If $A=\left[\begin{array}{lll}1 & 2 & i \ 1 & 1 & 1 \ 1 & 1 & 0\end{array} ight]$, then $[\operatorname{adj}(\operatorname{adj} A)]^{-1}=$

Question

If $A=\left[\begin{array}{lll}1 & 2 & i \ 1 & 1 & 1 \ 1 & 1 & 0\end{array}ight]$, then $[\operatorname{adj}(\operatorname{adj} A)]^{-1}=$

Aicharya · Accepted Answer

We have $|A|=\left[\begin{array}{lll}1 & 2 & 1 \ 1 & 1 & 1 \ 1 & 1 & 0\end{array}ight]$ $\quad=1(0-1)-2(0-1)+i(0)$ $\quad=-1+2=1$ $\begin{aligned} 	ext { Adj(Adj A) } &=|A|^{n-2} A \ 	ext { Now [adj (adj A)] }^{-1} &=\left[|A|^{p-2} Aight]^{-1} \ &=\left[(1)^{3-2} Aight]^{-1}=A^{-1} \end{aligned}$

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

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