If A=begin{bmatrix} 4 & 5 2 & 1 end{bmatrix} | vedclass

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(C) Given the characteristic equation $A^{2}-5A-6I=0$.Multiplying by $A^{-1}$ on both sides,we get:$A^{-1}(A^{2}-5A-6I) = A^{-1}(0)$$A - 5I - 6A^{-1}

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$\frac{1}{6}\begin{bmatrix} -1 & 5 \ 2 & -4 \end{bmatrix}$

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(C) Given the characteristic equation $A^{2}-5A-6I=0$.Multiplying by $A^{-1}$ on both sides,we get:$A^{-1}(A^{2}-5A-6I) = A^{-1}(0)$$A - 5I - 6A^{-1} = 0$$6A^{-1} = A - 5I$Substituting the matrices:$6A^{-1} = \begin{bmatrix} 4 & 5 \ 2 & 1 \end{bmatrix} - 5\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$6A^{-1} = \begin{bmatrix} 4-5 & 5-0 \ 2-0 & 1-5 \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 2 & -4 \end{bmatrix}$$A^{-1} = \frac{1}{6}\begin{bmatrix} -1 & 5 \ 2 & -4 \end{bmatrix}$Thus,the correct option is $C$.

If $A=\begin{bmatrix} 4 & 5 \\ 2 & 1 \end{bmatrix}$ and $A^{2}-5A-6I=0$,then $A^{-1}=$

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