Question

Find the matrix X for which 

\[\begin{bmatrix}3 & 2 \\ 7 & 5\end{bmatrix} X \begin{bmatrix}- 1 & 1 \\ - 2 & 1\end{bmatrix} = \begin{bmatrix}2 & - 1 \\ 0 & 4\end{bmatrix}\]

 

Answer 1

\[Let A = \begin{bmatrix} 3 & 2\\7 & 5 \end{bmatrix}, B = \begin{bmatrix} - 1 & 1\\ - 2 & 1 \end{bmatrix} \text{ and }C = \begin{bmatrix} 2 & - 1\\0 & 4 \end{bmatrix}\]
Now,
\[\left| A \right| = \begin{vmatrix} 3 & 2\\7 & 5 \end{vmatrix} = 15 - 14 = 1 \]
\[\left| B \right| = \begin{vmatrix} - 1 & 1\\ - 2 & 1 \end{vmatrix} = - 1 + 2 = 1 \]
\[\text{ Since, }\left| A \right| \neq 0\text{ and }\left| B \right| \neq 0\]
\[\text{ Hence, A & B are invertible, so }A^{- 1}\text{ and }B^{- 1}\text{ exist }. \]
Cofactors of matrix A are
\[ A_{11} = 5 A_{12} = - 7 A_{21} = - 2 A_{22} = 3 \]
Now, 
\[adj A = \begin{bmatrix} 5 & - 7\\ - 2 & 3 \end{bmatrix}T = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\]
\[ A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix} \]
Cofactors of matrix B are
\[ B_{11} = 1 B_{12} = 2 B_{21} = - 1 B_{22} = - 1\]
Now, 
\[adj B = \begin{bmatrix} 1 & 2\\ - 1 & - 1 \end{bmatrix}^T = \begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix}\]
\[ B^{- 1} = \frac{1}{\left| B \right|}adj B = \begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix} \]
The given equation becomes AXB = C
\[ \Rightarrow \left( A^{- 1} A \right)X\left( B B^{- 1} \right) = A^{- 1} C B^{- 1} \]
\[ \Rightarrow \left( I \right)X\left( I \right) = A^{- 1} C B^{- 1} \]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 2 & - 1\\0 & 4 \end{bmatrix}\begin{bmatrix} 1 & - 1\\ 2 & - 1 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 2 - 2 & - 2 + 1\\ 0 + 8 & 0 - 4 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} 5 & - 2\\ - 7 & 3 \end{bmatrix}\begin{bmatrix} 0 & - 1\\ 8 & - 4 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} - 16 & 3\\ 24 & - 5 \end{bmatrix}\]