Question

If $P=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$, $Q=\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]$ and $R=PQ{P}^{-1}$. Then which of the following options is/are correct?

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

Answer 1

Answer: (C), (D)

For $x=0$, if

$\begin{array}{l}R\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=6\left[\begin{array}{c}1\\ a\\ b\end{array}\right]\end{array}$, then $a+b=5$

Explanation for the correct options:

Finding the value of $a+b$:

Given that,

$P=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]$, $Q=\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]$

Calculating $P^{-1}$

Since it is diagonal matrix

so, $\left|P\right|=1\times 2\times 3=6$

$$\begin{gathered} Adj\left(P\right)={\left[\begin{array}{ccc}6& 6& 0\\ -3& 3& 0\\ 0& -2& 2\end{array}\right]}^T \\ =\left[\begin{array}{ccc}6& -3& 0\\ 6& 3& -2\\ 0& 0& 2\end{array}\right] \end{gathered}$$

$$\begin{gathered} P^{-1}=\frac{Adj\left(P\right)}{\left|P\right|} \\ =\frac{1}{6}\left[\begin{array}{ccc}6& -3& 0\\ 0& 3& -2\\ 0& 0& 2\end{array}\right] \\ \Rightarrow 6P^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6& -3& 0\\ 0& 3& -2\\ 0& 0& 2\end{array}\right] \end{gathered}$$

Now, $R=PQ{P}^{-1}$

$$\begin{gathered} \Rightarrow \left|R\right|=\left|P\right|\left|Q\right|\left|P^{-1}\right| \\ \Rightarrow \left|R\right|=\left(\left|P\right|\left|P^{-1}\right|\right)\left|Q\right| \\ \Rightarrow \left|R\right|=\left|Q\right| \\ =\left|\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right| \\ =\left|\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 5\end{array}\right|+\left|\begin{array}{ccc}2& x& 0\\ 0& 4& 0\\ x& x& 1\end{array}\right| \\ =\left|\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 5\end{array}\right|+8 \end{gathered}$$

Hence, $detR=\begin{array}{l}det\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 5\end{array}\right]+8\end{array}$

Now,

$\begin{array}{l}R\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=6\left[\begin{array}{c}1\\ a\\ b\end{array}\right]\end{array}$

$$\begin{gathered} \Rightarrow PQ{P}^{-1}\begin{array}{l}\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}6\\ 6a\\ 6b\end{array}\right]\end{array} \\ \Rightarrow \frac{1}{6}\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]\left[\begin{array}{ccc}2& 0& 0\\ 0& 4& 2\\ 0& 0& 6\end{array}\right]\left[\begin{array}{ccc}6& -3& 0\\ 0& 3& -2\\ 0& 0& 2\end{array}\right]\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}6\\ 6a\\ 6b\end{array}\right] \\ \Rightarrow \left[\begin{array}{ccc}12& 6& 4\\ 0& 24& 8\\ 0& 0& 36\end{array}\right]\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}36\\ 36a\\ 36b\end{array}\right] \\ \Rightarrow a=2\mathrm{and}b=3 \\ \Rightarrow a+b=5 \end{gathered}$$

Therefore, the correct answers are options (C) and (D).