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