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

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(A) First,we find the characteristic equation of matrix $A$ using $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & -2 \ 2 &

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$\begin{bmatrix} 1 & 4 & 0 \ 4 & -5 & -4 \ 0 & -2 & -7 \end{bmatrix}$

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(A) First,we find the characteristic equation of matrix $A$ using $|A - \lambda I| = 0$. $|A - \lambda I| = \begin{vmatrix} 1-\lambda & 2 & -2 \ 2 & -1-\lambda & 2 \ -1 & 1 & -2-\lambda \end{vmatrix} = 0$. Expanding the determinant: $(1-\lambda)[(-1-\lambda)(-2-\lambda) - 2] - 2[2(-2-\lambda) - (-2)] - 2[2 - (-1)(-1-\lambda)] = 0$. $(1-\lambda)[\lambda^2 + 3\lambda + 2 - 2] - 2[-4 - 2\lambda + 2] - 2[2 - 1 - \lambda] = 0$. $(1-\lambda)(\lambda^2 + 3\lambda) - 2(-2\lambda - 2) - 2(1 - \lambda) = 0$. $\lambda^2 + 3\lambda - \lambda^3 - 3\lambda^2 + 4\lambda + 4 - 2 + 2\lambda = 0$. $-\lambda^3 - 2\lambda^2 + 9\lambda + 2 = 0 \implies \lambda^3 + 2\lambda^2 - 9\lambda - 2 = 0$. By Cayley-Hamilton theorem,$A^3 + 2A^2 - 9A - 2I = 0$. Multiply by $A^{-1}$: $A^2 + 2A - 9I - 2A^{-1} = 0$. So,$2A^{-1} = A^2 + 2A - 9I$. Then $A + 2A^{-1} = A + A^2 + 2A - 9I = A^2 + 3A - 9I$. Calculating $A^2 = \begin{bmatrix} 1 & 2 & -2 \ 2 & -1 & 2 \ -1 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \ 2 & -1 & 2 \ -1 & 1 & -2 \end{bmatrix} = \begin{bmatrix} 7 & -2 & 6 \ -2 & 7 & -10 \ 3 & -5 & 8 \end{bmatrix}$. $A^2 + 3A - 9I = \begin{bmatrix} 7 & -2 & 6 \ -2 & 7 & -10 \ 3 & -5 & 8 \end{bmatrix} + \begin{bmatrix} 3 & 6 & -6 \ 6 & -3 & 6 \ -3 & 3 & -6 \end{bmatrix} - \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix} = \begin{bmatrix} 1 & 4 & 0 \ 4 & -5 & -4 \ 0 & -2 & -7 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 & -2 \\ 2 & -1 & 2 \\ -1 & 1 & -2 \end{bmatrix}$,then $A + 2A^{-1} =$

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