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

Given the matrix equation $A^2 - 4A + 3I = 0$, we seek the inverse of $A + 3I$.

Let $X = A + 3I$, so that $A = X - 3I$. Substituting into the given equation:

$(X - 3I)^2 - 4(X - 3I) + 3I = 0$

Expanding, since $X$ and $I$ commute:

$X^2 - 6X + 9I - 4X + 12I + 3I = 0$

Combining terms:

$X^2 - 10X + 24I = 0$

Rewriting to isolate $24I$:

$24I = 10X - X^2$

Factoring the right side:

$24I = X(10I - X)$

Multiplying both sides on the right by $X^{-1}$:

$24X^{-1} = 10I - X$

Solving for $X^{-1}$:

$X^{-1} = \frac{1}{24}(10I - X)$

Substituting $X = A + 3I$:

$(A + 3I)^{-1} = \frac{1}{24}(7I - A)$

This matches option C.