If A = begin{bmatrix} 1 & 2 & i 1 & 1 & 1 1 | vedclass

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(C) We know that for a square matrix $A$ of order $n$,the property of the adjoint of an adjoint is given by $\operatorname{adj}(\operatorname{adj} A)

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$A^{-1}$

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(C) We know that for a square matrix $A$ of order $n$,the property of the adjoint of an adjoint is given by $\operatorname{adj}(\operatorname{adj} A) = |A|^{n-2} A$.Here,the order of matrix $A$ is $n = 3$.First,calculate the determinant $|A|$:$|A| = \begin{vmatrix} 1 & 2 & i \ 1 & 1 & 1 \ 1 & 1 & 0 \end{vmatrix} = 1(0 - 1) - 2(0 - 1) + i(1 - 1) = -1 + 2 + 0 = 1$.Now,substitute $n = 3$ and $|A| = 1$ into the formula:$\operatorname{adj}(\operatorname{adj} A) = (1)^{3-2} A = (1)^1 A = A$.Therefore,$[\operatorname{adj}(\operatorname{adj} A)]^{-1} = (A)^{-1} = A^{-1}$.

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

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