& -\frac{1}{7}\end{array}\right]\cdot \left[\begin{array}{cc}1& 5\\ 1& -2\end{array}\right]\\ =\left[\begin{array}{cc}\left(\frac{2}{7}\right)\left(1\right)+\left(\frac{5}{7}\right)\left(1\right)& \left(\frac{2}{7}\right)\left(5\right)+\left(\frac{5}{7}\right)(-2)\\ \left(\frac{1}{7}\right)\left(1\right)+(-\frac{1}{7})\left(1\right)& \left(\frac{1}{7}\right)\left(5\right)+(-\frac{1}{7})(-2)\end{array}\right]\\ =\left[\begin{array}{cc}\frac{2}{7}+\frac{5}{7}& \frac{10}{7}-\frac{10}{7}\\ \frac{1}{7}-\frac{1}{7}& \frac{5}{7}+\frac{2}{7}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{7}{7}& 0\\ 0& \frac{7}{7}\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

As the product of the given matrices is equal to the identity matrix, therefore the given matrices are inverse of each other.

Step 3 ­- Write the conclusion.

The given matrices are inverse of each other.