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

We are told that the square of the matrix $$A$$ satisfies the quadratic relation

$$A^2-5A+7I = O,$$

where $$I$$ is the $$3\times3$$ identity matrix and $$O$$ is the $$3\times3$$ zero matrix. This single equation is the only information we need; every other result will follow just by ordinary algebra on matrices.

First, we rearrange this relation so that the constant term is on the left:

$$A^2-5A = -\,7I.$$

Multiplying both sides by $$-1$$ gives a slightly friendlier form:

$$-A^2+5A = 7I.$$

and, equivalently,

$$5A-A^2 = 7I.$$

Now we investigate Statement I, which concerns the inverse of $$A$$. Because the right‐hand side is $$7I$$, whose determinant is $$7^3\neq0,$$ we already know that $$A$$ is non-singular and an inverse exists. To display that inverse explicitly, we right-multiply both sides by $$A^{-1}$$ (allowed because $$A^{-1}$$ exists):

$$\bigl(5A-A^2\bigr)A^{-1}=7IA^{-1}.$$

We now perform the right multiplication term by term:

$$5AA^{-1}-A^2A^{-1}=7A^{-1}.$$

Using the basic facts $$AA^{-1}=I$$ and $$A^2A^{-1}=A(AA^{-1})=AI=A,$$ we simplify each term:

$$5I-A = 7A^{-1}.$$

Finally we isolate $$A^{-1}$$ by dividing by the scalar $$7$$:

$$A^{-1}= \frac{1}{7}\bigl(5I-A\bigr).$$

This matches Statement I exactly, so Statement I is true.

Next, we verify Statement II. We wish to simplify the cubic polynomial

$$A^3-2A^2-3A+I.$$

The idea is to eliminate every occurrence of $$A^2$$ or higher powers in favour of lower ones using the basic relation $$A^2=5A-7I.$$

We start by writing $$A^3$$ as $$A\cdot A^2$$ and then substituting for $$A^2$$:

$$A^3 = A\bigl(5A-7I\bigr)=5A^2-7A.$$

Substituting this expression for $$A^3$$ into the cubic polynomial, we get

$$A^3-2A^2-3A+I = \bigl(5A^2-7A\bigr)-2A^2-3A+I.$$

Now we gather the $$A^2$$ terms together and the $$A$$ terms together:

$$\bigl(5A^2-2A^2\bigr)+\bigl(-7A-3A\bigr)+I = 3A^2-10A+I.$$

We still have an $$A^2$$ term, so we substitute once more using $$A^2=5A-7I$$:

$$3A^2=3\bigl(5A-7I\bigr)=15A-21I.$$

Replacing $$3A^2$$ with this result in the expression $$3A^2-10A+I$$ yields

$$\bigl(15A-21I\bigr)-10A+I = 15A-10A-21I+I.$$

Combining like terms produces

$$5A-20I.$$

A common factor of $$5$$ can be taken out:

$$5A-20I = 5\bigl(A-4I\bigr).$$

This is exactly the reduction claimed in Statement II, so Statement II is also true.

We have demonstrated that both statements are correct.

Hence, the correct answer is Option A.