Matrix Product Property Involving Adjoint and Determinant

To determine which of the given relations are true, we start by analyzing the matrix equation $PQ = kI$, where $P, Q \in M_3(\mathbb{R})$ and $k \neq 0$.

Step 1: Relation between $\text{det}(P)$ and $k$

Given the matrix:

$$P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$$

First, we calculate the determinant of $P$: $\text{det}(P) = 3(0 - (-5\alpha)) - (-1)(0 - 3\alpha) + (-2)(-10 - 0)$ $\text{det}(P) = 15\alpha - 3\alpha + 20 = 12\alpha + 20$

Taking the determinant on both sides of $PQ = kI$: $\text{det}(PQ) = \text{det}(kI)$ $\text{det}(P) \cdot \text{det}(Q) = k^3$

We are given that $\text{det}(Q) = \frac{k^2}{2}$. Substituting this into the equation above: $\text{det}(P) \cdot \frac{k^2}{2} = k^3$

Since $k \neq 0$, we can divide both sides by $k^2$: $\text{det}(P) = 2k$

Equating the two expressions for $\text{det}(P)$: $12\alpha + 20 = 2k \implies k = 6\alpha + 10 \quad \text{--- (Equation 1)}$

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Step 2: Finding $q_{23}$ and solving for $\alpha$ and $k$

From $PQ = kI$, we can express $Q$ in terms of the adjugate of $P$: $Q = k P^{-1} = \frac{k}{\text{det}(P)} \text{adj}(P)$

Since $\text{det}(P) = 2k$: $Q = \frac{k}{2k} \text{adj}(P) = \frac{1}{2} \text{adj}(P)$

The element $q_{23}$ of matrix $Q$ corresponds to the entry in the $2^{\text{nd}}$ row and $3^{\text{rd}}$ column, which is given by: $q_{23} = \frac{1}{2} C_{32}(P)$ where $C_{32}(P)$ is the cofactor of the entry $p_{32}$ in matrix $P$.

The entry $p_{32} = -5$. Its cofactor $C_{32}(P)$ is:

$$C_{32}(P) = (-1)^{3+2} \text{det} \begin{bmatrix} 3 & -2 \\ 2 & \alpha \end{bmatrix} = -(3\alpha - (-4)) = -(3\alpha + 4)$$

Thus: $q_{23} = -\frac{3\alpha + 4}{2}$

We are given $q_{23} = -\frac{k}{8}$: $-\frac{3\alpha + 4}{2} = -\frac{k}{8} \implies k = 4(3\alpha + 4) = 12\alpha + 16 \quad \text{--- (Equation 2)}$

Equating Equation 1 and Equation 2: $6\alpha + 10 = 12\alpha + 16$ $6\alpha = -6 \implies \alpha = -1$

Substituting $\alpha = -1$ back into Equation 1: $k = 6(-1) + 10 = 4$

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Step 3: Checking the Options

1. Option A: $\alpha = 0, k = 8$ Since $\alpha = -1$ and $k = 4$, Option A is incorrect.

2. Option B: $4\alpha - k + 8 = 0$ Substituting $\alpha = -1$ and $k = 4$: $4(-1) - 4 + 8 = -4 - 4 + 8 = 0$ Thus, Option B is correct.

3. Option C: $\text{det}(P \text{ adj}(Q)) = 2^9$ First, find $\text{det}(P)$ and $\text{det}(Q)$: $\text{det}(P) = 2k = 2(4) = 8 = 2^3$ $\text{det}(Q) = \frac{k^2}{2} = \frac{4^2}{2} = 8 = 2^3$

   For a $3 \times 3$ matrix $Q$, $\text{det}(\text{adj}(Q)) = (\text{det}(Q))^{3-1} = (\text{det}(Q))^2$. $\text{det}(P \text{ adj}(Q)) = \text{det}(P) \cdot (\text{det}(Q))^2 = 2^3 \cdot (2^3)^2 = 2^3 \cdot 2^6 = 2^9$ Thus, Option C is correct.

4. Option D: $\text{det}(Q \text{ adj}(P)) = 2^{13}$ $\text{det}(Q \text{ adj}(P)) = \text{det}(Q) \cdot (\text{det}(P))^2 = 2^3 \cdot (2^3)^2 = 2^9 \neq 2^{13}$ Thus, Option D is incorrect.

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Conclusion

The correct options are B and C.