If $A=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 2 & 3 \ 1 & 2 & 1\end{array} ight]$, then the value of determinant of $A^{-1}$ is

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If $A=\left[\begin{array}{lll}1 & 0 & 1 \ 0 & 2 & 3 \ 1 & 2 & 1\end{array}ight]$, then the value of determinant of $A^{-1}$ is

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { We have } A=\left[\begin{array}{lll} 1 & 0 & 1 \ 0 & 2 & 3 \ 1 & 2 & 1 \end{array}ight] \ & 	herefore|A|=(2-6)+(0-2)=-6 \end{aligned}$ We know that $\left|\mathrm{A}^{-1}ight|=\frac{1}{|\mathrm{~A}|}$ $	herefore\left|\mathrm{A}^{-1}ight|=\frac{1}{-6}$

If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$, then the value of…

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