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If $A=\left[\begin{array}{lll}1 & 2 & i \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right]$, then $[\operatorname{adj}(\operatorname{adj} A)]^{-1}=$
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The correct answer is:
$\mathrm{A}^{-1}$
We have $|A|=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 0\end{array}\right]$
$\quad=1(0-1)-2(0-1)+i(0)$
$\quad=-1+2=1$
$\begin{aligned} \text { Adj(Adj A) } &=|A|^{n-2} A \\ \text { Now [adj (adj A)] }^{-1} &=\left[|A|^{p-2} A\right]^{-1} \\ &=\left[(1)^{3-2} A\right]^{-1}=A^{-1} \end{aligned}$
$\quad=1(0-1)-2(0-1)+i(0)$
$\quad=-1+2=1$
$\begin{aligned} \text { Adj(Adj A) } &=|A|^{n-2} A \\ \text { Now [adj (adj A)] }^{-1} &=\left[|A|^{p-2} A\right]^{-1} \\ &=\left[(1)^{3-2} A\right]^{-1}=A^{-1} \end{aligned}$
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