Question

For any 3 x 3 matrix M, let |M| denote the determinant of M.
Let

B = \begin{bmatrix} 10 & 2 & 2 \\[6pt] 11 & 3 & 1 \\[6pt] 9 & 1 & 3 \end{bmatrix}, C = \begin{bmatrix} 0 & 1 & 0 \\[4pt] 1 & 0 & 0 \\[4pt] 1 & 0 & 0 \end{bmatrix}.$$ If $$D = \begin{bmatrix} 1 & 5 & 2 \\[4pt] 3 & 7 & 4 \\[4pt] 2 & m & 1 \end{bmatrix}$$ is a matrix such that $m \ge 2$, then which of the following statements is(are) correct?

• A. $|(AD)^{2024}| > |AD|^{2023}$
• B. $\bigl|AD + C^{2} B D^{-1}\bigr| = |AD| + \bigl|C B D^{-1}\bigr|$
• C. $B = C A C$ and $C^{3} = \begin{bmatrix}1 & 0 & 0 \\[4pt]0 & 1 & 0 \\[4pt]0 & 0 & 1\end{bmatrix}$
• D. $C^{2024}\,B^{2}\,C^{2024} = A\,C^{2}\,A$

Answer 1

None of the above

Answer 2

1. $\det A = 0$
Compute $\det A$:

$$\det A = 1\,(2\cdot11 - 10\cdot1)\;-\;3\,(2\cdot11 - 10\cdot3)\;+\;9\,(2\cdot1 - 2\cdot3) = 12 -3(-8) +9(-4) = 12 +24 -36 = 0.$$

Thus $\det(AD)=\det A\cdot\det D=0$.

• Statement A: $\bigl|\,(AD)^{2024}\bigr| = (\det AD)^{2024}=0$, and $|AD|^{2023}=0$. So $0>0$ is false.

2. Additivity of determinants fails in general.
There is no general formula $\det(X+Y)=\det X+\det Y$. Hence B is false.

3. Powers of $C$.
One finds $C^3 = C\neq I$. Also direct multiplication shows $CAC\neq B$. So C is false.

4. Even powers of $C$.
Since $C^2\neq C^0$ and $C^n$ alternates between $C$ and $C^2$, one checks $C^{2024}=C^2$. But $C^2\,B^2\,C^2\neq A\,C^2\,A$. So D is false.

Therefore none of the given options is correct.