Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{QR}=\mathrm{RP}\\ & P=\left(\begin{array}{ll}2& 0\\ 0& 3\end{array}\right)Q=\left(\begin{array}{ll}x& y\\ z& 4\end{array}\right)\\ & \left(\begin{array}{ll}x& y\\ z& 4\end{array}\right)\left(\begin{array}{ll}{r}_1& r_2\\ r_3& r_4\end{array}\right)=\left(\begin{array}{ll}{r}_1& r_2\\ r_3& r_4\end{array}\right)\left(\begin{array}{ll}2& 0\\ 0& 3\end{array}\right)\\ & \begin{array}{l}{\mathrm{xr}}_1+{\mathrm{yr}}_3=2{\mathrm{r}}_1\\ {\mathrm{zr}}_1+4{\mathrm{r}}_3=2{\mathrm{r}}_3\end{array}\}\to \frac{\mathrm{x}-2}{-\mathrm{y}}=\frac{{\mathrm{r}}_3}{{\mathrm{r}}_1}=\frac{\mathrm{z}}{-2}\end{array}$ $2x-4=yz$
$\begin{array}{l}{\mathrm{xr}}_2+{\mathrm{yr}}_4=3{\mathrm{r}}_2\\ {\mathrm{zr}}_2+4{\mathrm{r}}_4=3{\mathrm{r}}_4\end{array}\}\to \frac{\mathrm{x}-3}{-\mathrm{y}}=\frac{{\mathrm{r}}_4}{{\mathrm{r}}_2}=-\mathrm{z}$ $\mathrm{x}-3=\mathrm{yz}$ $\Rightarrow 2x-4=x-3$ $\Rightarrow x=1\&yz=-2$ $\begin{array}{rl}& \Rightarrow \mathrm{Q}-\lambda \mathrm{I}=\left(\begin{array}{cc}\mathrm{x}-\lambda & \mathrm{y}\\ \mathrm{z}& 4-\lambda \end{array}\right)\\ & |\mathrm{Q}-\lambda \mathrm{I}|=(\lambda -\mathrm{x})(\lambda -4)-\mathrm{yz}\\ & ={\lambda }^2-(\mathrm{x}+4)\lambda +4\mathrm{x}-\mathrm{yz}\\ & |\mathrm{Q}-\lambda \mathrm{I}|={\lambda }^2-5\lambda +6\end{array}$ Now verify the option