General Analysis

Given matrices are: $P = \left[ {\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \\ \end{matrix} } \right]$, $Q = \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \\ \end{matrix} } \right]$, and $R = PQP^{-1}$.

The relation $R = PQP^{-1}$ means that matrices $R$ and $Q$ are similar. Similar matrices have several important properties, including:

1. They have the same determinant: $\det(R) = \det(Q)$.
2. They have the same trace.
3. They have the same characteristic polynomial and hence the same eigenvalues.

Let's calculate the determinant of $Q$: \det(Q) = 2(4 \cdot 6 - 0 \cdot x) - x(0 \cdot 6 - 0 \cdot x) + x(0 \cdot x - 4 \cdot x) $$$$ \det(Q) = 2(24) - x(0) + x(-4x) = 48 - 4x^2 Therefore, $\det(R) = \det(Q) = 48 - 4x^2$.

Now, let's evaluate each option.

Option A: There exists a real number x such that PQ = QP

We need to check if the equation $PQ = QP$ has a solution for $x \in \mathbb{R}$.

Calculate $PQ$: $PQ = \left[ {\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \\ \end{matrix} } \right] \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 2+x & 2x+4 & x+6 \\ 2x & 8+2x & 12 \\ 3x & 3x & 18 \\ \end{matrix} } \right]$

Calculate $QP$: $QP = \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \\ \end{matrix} } \right] \left[ {\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 2 & 2+2x & 2+5x \\ 0 & 8 & 8 \\ x & 3x & 3x+18 \\ \end{matrix} } \right]$

For $PQ = QP$, all corresponding entries must be equal. Let's compare the entry at position (2,1): $(PQ)_{21} = 2x$ $(QP)_{21} = 0$ This implies $2x = 0$, so $x=0$.

Now let's check if for $x=0$, the matrices are equal. For $x=0$: PQ = \left[ {\begin{matrix} 2 & 4 & 6 \\ 0 & 8 & 12 \\ 0 & 0 & 18 \\ \end{matrix} } \right] $$$$ QP = \left[ {\begin{matrix} 2 & 2 & 2 \\ 0 & 8 & 8 \\ 0 & 0 & 18 \\ \end{matrix} } \right] Since $PQ \neq QP$ for $x=0$, there is no real number $x$ for which $PQ=QP$. Thus, option A is incorrect.

Option B: For $x = 0$, if $R \left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right] = 6\left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right]$, then a + b = 5

The given equation is an eigenvalue equation $R\mathbf{v} = \lambda\mathbf{v}$ with eigenvalue $\lambda=6$ and eigenvector $\mathbf{v} = \left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right]$.

For $x=0$, the matrix $Q$ becomes: $Q = \left[ {\begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} } \right]$ Since $R$ and $Q$ are similar, they have the same eigenvalues. The eigenvalues of the diagonal matrix $Q$ are its diagonal entries: 2, 4, and 6. So, 6 is indeed an eigenvalue of $R$.

Let $\mathbf{u}$ be the eigenvector of $Q$ corresponding to the eigenvalue 6. Then $Q\mathbf{u} = 6\mathbf{u}$. The corresponding eigenvector $\mathbf{v}$ of $R$ is given by $\mathbf{v} = P\mathbf{u}$.

Let's find $\mathbf{u} = \left[ {\begin{matrix} u_1 \\ u_2 \\ u_3 \\ \end{matrix} } \right]$: $\left[ {\begin{matrix} 2 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 6 \\ \end{matrix} } \right] \left[ {\begin{matrix} u_1 \\ u_2 \\ u_3 \\ \end{matrix} } \right] = 6 \left[ {\begin{matrix} u_1 \\ u_2 \\ u_3 \\ \end{matrix} } \right]$ This gives $2u_1 = 6u_1 \Rightarrow u_1=0$, and $4u_2 = 6u_2 \Rightarrow u_2=0$. $6u_3=6u_3$ is true for any $u_3$. So, the eigenvector is of the form $k\left[ {\begin{matrix} 0 \\ 0 \\ 1 \\ \end{matrix} } \right]$ for any non-zero constant $k$. We can choose $k=1$, so $\mathbf{u} = \left[ {\begin{matrix} 0 \\ 0 \\ 1 \\ \end{matrix} } \right]$.

Now, we find the eigenvector $\mathbf{v}$ of $R$: $\mathbf{v} = P\mathbf{u} = \left[ {\begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \\ \end{matrix} } \right] \left[ {\begin{matrix} 0 \\ 0 \\ 1 \\ \end{matrix} } \right] = \left[ {\begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} } \right]$ We are given that the eigenvector is $\left[ {\begin{matrix} 1 \\ a \\ b \\ \end{matrix} } \right]$. Comparing this with our result, we get $a=2$ and $b=3$. Therefore, $a+b = 2+3 = 5$. Thus, option B is correct.

Option C: For $x = 1$, there exists a unit vector $\alpha \widehat i + \beta \widehat j + \gamma \widehat k$ for which $R\left[ {\begin{matrix} \alpha \\ \beta \\ \gamma \\ \end{matrix} } \right] = \left[ {\begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} } \right]$

The equation $R\mathbf{v} = \mathbf{0}$ for a non-zero vector $\mathbf{v}$ implies that 0 is an eigenvalue of $R$. A unit vector is a non-zero vector. A matrix has an eigenvalue of 0 if and only if its determinant is 0.

We need to check if $\det(R) = 0$ for $x=1$. We know $\det(R) = 48 - 4x^2$. For $x=1$: $\det(R) = 48 - 4(1)^2 = 44$ Since $\det(R) = 44 \neq 0$, the matrix $R$ is invertible and 0 is not an eigenvalue. The only solution to $R\mathbf{v} = \mathbf{0}$ is the trivial solution $\mathbf{v} = \mathbf{0}$. A unit vector cannot be the zero vector. Thus, option C is incorrect.

Option D: $\det R = \det \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \\ \end{matrix} } \right] + 8$, for all x $\in$ R

Let's evaluate both sides of the equation. Left Hand Side (LHS): $\det(R) = \det(Q) = 48 - 4x^2$.

Right Hand Side (RHS): We need to compute the determinant of the given matrix and add 8. \det \left[ {\begin{matrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \\ \end{matrix} } \right] = 2(4 \cdot 5 - 0 \cdot x) - x(0 \cdot 5 - 0 \cdot x) + x(0 \cdot x - 4 \cdot x) $$$$ = 2(20) - x(0) + x(-4x) = 40 - 4x^2 So, the RHS is $(40 - 4x^2) + 8 = 48 - 4x^2$.

Since LHS = $48 - 4x^2$ and RHS = $48 - 4x^2$, the equation holds true for all $x \in \mathbb{R}$. Thus, option D is correct.

Final Conclusion: The correct options are B and D.