Properties and Invertibility of Matrix Similarity Transformation

To determine which of the given statements are correct, let us analyze each option step by step.

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Option (A): Existence of $x \in \mathbb{R}$ such that $PQ = QP$

Let us compute the matrix products $PQ$ and $QP$:

$$PQ = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix} = \begin{bmatrix} 2 + x & 2x + 4 & x + 6 \\ 2x & 2x + 8 & 12 \\ 3x & 3x & 18 \end{bmatrix}$$ $$QP = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 2x + 2 & 5x + 2 \\ 0 & 8 & 8 \\ x & 3x & 3x + 18 \end{bmatrix}$$

Equating the corresponding entries of $PQ$ and $QP$:

• From entry $(1,1)$: $2 + x = 2 \implies x = 0$

Substituting $x = 0$ into both matrices:

$$PQ = \begin{bmatrix} 2 & 4 & 6 \\ 0 & 8 & 12 \\ 0 & 0 & 18 \end{bmatrix} \quad \text{and} \quad QP = \begin{bmatrix} 2 & 2 & 2 \\ 0 & 8 & 8 \\ 0 & 0 & 18 \end{bmatrix}$$

Clearly, $PQ \neq QP$ (e.g., $(PQ)_{12} = 4 \neq 2 = (QP)_{12}$).
Hence, there is no real value of $x$ for which $PQ = QP$.
Option (A) is incorrect.

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Option (B): Determinant of $R$

Since $R = PQP^{-1}$, we have:

$$\det R = \det(PQP^{-1}) = \det(P)\det(Q)\det(P^{-1}) = \det Q$$

Let us evaluate $\det Q$:

$$\det Q = \begin{vmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{vmatrix} = 4 \begin{vmatrix} 2 & x \\ x & 6 \end{vmatrix} = 4(12 - x^2) = 48 - 4x^2$$

Now, consider the matrix given on the right-hand side of Option (B):

$$\det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} = 4 \begin{vmatrix} 2 & x \\ x & 5 \end{vmatrix} = 4(10 - x^2) = 40 - 4x^2$$

Adding $8$ to this determinant:

$$(40 - 4x^2) + 8 = 48 - 4x^2 = \det R \quad \text{for all } x \in \mathbb{R}$$

Option (B) is correct.

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Option (C): Eigenvector relationship for $x = 0$

For $x = 0$, the matrix $Q$ becomes:

$$Q = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \end{bmatrix}$$

The given equation is:

$$R \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix}$$

Since $R = PQP^{-1}$, this can be rewritten as:

$$P Q P^{-1} \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = 6 \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} \implies Q \left( P^{-1} \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} \right) = 6 \left( P^{-1} \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} \right)$$

Let $\mathbf{w} = P^{-1} \begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}$. Then:

$$Q \mathbf{w} = 6 \mathbf{w} \implies \begin{bmatrix} 2w_1 \\ 4w_2 \\ 6w_3 \end{bmatrix} = \begin{bmatrix} 6w_1 \\ 6w_2 \\ 6w_3 \end{bmatrix} \implies w_1 = 0, \quad w_2 = 0, \quad w_3 = k \quad (k \neq 0)$$

Now, using $\begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = P \mathbf{w}$:

$$\begin{bmatrix} 1 \\ a \\ b \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ k \end{bmatrix} = \begin{bmatrix} k \\ 2k \\ 3k \end{bmatrix}$$

Equating the components:

$$k = 1 \implies a = 2, \quad b = 3$$

Thus, $a + b = 2 + 3 = 5$.
Option (C) is correct.

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Option (D): Non-trivial null space for $x = 1$

The equation $R \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ has a non-zero (and hence unit vector) solution if and only if $\det R = 0$.

For $x = 1$:

$$\det R = \det Q = 48 - 4(1)^2 = 44 \neq 0$$

Since $\det R \neq 0$, the matrix $R$ is non-singular and invertible, which implies the only solution is the trivial solution $\alpha = \beta = \gamma = 0$. Hence, no unit vector exists.
Option (D) is incorrect.

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Conclusion:

The correct options are B and C.