Question

If $A$ is a $3\times 3$ invertible matrix, then what will be the value of $k$ if $det\left(A^{-1}\right)=(detA{)}^k$

Answer 1

As we know that,

$A^{-1}=\frac{adj.A}{\left|\begin{array}{c}A\end{array}\right|}$

$\Rightarrow \left|\begin{array}{c}{A}^{-1}\end{array}\right|=\frac{\left|\begin{array}{c}Adj.A\end{array}\right|}{\left|\begin{array}{c}A\end{array}\right|}$

Also,

$\left|\begin{array}{c}adj.A\end{array}\right|={\left|\begin{array}{c}A\end{array}\right|}^{n-1}$

Where $n$ is the order of the singular matrix

Therefore,

For the given matrix,

$\left|\begin{array}{c}{A}^{-1}\end{array}\right|=\frac{{\left|\begin{array}{c}A\end{array}\right|}^{3-1}}{\left|\begin{array}{c}A\end{array}\right|}$

$\Rightarrow \left|\begin{array}{c}{A}^{-1}\end{array}\right|=\frac{{\left|\begin{array}{c}A\end{array}\right|}^2}{\left|\begin{array}{c}A\end{array}\right|}=\left|\begin{array}{c}A\end{array}\right|$

Hence the value of $k$ is $1$