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(B) The characteristic equation of a square matrix $A$ is given by $|A-\lambda I|=0$.$\begin{vmatrix} 1-\lambda & 0 & -2 \ -2 & -1-\lambda & 2 \ 3 &

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$A^2-A-3I$

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(B) The characteristic equation of a square matrix $A$ is given by $|A-\lambda I|=0$.$\begin{vmatrix} 1-\lambda & 0 & -2 \ -2 & -1-\lambda & 2 \ 3 & 4 & 1-\lambda \end{vmatrix}=0$Expanding the determinant:$(1-\lambda)[(-1-\lambda)(1-\lambda)-8] - 2[-8 - 3(-1-\lambda)] = 0$$(1-\lambda)[-(1-\lambda^2)-8] - 2[-8 + 3 + 3\lambda] = 0$$(1-\lambda)(\lambda^2-9) - 2(3\lambda-5) = 0$$\lambda^2 - 9 - \lambda^3 + 9\lambda - 6\lambda + 10 = 0$$-\lambda^3 + \lambda^2 + 3\lambda + 1 = 0 \Rightarrow \lambda^3 - \lambda^2 - 3\lambda - 1 = 0$By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation:$A^3 - A^2 - 3A - I = 0$Multiplying by $A^{-1}$ on both sides:$A^{-1}(A^3 - A^2 - 3A - I) = 0$$A^2 - A - 3I - A^{-1} = 0$$A^{-1} = A^2 - A - 3I$

If $A=\begin{bmatrix} 1 & 0 & -2 \\ -2 & -1 & 2 \\ 3 & 4 & 1 \end{bmatrix}$,then $A^{-1}=$

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