Question

Find the matrix X satisfying the matrix equation \[X\begin{bmatrix}5 & 3 \\ - 1 & - 2\end{bmatrix} = \begin{bmatrix}14 & 7 \\ 7 & 7\end{bmatrix}\]

Answer 1

Let: 
\[A = \begin{bmatrix} 5 & 3\\ - 1 & - 2 \end{bmatrix} \]
\[ \Rightarrow \left| A \right| = \begin{vmatrix} 5 & 3\\ - 1 & - 2 \end{vmatrix} = - 10 + 3 = - 7 \neq 0 \]
Hence, A is invertible .
\[\text{ If }C_{ij}\text{ is a cofactor of }a_{ij}\text{ in A, then }C_{11} = - 2, C_{12} = 1, C_{21} = - 3\text{ and }C_{22} = 5 . \]
Now, 
\[adj A = \begin{bmatrix} - 2 & 1\\ - 3 & 5 \end{bmatrix}^T = \begin{bmatrix} - 2 & - 3\\ 1 & 5 \end{bmatrix}\]
\[ \Rightarrow A^{- 1} = \frac{1}{\left| A \right|}adj A = \frac{- 1}{7}\begin{bmatrix} - 2 & - 3\\ 1 & 5 \end{bmatrix} \]
Let:
\[B = \begin{bmatrix} 14 & 7\\7 & 7 \end{bmatrix}\]
\[ \Rightarrow \left| B \right| = \begin{bmatrix} 14 & 7\\7 & 7 \end{bmatrix} = 98 - 49 = 49 \neq 0 \]
Hence, B is invertible .
The given matrix equation becomes XA = B . 
\[ \Rightarrow \left( XA \right) A^{- 1} = B A^{- 1} \]
\[ \Rightarrow X\left( A A^{- 1} \right) = \begin{bmatrix} 14 & 7\\7 & 7 \end{bmatrix} \times \frac{- 1}{7} \times \begin{bmatrix} - 2 & - 3\\ 1 & 5 \end{bmatrix}\]
\[ \Rightarrow X = \frac{- 1}{7}\begin{bmatrix} - 28 + 7 & - 42 + 35\\ - 14 + 7 & - 21 + 35 \end{bmatrix}\]
\[ \Rightarrow X = \frac{- 1}{7}\begin{bmatrix} - 21 & - 7\\ - 7 & 14 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} 3 & 1\\ 1 & - 2 \end{bmatrix}\]