Question

Solve the matrix equation \[\begin{bmatrix}5 & 4 \\ 1 & 1\end{bmatrix}X = \begin{bmatrix}1 & - 2 \\ 1 & 3\end{bmatrix}\], where X is a 2 × 2 matrix.

Answer 1

Let: 
\[A = \begin{bmatrix} 5 & 4 \\ 1 & 1 \end{bmatrix} \]
\[B = \begin{bmatrix} 1 &- 2\\1 & 3 \end{bmatrix}\]
Now, 
\[\left| A \right| = \begin{bmatrix} 5 & 4 \\1 & 1 \end{bmatrix} = 5 - 4 = 1 \neq 0 \]
Hence, A is invertible . 
\[\text{ If }C_{ij}\text{ is cofactor of }a_{ij} \text{ in A, then }C_{11} = 1, C_{12} = - 1, C_{21} = - 4\text{ and }C_{22} = 5 . \]
\[ \Rightarrow adj A = \begin{bmatrix} 1 & - 1\\ - 4 & - 5 \end{bmatrix}^T = \begin{bmatrix} 1 & - 4 \\ - 1 & 5 \end{bmatrix}\]
\[ \therefore A^{- 1} = \frac{1}{\left| A \right|}adj A = \begin{bmatrix} 1 & - 4 \\ - 1 & 5 \end{bmatrix} \]
Now, the given equation becomes AX = B . 
\[ \Rightarrow A^{- 1} \left( AX \right) = A^{- 1} \times B\]
\[ \Rightarrow \left( A^{- 1} A \right)X = A^{- 1} \times B\]
\[ \Rightarrow X = A^{- 1} \times B \]
\[ \Rightarrow X = \begin{bmatrix} 1 & - 4 \\ - 1 & 5 \end{bmatrix} \times \begin{bmatrix} 1 & - 2\\1 & 3 \end{bmatrix} \]
\[ \Rightarrow X = \begin{bmatrix} 1 - 4 & - 2 - 12\\ - 1 + 5 & 2 + 15 \end{bmatrix}\]
\[ \Rightarrow X = \begin{bmatrix} - 3 & - 14\\ 4 & 17 \end{bmatrix}\]