# Find the matrix X satisfying the equation

Question:

Find the matrix X satisfying the equation

$\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right] X\left[\begin{array}{ll}5 & 3 \\ 3 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

Solution:

Let $A=\left[\begin{array}{ll}2 & 1\end{array}\right.$

$5 \quad 3], B=\left[\begin{array}{ll}5 & 3\end{array}\right]$

$\left.\begin{array}{ll}3 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0\end{array}\right.$

$\left.\begin{array}{ll}0 & 1\end{array}\right]$

$\Rightarrow|\mathrm{A}|=\mid 2 \quad 1$

$5 \quad 3 \mid=6-5=1$

Since, $|\mathrm{A}| \neq 0$

Thus, $A$ is invertible.

Also, $|\mathrm{B}|=\mid \begin{array}{ll}5 & 3\end{array}$

$3 \quad 2 \mid=10-9=1$

Thus, $B$ is invertible.

Cofactors of matrices $A \& B$ are

$\begin{array}{cccc}A_{11}=3 & A_{12}=-5 & A_{21}=-1 & A_{22}=2 \\ B_{11}=2 & B_{12}=-3 & B_{21}=-3 & B_{22}=5\end{array}$

Now,

$\operatorname{adj} A=\left[\begin{array}{ll}3 & -5\end{array}\right.$

$-1 \quad 2]^{\mathrm{T}}=\left[\begin{array}{ll}3 & -1\end{array}\right.$

$\left.\begin{array}{ll}-5 & 2\end{array}\right]$

$\operatorname{adj} B=\left[\begin{array}{ll}2 & -3\end{array}\right.$

$-3 \quad 5]^{\mathrm{T}}=\left[\begin{array}{ll}2 & -3\end{array}\right.$

$\left.\begin{array}{ll}-3 & 5\end{array}\right]$

$A^{-1}=\frac{1}{|\mathrm{~A}|} \mathrm{adj} A=\left[\begin{array}{ll}3 & -1\end{array}\right.$

$\left.\begin{array}{ll}-5 & 2\end{array}\right]$

$B^{-1}=\frac{1}{|\mathrm{~B}|}$ adj $B=\left[\begin{array}{ll}2 & -3\end{array}\right.$

$-3 \quad 5]$

The given matrix equation becomes $A X B=I$

$\Rightarrow A^{-1} A X B B^{-1}=I A^{-1} B^{-1}$

$\Rightarrow\left(A^{-1} A\right) X\left(B B^{-1}\right)=A^{-1} B^{-1}$

$\Rightarrow I X I=A^{-1} B^{-1}$

$\Rightarrow X=A^{-1} B^{-1}$

$\Rightarrow X=\left[\begin{array}{ll}3 & -1\end{array}\right.$

$-5 \quad 2]\left[\begin{array}{ll}2 & -3\end{array}\right.$

$-3 \quad 5]=\left[\begin{array}{ll}6+3 & -9-5\end{array}\right.$

$-10-6 \quad 15+10]=\left[\begin{array}{ll}9 & -14\end{array}\right.$

$\left.\begin{array}{ll}-16 & 25\end{array}\right]$