Solve this
Question:

If $A=\left[\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}\right]$ be such that $A^{-1}=k A$, then $k$ equals

(a) 19

(b) $1 / 19$

(c) $-19$

(d) $-1 / 19$

Solution:

(b) $1 / 19$

$\operatorname{adj} A=\left[\begin{array}{rr}-2 & -3 \\ -5 & 2\end{array}\right]$

$|A|=-19$

$\therefore A^{-1}=\frac{1}{|A|} \operatorname{adj} A$

$\Rightarrow A^{-1}=-\frac{1}{19}\left[\begin{array}{rr}-2 & -3 \\ -5 & 2\end{array}\right]$

Now,

$A^{-1}=k A$

$\Rightarrow-\frac{1}{19}\left[\begin{array}{rr}-2 & -3 \\ -5 & 2\end{array}\right]=k A$

$\Rightarrow \frac{1}{19}\left[\begin{array}{rr}2 & 3 \\ 5 & -2\end{array}\right]=k A$

$\Rightarrow \frac{1}{19} A=k A$

$\Rightarrow k=\frac{1}{19}$