Matrix A is a non-singular matrix and (A-3I)( | vedclass

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(C) Given the equation $(A-3I)(A-5I) = 0$. Expanding the expression,we get $A^2 - 5A - 3A + 15I = 0$,which simplifies to $A^2 - 8A + 15I = 0$. Since $

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$I - \frac{1}{8} A$

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(C) Given the equation $(A-3I)(A-5I) = 0$. Expanding the expression,we get $A^2 - 5A - 3A + 15I = 0$,which simplifies to $A^2 - 8A + 15I = 0$. Since $A$ is a non-singular matrix,we can multiply the entire equation by $A^{-1}$ on both sides: $A^{-1}(A^2 - 8A + 15I) = A^{-1}(0)$. This gives $A - 8I + 15A^{-1} = 0$. Rearranging the terms to isolate $15A^{-1}$,we get $15A^{-1} = 8I - A$. Dividing both sides by $8$,we obtain $\frac{15}{8} A^{-1} = \frac{8I - A}{8} = I - \frac{1}{8} A$.

Matrix $A$ is a non-singular matrix and $(A-3I)(A-5I)=0$. Then,$\frac{15}{8} A^{-1} =$ . . . . . .

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