Question

Find the matrix X for which

$\left[\begin{array}{cc}3& 2\\ 7& 5\end{array}\right]X\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}2& -1\\ 0& 4\end{array}\right]$

Answer 1

$$\begin{gathered} \mathrm{Let}A=\left[32 \\ 75\right],B=\left[-11 \\ -21\right]\mathrm{and}C=\left[2-1 \\ 04\right] \\ \mathrm{Now}, \\ \left|A\right|=\left|32 \\ 75\right|=15-14=1 \\ \left|B\right|=\left|-11 \\ -21\right|=-1+2=1 \\ \mathrm{Since},\left|A\right|\ne 0\mathrm{and}\left|B\right|\ne 0 \\ \mathrm{Hence},A\&B\mathit{}\mathrm{are}\mathrm{invertible},\mathrm{so}{A}^{-1}\mathrm{and}{B}^{-1}\mathrm{exist}. \\ \mathrm{Cofactors}\mathrm{of}\mathrm{matrix}A\mathrm{are} \\ {\mathrm{A}}_{11}=5{\mathrm{A}}_{12}=-7{\mathrm{A}}_{21}=-2{\mathrm{A}}_{22}=3 \\ \mathrm{Now}, \\ \mathrm{adj}A={\left[5-7 \\ -23\right]}^T=\left[5-2 \\ -73\right] \\ {A}^{-1}=\frac{1}{\left|\mathit{A}\right|}\mathrm{adj}A=\left[5-2 \\ -73\right] \\ \mathrm{Cofactors}\mathrm{of}\mathrm{matrix}B\mathrm{are} \\ {B}_{11}=1B_{12}=2B_{21}=-1B_{22}=-1 \\ \mathrm{Now}, \\ \mathrm{adj}\mathit{}B={\left[12 \\ -1-1\right]}^T=\left[1-1 \\ 2-1\right] \\ {B}^{-1}=\frac{1}{\left|B\right|}\mathrm{adj}B=\left[1-1 \\ 2-1\right] \\ \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{becomes}AXB=C \\ \Rightarrow \left({\mathit{A}}^{\mathit{-}\mathit{1}}\mathit{A}\right)X\left(B{B}^{-1}\right)=A^{-1}C{B}^{-1} \\ \Rightarrow \left(I\right)X\left(I\right)=A^{-1}C{B}^{-1} \\ \Rightarrow X=\left[5-2 \\ -73\right]\left[2-1 \\ 04\right]\left[1-1 \\ 2-1\right] \\ \Rightarrow \mathrm{X}=\left[5-2 \\ -73\right]\left[2-2-2+1 \\ 0+80-4\right] \\ \Rightarrow \mathrm{X}=\left[5-2 \\ -73\right]\left[0-1 \\ 8-4\right] \\ \Rightarrow X=\left[-163 \\ 24-5\right] \end{gathered}$$