Mathematics - Determinant Question with Solution | TestHub

$\begin{array}{cc}P=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right]& Q=\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]\end{array}$
Option$\left(\mathrm{C}\right)$
Now$R=PQ{P}^{-1}$
$\det \left(R\right)=\det \left(R\right)\det \left(Q\right)\det \left(P^{-1}\right)$
$\begin{array}{cc}\det \left(R\right)=\det \left(Q\right)& \left(\det \left(P^{-1}\right)=\frac{1}{\det \left(P\right)}\right)\end{array}$
$\det \left(R\right)=\det \left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]$
$\det \left(R\right)=48-4x^2$
now$\det \left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 5\end{array}\right]=40-4x^2$
$\det R=\det \left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 5\end{array}\right]+8$
Option$\left(\mathrm{A}\right)$
$R\left[\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$must have not trivial solution
So$\det \left(R\right)=0$
$48-4x^2=0\Rightarrow x=\pm 2\sqrt{3}$
Option$\left(\mathrm{D}\right)$
$R\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}6\\ 6a\\ 6b\end{array}\right]$
$PQ{P}^{-1}\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}6\\ 6a\\ 6b\end{array}\right]$…(i)
$P^{-1}=\frac{1}{6}\left[\begin{array}{ccc}6& -3& 0\\ 0& 3& -2\\ 0& 0& 2\end{array}\right]$
Putting$P, Q, P^{-1}$in equation (i)
$\frac{1}{6}\left[\begin{array}{ccc}12& 6& 4\\ 0& 24& 8\\ 0& 0& 36\end{array}\right]\left[\begin{array}{c}1\\ a\\ b\end{array}\right]=\left[\begin{array}{c}6\\ 6a\\ 6b\end{array}\right]$
$\Rightarrow 12+6a+4b=36$
$24a+8b=36a$
$a=2$and$b=3$
$a+b=5$
Option$\left(\mathrm{B}\right)$
$PQ=QP$
$PQ=QP\Rightarrow PQ{P}^{-1}=Q$
$R=Q$
Not possible for any value of$x.$