Let $A$ be a $4 imes 4$ matrix and $P$ be its adjoint matrix. If $|P|=\left|\frac{A}{2} ight|$, then $\left|A^{-1} ight|=$

Question

Let $A$ be a $4 	imes 4$ matrix and $P$ be its adjoint matrix. If $|P|=\left|\frac{A}{2}ight|$, then $\left|A^{-1}ight|=$

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { } \mathrm{P}=\operatorname{adj}(\mathrm{A}) \ & |\mathrm{P}|=|\operatorname{adj}(\mathrm{A})|=|\mathrm{A}|^{n-1}=|\mathrm{A}|^3 \ & |\mathrm{P}|=\left|\frac{\mathrm{A}}{2}ight|=\frac{1}{2^4}|\mathrm{~A}| \ & |\mathrm{A}|^3=\frac{1}{2^4}|\mathrm{~A}| \Rightarrow|\mathrm{A}|^2=\frac{1}{16} \Rightarrow|\mathrm{~A}|= \pm \frac{1}{4}\end{aligned}$ $\left|\mathrm{A}^{-1}ight|=\frac{1}{|\mathrm{~A}|}= \pm 4$

Let $A$ be a $4 \times 4$ matrix and $P$ be its adjoint matrix. If $|P|=\left|\frac{A}{2}\right|$, then…

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