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