If A = begin{bmatrix} 1 & 0 & 1 0 & 1 & 1 0 | vedclass

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(B) Given $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix}$.The characteristic equation is given by $|A - \lambda I| = 0$.$\begi

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$2A - A^2$

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(B) Given $A = \begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix}$.The characteristic equation is given by $|A - \lambda I| = 0$.$\begin{vmatrix} 1-\lambda & 0 & 1 \ 0 & 1-\lambda & 1 \ 0 & 1 & -\lambda \end{vmatrix} = 0$$(1-\lambda) [-\lambda(1-\lambda) - 1] = 0$$(1-\lambda) [-\lambda + \lambda^2 - 1] = 0$$-\lambda + \lambda^2 - 1 + \lambda^2 - \lambda^3 + \lambda = 0$$\lambda^3 - 2\lambda^2 + 1 = 0$By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation:$A^3 - 2A^2 + I = 0$Multiplying both sides by $A^{-1}$:$A^3 A^{-1} - 2A^2 A^{-1} + I A^{-1} = 0$$A^2 - 2A + A^{-1} = 0$$A^{-1} = 2A - A^2$.

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$,then $A^{-1} =$

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