Let $x \in R$ and let $P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}$, $Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix}$ and $R = PQP^{-1}$. Then which of the following options is/are correct?

$P = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix},\quad Q = \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 6 \end{bmatrix}$ $R = PQP^{-1}$ $\text{Check options (1), (2), (3), (4)}$ $R \sim Q \Rightarrow \det R = \det Q,\ \ \operatorname{tr}R = \operatorname{tr}Q$ $1.\ \det P,\ \det Q \Rightarrow \det R$ $2.\ x=1:\ \exists\ \text{unit } v,\ Rv=0\ ?$ $3.\ PQ=QP\ ?\qquad 4.\ x=0:\ Rv=6v$ $\det P = 1\cdot 2\cdot 3 = 6 \neq 0$ $R = PQP^{-1} \Rightarrow R \sim Q \Rightarrow \det R = \det Q$ $\det Q = 2(4\cdot 6) - x(0) + x(0\cdot x - 4x)$ $= 48 - 4x^2 = \det R$ $\det \begin{bmatrix} 2 & x & x \\ 0 & 4 & 0 \\ x & x & 5 \end{bmatrix} = 40 - 4x^2$ $40 - 4x^2 + 8 = 48 - 4x^2 = \det R \Rightarrow \text{TRUE}$ $x=1:\ \det R = \det Q = 48 - 4 = 44 \neq 0$ $R \text{ invertible} \Rightarrow Rv=0 \text{ only for } v=0 \Rightarrow \text{FALSE}$ $PQ = \begin{bmatrix} 2+x & 4+2x & 6+x \\ 2x & 8+2x & 12 \\ 3x & 3x & 18 \end{bmatrix}$ $QP = \begin{bmatrix} 2 & 2+2x & 2+5x \\ 0 & 8 & 8 \\ x & 3x & 3x+18 \end{bmatrix}$ $(PQ)_{12}=4+2x \neq 2+2x=(QP)_{12} \Rightarrow \text{FALSE}$ $x=0:\ Q = \operatorname{diag}(2,4,6);\quad Qe_3 = 6e_3$ $R(Pe_3) = P(Qe_3) = 6(Pe_3),\quad Pe_3 = (1,2,3)^T$ $\Rightarrow (a,b)=(2,3),\ a+b=5 \Rightarrow \text{TRUE}$ $\text{True: (1) and (4)}$ $\boxed{\text{Answer: (1), (4)}}$