Mathematics - Matrices Question with Solution | TestHub

$$\text{ Given }A\left[\begin{array}{lll}1& 2& 3\\ 0& 2& 3\\ 0& 1& 1\end{array}\right]=\left[\begin{array}{lll}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$$ Applying ${\mathrm{C}}_1\leftrightarrow {\mathrm{C}}_3$ $$\mathrm{A}\left[\begin{array}{lll}3& 2& 1\\ 3& 2& 0\\ 1& 1& 0\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$$ Again Applying ${\mathrm{C}}_2\leftrightarrow {\mathrm{C}}_3$ $$\text{ A }\left[\begin{array}{lll}3& 1& 2\\ 3& 0& 2\\ 1& 0& 1\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$ pre-multiplying both sides by ${\mathrm{A}}^{-1}$ $$\begin{array}{rl}& {\mathrm{A}}^{-1}\mathrm{A}\left[\begin{array}{lll}3& 1& 2\\ 3& 0& 2\\ 1& 0& 1\end{array}\right]={\mathrm{A}}^{-1}\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & \mathrm{I}\left[\begin{array}{lll}3& 1& 2\\ 3& 0& 2\\ 1& 0& 1\end{array}\right]={\mathrm{A}}^{-1}\mathrm{I}={\mathrm{A}}^{-1}\\ & \left(\because {\mathrm{A}}^{-1}\mathrm{A}=\mathrm{I}\text{ and }\mathrm{I}=\text{ Identity matrix }\right)\end{array}$$ $$\text{ Hence, }{A}^{-1}=\left[\begin{array}{lll}3& 1& 2\\ 3& 0& 2\\ 1& 0& 1\end{array}\right]$$