Question

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

• A. There exists a real number x such that PQ = QP
• B. detR = det $\begin{bmatrix}2&x&x\\ 0&4&0\\ x&x&5\end{bmatrix} +8,$ for all $x \, \in \, ℝ$
• C. For x = 0,if R $\begin{bmatrix}1\\ a\\ b\end{bmatrix} = 6 \begin{bmatrix}1\\ a\\ b\end{bmatrix}, $ then a + b = 5
• D. For x = 1, there exists a unit vector $\alpha\hat{i} + \beta\hat{j} +\gamma\hat{k}$ for which $R\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix} = \begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

Answer 1

Answer: (B), (C)

det($R$) = det($PQP^{-1}$) = (det $P$)(det$Q$) $\left(\frac{1}{det\,P}\right)$
$=$ det $Q$
$=48-4x^{2}$
Option-1 :
for $x = 1$ det $\left(R\right)-44 \ne 0$
$\therefore $ for equation $R\begin{bmatrix}\alpha\\ \beta\\ \gamma\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$
We will have trivial solution
$\alpha=\beta=\gamma$
Option-2 :
$PQ = QP$
$PQP^{-1} = Q$
$R = Q$
No value of x.
Option-3 :
det $\begin{bmatrix}2&x&x\\ 0&4&0\\ x&x&5\end{bmatrix}+8$
$=\left(40-4x^{2}\right)+8=48-4x^{2}-det R\,\forall\,x\,\in\,R$
Option-4 :
$R=\begin{bmatrix}2&1&2/3\\ 0&4&4/3\\ 0&0&6\end{bmatrix}$
$\left(R-6I\right)\begin{bmatrix}1\\ a\\ b\end{bmatrix}=O$
$\Rightarrow -4+a+\frac{2b}{3}=0$
$-2a+\frac{4b}{3}=0$
$\Rightarrow a=2\,b=3$
$a+b=5$