Determine whether each pair of matrices are inverses.15.

$\mathbf{J}\mathbf{=}\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 1& 1& 2\end{array}\right]$

$\mathbf{K}\mathbf{=}\left[\begin{array}{ccc}\frac{-5}{4}& \frac{1}{4}& \frac{7}{4}\\ \frac{3}{4}& \frac{1}{4}& \mathbf{-}\frac{5}{4}\\ \frac{1}{4}& \mathbf{-}\frac{1}{4}& \frac{1}{4}\end{array}\right]$

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

The product of the given matrices is:

$\begin{array}{c}K\cdot J=\left[\begin{array}{ccc}\frac{-5}{4}& \frac{1}{4}& \frac{7}{4}\\ \frac{3}{4}& \frac{1}{4}& -\frac{5}{4}\\ \frac{1}{4}& -\frac{1}{4}& \frac{1}{4}\end{array}\right]\cdot \left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\\ 1& 1& 2\end{array}\right]\\ =\left[\begin{array}{ccc}\left(\frac{-5}{4}\right)\left(1\right)+\left(\frac{1}{4}\right)\left(2\right)+\left(\frac{7}{4}\right)\left(1\right)& \left(\frac{-5}{4}\right)\left(2\right)+\left(\frac{1}{4}\right)\left(3\right)+\left(\frac{7}{4}\right)\left(1\right)& \left(\frac{-5}{4}\right)\left(3\right)+\left(\frac{1}{4}\right)\left(1\right)+\left(\frac{7}{4}\right)\left(2\right)\\ \left(\frac{3}{4}\right)\left(1\right)+\left(\frac{1}{4}\right)\left(2\right)+(-\frac{5}{4})\left(1\right)& \left(\frac{3}{4}\right)\left(2\right)+\left(\frac{1}{4}\right)\left(3\right)+(-\frac{5}{4})\left(1\right)& \left(\frac{3}{4}\right)\left(3\right)+\left(\frac{1}{4}\right)\left(1\right)+(-\frac{5}{4})\left(2\right)\\ \left(\frac{1}{4}\right)\left(1\right)+(-\frac{1}{4})\left(2\right)+\left(\frac{1}{4}\right)\left(1\right)& \left(\frac{1}{4}\right)\left(2\right)+(-\frac{1}{4})\left(3\right)+\left(\frac{1}{4}\right)\left(1\right)& \left(\frac{1}{4}\right)\left(3\right)+(-\frac{1}{4})\left(1\right)+\left(\frac{1}{4}\right)\left(2\right)\end{array}\right]\\ =\left[\begin{array}{ccc}-\frac{5}{4}+\frac{2}{4}+\frac{7}{4}& -\frac{10}{4}+\frac{3}{4}+\frac{7}{4}& -\frac{15}{4}+\frac{1}{4}+\frac{14}{4}\\ \frac{3}{4}+\frac{2}{4}-\frac{5}{4}& \frac{6}{4}+\frac{3}{4}-\frac{5}{4}& \frac{9}{4}+\frac{1}{4}-\frac{10}{4}\\ \frac{1}{4}-\frac{2}{4}+\frac{1}{4}& \frac{2}{4}-\frac{3}{4}+\frac{1}{4}& \frac{3}{4}-\frac{1}{4}+\frac{2}{4}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{-5+2+7}{4}& \frac{-10+3+7}{4}& \frac{-15+1+14}{4}\\ \frac{3+2-5}{4}& \frac{6+3-5}{4}& \frac{9+1-10}{4}\\ \frac{1-2+1}{4}& \frac{2-3+1}{4}& \frac{3-1+2}{4}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{4}{4}& \frac{0}{4}& \frac{0}{4}\\ \frac{0}{4}& \frac{4}{4}& \frac{0}{4}\\ \frac{0}{4}& \frac{0}{4}& \frac{4}{4}\end{array}\right]\\ =\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 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.

The given matrices are inverse of each other.