91Ó°ÊÓ

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}Kâ‹\dots 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]â‹\dots \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.