Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ A=\left[54 \\ 11\right] \\ B=\left[1-2 \\ 13\right] \\ \mathrm{Now}, \\ \left|\mathrm{A}\right|=\left|54 \\ 11\right|=5-4=1\ne 0 \\ \mathrm{Hence},A\mathrm{is}\mathrm{invertible}. \\ \mathrm{If}{C}_{ij}\mathrm{is}\mathrm{cofactor}\mathrm{of}{a}_{ij\mathit{}}\mathrm{in}A,\mathrm{then}{C}_{11}=1,C_{12}=-1,C_{21}=-4\mathrm{and}{C}_{22}=5. \\ \Rightarrow \mathrm{adj}A={\left[1-1 \\ -4-5\right]}^T=\left[1-4 \\ -15\right] \\ \therefore A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A=\left[1-4 \\ -15\right] \\ \mathrm{Now},\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{becomes}AX=B. \\ \Rightarrow A^{-1}\left(AX\right)=\mathit{}{A}^{-1}\times B \\ \Rightarrow \left(A^{-1}A\right)X\mathit{}=A^{-1}\times B \\ \Rightarrow X=\mathit{}{A}^{-1}\times B \\ \Rightarrow X=\left[1-4 \\ -15\right]\times \left[1-2 \\ 13\right] \\ \Rightarrow X=\left[1-4-2-12 \\ -1+52+15\right] \\ \Rightarrow X=\left[-3-14 \\ 417\right] \end{gathered}$$