Inverse Using Adjoint

 For any non-singular matrix $$M$$, we know that $$M^{-1} = \frac{1}{|M|}\operatorname{adj}(M)$$, which can be rewritten as:
$$\operatorname{adj}(M) = |M|M^{-1}$$
Substitute $$M = A^{-1}$$ and $$M = B^{-1}$$ into the property:
$$\operatorname{adj}(A^{-1}) = |A^{-1}|(A^{-1})^{-1} = \frac{1}{|A|}A$$
$$\operatorname{adj}(B^{-1}) = |B^{-1}|(B^{-1})^{-1} = \frac{1}{|B|}B$$

$$X = A \left( \frac{1}{|A|}A + \frac{1}{|B|}B \right)^{-1} B$$
 ($$(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$$):
$$X^{-1} = \left[ A \left( \frac{1}{|A|}A + \frac{1}{|B|}B \right)^{-1} B \right]^{-1}$$
$$X^{-1} = B^{-1} \left[ \left( \frac{1}{|A|}A + \frac{1}{|B|}B \right)^{-1} \right]^{-1} A^{-1}$$
$$X^{-1} = B^{-1} \left( \frac{1}{|A|}A + \frac{1}{|B|}B \right) A^{-1}$$
$$X^{-1} = \frac{1}{|A|} B^{-1} A A^{-1} + \frac{1}{|B|} B^{-1} B A^{-1}$$
Since $$A A^{-1} = I$$ and $$B^{-1} B = I$$:
$$X^{-1} = \frac{1}{|A|} B^{-1} + \frac{1}{|B|} A^{-1}$$
Replace $$B^{-1}$$ with $$\frac{\operatorname{adj}(B)}{|B|}$$ and $$A^{-1}$$ with $$\frac{\operatorname{adj}(A)}{|A|}$$:
$$X^{-1} = \frac{1}{|A|} \left( \frac{\operatorname{adj}(B)}{|B|} \right) + \frac{1}{|B|} \left( \frac{\operatorname{adj}(A)}{|A|} \right)$$
$$X^{-1} = \frac{\operatorname{adj}(B)}{|A||B|} + \frac{\operatorname{adj}(A)}{|A||B|}$$
$$X^{-1} = \frac{1}{|AB|} (\operatorname{adj}(B) + \operatorname{adj}(A))$$

Given

$$A^3-3A^2+4I=\mathcal O$$

Since

$$\det(A)=-4\ne0$$

the matrix

$$A$$

is invertible.

Multiplying the given equation by

$$A^{-1}$$

gives

$$A^2-3A+4A^{-1}=\mathcal O$$

Hence

$$A^2-3A=-4A^{-1}$$

Now use

$$\operatorname{adj}(A)=\det(A)\,A^{-1}$$

Since

$$\det(A)=-4$$

we get

$$\operatorname{adj}(A)=-4A^{-1}$$

Therefore

$$\operatorname{adj}(A)=A^2-3A$$

Hence statement (A) is correct.

Taking trace on both sides,

$$\operatorname{tr}(A^2-3A)=\operatorname{tr}(\operatorname{adj}(A))$$

To find this trace, first determine the eigenvalues.

Let

$$\lambda$$

be an eigenvalue of

$$A$$

Then

$$\lambda^3-3\lambda^2+4=0$$

Factorizing,

$$\lambda^3-3\lambda^2+4=(\lambda-2)^2(\lambda+1)$$

Thus the possible eigenvalues are

$$2,\;2,\;-1$$

Since

$$\det(A)=-4$$

and

$$2\cdot2\cdot(-1)=-4$$

the eigenvalues of

$$A$$

must be exactly

$$2,2,-1$$

Therefore the eigenvalues of

$$A^{-1}$$

are

$$\frac12,\frac12,-1$$

Hence

$$\operatorname{tr}(A^{-1})=\frac12+\frac12-1$$

$$=0$$

So statement (C) is correct.

Using

$$A^2-3A=-4A^{-1}$$

we obtain

$$\operatorname{tr}(A^2-3A)=-4\operatorname{tr}(A^{-1})$$

$$=-4(0)$$

$$=0$$

Hence statement (B) is correct.

Finally,

$$\det(\operatorname{adj}(A))=\det(A)^{n-1}$$

for an

$$n\times n$$

matrix.

Here

$$n=3$$

Therefore

$$\det(\operatorname{adj}(A))=\det(A)^2$$

$$=(-4)^2$$

$$=16$$

Thus statement (D) is false.

Therefore the correct statements are

$$(A),\ (B),\ (C)$$