Question

If $A=\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]$ be such that $A^{-1}=kA$, then k equals
(a) 19
(b) 1/19
(c) − 19
(d) − 1/19

Answer 1

(b) 1/19

$$\begin{gathered} \mathrm{adj}A=\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right] \\ \left|A\right|=-19 \\ \therefore A^{-1}=\frac{1}{\left|A\right|}\mathrm{adj}A \\ \Rightarrow A^{-1}=-\frac{1}{19}\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right] \\ \mathrm{Now}, \\ {A}^{-1}=kA \\ \Rightarrow -\frac{1}{19}\left[\begin{array}{cc}-2& -3\\ -5& 2\end{array}\right]=kA \\ \Rightarrow \frac{1}{19}\left[\begin{array}{cc}2& 3\\ 5& -2\end{array}\right]=kA \\ \Rightarrow \frac{1}{19}A=kA \\ \Rightarrow k=\frac{1}{19} \end{gathered}$$