Fundamental Adjoint Identity

To solve this problem, we need to apply three key properties of determinants:

1. Scalar Multiplication Property: For any $$n \times n$$ matrix $$M$$ and a scalar $$k$$:$$\det(kM) = k^n \det(M)$$
2. Product Property: For any square matrices $$M$$ and $$N$$ of the same size:$$\det(MN) = \det(M) \cdot \det(N)$$
3. Adjoint Property: For any $$n \times n$$ matrix $$M$$:$$\det(\text{adj}(M)) = (\det(M))^{n-1}$$

Step 1: Set up the determinant expression

We are given the equation:

$$B = 2A \cdot \text{adj}(A)$$

Taking the determinant on both sides:

$$\det(B) = \det(2A \cdot \text{adj}(A))$$

Step 2: Extract the scalar factor

The matrix inside is a $$4 \times 4$$ matrix (since $$A$$ is $$4 \times 4$$, its adjoint is also $$4 \times 4$$, and their product is $$4 \times 4$$).

Using the property $$\det(kM) = k^n \det(M)$$ with scalar $$k = 2$$ and dimension $$n = 4$$:

$$\det(B) = 2^4 \cdot \det(A \cdot \text{adj}(A))$$ $$\det(B) = 16 \cdot \det(A \cdot \text{adj}(A))$$

Step 3: Split the matrix product

Using the product property $$\det(MN) = \det(M) \cdot \det(N)$$:

$$\det(B) = 16 \cdot \det(A) \cdot \det(\text{adj}(A))$$

Step 4: Substitute the given values

We know that $$\det(A) = 3$$. Using the adjoint identity $$\det(\text{adj}(A)) = (\det(A))^{n-1}$$ where $n = 4$:

$$\det(\text{adj}(A)) = 3^{4-1} = 3^3 = 27$$

Now, substitute $$\det(A) = 3$$ and $$\det(\text{adj}(A)) = 27$$ back into our equation:

$$\det(B) = 16 \cdot 3 \cdot 27$$ $$\det(B) = 48 \cdot 27$$ $$\det(B) = 1296$$

This perfectly matches Option B.

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

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