Trace of a Sum

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

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