Let A be a 3 times 3 matrix such that A^2 - 5 | vedclass

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(A) Given $A^2 - 5A + 7I = 0$.For Statement-$I$:$A^2 - 5A = -7I$Multiply by $A^{-1}$ on both sides:$A(A A^{-1}) - 5(A A^{-1}) = -7(I A^{-1})$$A(I) - 5

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Both the statements are true

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(A) Given $A^2 - 5A + 7I = 0$.For Statement-$I$:$A^2 - 5A = -7I$Multiply by $A^{-1}$ on both sides:$A(A A^{-1}) - 5(A A^{-1}) = -7(I A^{-1})$$A(I) - 5(I) = -7 A^{-1}$$A - 5I = -7 A^{-1}$$A^{-1} = \frac{1}{7}(5I - A)$.So,Statement-$I$ is true.For Statement-$II$:We have $A^2 = 5A - 7I$.Then $A^3 = A(5A - 7I) = 5A^2 - 7A = 5(5A - 7I) - 7A = 25A - 35I - 7A = 18A - 35I$.Now,substitute these into $A^3 - 2A^2 - 3A + I$:$(18A - 35I) - 2(5A - 7I) - 3A + I$$= 18A - 35I - 10A + 14I - 3A + I$$= (18 - 10 - 3)A + (-35 + 14 + 1)I$$= 5A - 20I$$= 5(A - 4I)$.So,Statement-$II$ is true.

Let $A$ be a $3 \times 3$ matrix such that $A^2 - 5A + 7I = 0$.
Statement-$I$: ${A^{-1}} = \frac{1}{7}(5I - A)$.
Statement-$II$: The polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5(A - 4I)$.

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Let $A$ be a $3 \times 3$ matrix such that $A^2 - 5A + 7I = 0$.Statement-$I$: ${A^{-1}} = \frac{1}{7}(5I - A)$.Statement-$II$: The polynomial $A^3 - 2A^2 - 3A + I$ can be reduced to $5(A - 4I)$.