Question:
If $A$ is a $3 \times 3$ invertible matrix, then what will be the value of $k$ if $\operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{k}$.
Solution:
As we know that
$A^{-1}=\frac{\operatorname{Adj} A}{|\mathrm{~A}|}$
$\therefore\left|A^{-1}\right|=\frac{|\operatorname{Adj} A|}{|A|}$
$=\frac{|A|^{2-1}}{|A|} \quad\left[\because\right.$ If $A$ is a non singular matrix of order $n$, then $\left.|\operatorname{adj}(A)|=|\mathrm{A}|^{n-1}\right]$
$=\frac{|A|^{2}}{|A|}=|A|$
As we are given that $\left|A^{-1}\right|=\mid A^{k}$
$\therefore k=1$
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