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Let $A=[a_{ij}]_{3\times3}$ such that $\det(A)=8$ and it satisfy the matrix equation $A^3-6A^2+7A-8I=0$ where $I$ is $3\times3$ identity matrix.

Find the following:

1) Value of $\det(adj(I-2A^{-1}))$

2) Value of $\det(A-2I)$

I multiplied the equation by $A^{-1}$ on both sides to get $A^2-6A+7I-8A^{-1}=0$ and then I replaced $A^{-1}$ by $\frac{adj(A)}{\det(A)}$ and modified the equation to $adj(A)=A^2-6A+7I$. Now how should I proceed?

Any help will be appreciated. THANKS

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Hint : factorize the given equation as :

$$(A-2I)^3=5A$$ Now take determinant both the sides.

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