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(C) Given the characteristic equation $A^2 - 4A + 3I = 0$. We need to find $(A + 3I)^{-1}$. Let $f(A) = A^2 - 4A + 3I = 0$. We can write $A^2 - 4A = -

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$\frac{7I}{24} - \frac{A}{24}$

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(C) Given the characteristic equation $A^2 - 4A + 3I = 0$. We need to find $(A + 3I)^{-1}$. Let $f(A) = A^2 - 4A + 3I = 0$. We can write $A^2 - 4A = -3I$. Since $A^2 - 4A + 3I = 0$,we have $A(A - 4I) = -3I$,which implies $A(4I - A) = 3I$. However,we need the inverse of $(A + 3I)$. Let $P(x) = x^2 - 4x + 3 = (x - 1)(x - 3) = 0$. By Cayley-Hamilton theorem,$A$ satisfies its characteristic equation. We want to express $(A + 3I)^{-1}$ in terms of $A$ and $I$. Let $(A + 3I)^{-1} = xA + yI$. Then $(A + 3I)(xA + yI) = I$. $xA^2 + yA + 3xA + 3yI = I$. $xA^2 + (y + 3x)A + 3yI = I$. Substitute $A^2 = 4A - 3I$: $x(4A - 3I) + (y + 3x)A + 3yI = I$. $(4x + y + 3x)A + (-3x + 3y)I = I$. $(7x + y)A + (3y - 3x)I = I$. Comparing coefficients: $7x + y = 0 \implies y = -7x$. $3y - 3x = 1$. Substitute $y = -7x$ into the second equation: $3(-7x) - 3x = 1 \implies -21x - 3x = 1 \implies -24x = 1 \implies x = -\frac{1}{24}$. Then $y = -7(-\frac{1}{24}) = \frac{7}{24}$. Thus,$(A + 3I)^{-1} = -\frac{1}{24}A + \frac{7}{24}I = \frac{7I}{24} - \frac{A}{24}$. Therefore,the correct option is $C$.

If $A$ is a matrix of order $2$ and $I$ is the identity matrix of order $2$ such that $A^2 - 4A + 3I = 0$,then $(A + 3I)^{-1} =$

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