Question

Find the matrix X satisfying the equation 

\[\begin{bmatrix}2 & 1 \\ 5 & 3\end{bmatrix} X \begin{bmatrix}5 & 3 \\ 3 & 2\end{bmatrix} = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} .\]

Answer 1

\[\text{ Let } A = \begin{bmatrix} 2 & 1\\5 & 3 \end{bmatrix} , B = \begin{bmatrix} 5 & 3\\3 & 2 \end{bmatrix}\text{ and }I = \begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix}\]
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 2 & 1\\5 & 3 \end{vmatrix} = 6 - 5 = 1 \]
\[\text{ Since, }\left| A \right| \neq 0\]
Thus, A is invertible.
\[\text{ Also, }\left| B \right| = \begin{vmatrix} 5 & 3\\3 & 2 \end{vmatrix} = 10 - 9 = 1\]
Thus, B is invertible.
Cofactors of matrices A & B are
\[ A_{11} = 3, A_{12} = - 5, A_{21} = - 1, A_{22} = 2\]
\[ B_{11} = 2, B_{12} = - 3, B_{21} = - 3, B_{22} = 5\]
Now, 
\[adj A = \begin{bmatrix} 3 & - 5\\ - 1 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & - 1\\ - 5 & 2 \end{bmatrix} \]
\[adj B = \begin{bmatrix} 2 & - 3 \\ - 3 & 5 \end{bmatrix}^T = \begin{bmatrix} 2 & - 3 \\ - 3 & 5 \end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix} 3 & - 1\\ - 5 & 2 \end{bmatrix}\]
\[ B^{- 1} = \frac{1}{\left| B \right|}adj B = \begin{bmatrix} 2 & - 3 \\ - 3 & 5 \end{bmatrix}\]
The given matrix equation becomes AXB = I
\[ \Rightarrow A^{- 1} AXB B^{- 1} = I A^{- 1} B^{- 1} \]
\[ \Rightarrow \left( A^{- 1} A \right)X\left( B B^{- 1} \right) = A^{- 1} B^{- 1} \]
\[ \Rightarrow IXI = A^{- 1} B^{- 1} \]
\[ \Rightarrow X = A^{- 1} B^{- 1} \]
\[ \Rightarrow X = \begin{bmatrix} 3 & -1\\ - 5 & 2\end{bmatrix}\begin{bmatrix} 2 & - 3 \\ - 3 & 5 \end{bmatrix} = \begin{bmatrix} 6 + 3 & - 9 - 5\\ - 10 - 6 & 15 + 10 \end{bmatrix} = \begin{bmatrix} 9 & - 14\\ - 16 & 25 \end{bmatrix}\]