Let $A=\left[\begin{array}{l}a_{1} \\ a_{2}\end{array}\right]$ and $B=\left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right]$ be two $2 \times 1$
matrices with real entries such that $\mathrm{A}=\mathrm{XB}$,
where $\mathrm{X}=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & \mathrm{k}\end{array}\right]$, and $\mathrm{k} \in \mathrm{R} .$ If
$\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}=\frac{2}{3}\left(\mathrm{~b}_{1}^{2}+\mathrm{b}_{2}^{2}\right)$ and $\left(\mathrm{k}^{2}+1\right) \mathrm{b}_{2}^{2} \neq-2 \mathrm{~b}_{1} \mathrm{~b}_{2}$
then the value of $\mathrm{k}$ is__________.
$\mathrm{A}=\mathrm{XB}$
$\left[\begin{array}{l}\mathrm{a}_{1} \\ \mathrm{a}_{2}\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & \mathrm{k}\end{array}\right]\left[\begin{array}{l}\mathrm{b}_{1} \\ \mathrm{~b}_{2}\end{array}\right]$
$\left[\begin{array}{c}\sqrt{3} \mathrm{a}_{1} \\ \sqrt{3} \mathrm{a}_{2}\end{array}\right]=\left[\begin{array}{l}\mathrm{b}_{1}-\mathrm{b}_{2} \\ \mathrm{~b}_{1}+\mathrm{k} \mathrm{b}_{2}\end{array}\right]$
$\mathrm{b}_{1}-\mathrm{b}_{2}=\sqrt{3} \mathrm{a}_{1}$
$\mathrm{~b}_{1}+\mathrm{kb}_{2}=\sqrt{3} \mathrm{a}_{2}$
Given, $a_{1}^{2}+a_{2}^{2}=\frac{2}{3}\left(b_{1}^{2}+b_{2}^{2}\right)$
$(1)^{2}+(2)^{2}$
$\left(b_{1}+b_{2}\right)^{2}+\left(b_{1}+k b_{2}\right)^{2}=3\left(a_{1}^{2}+a_{2}^{2}\right)$
$\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}=\frac{2}{3} \mathrm{~b}_{1}^{2}+\frac{\left(1+\mathrm{k}^{2}\right)}{3} \mathrm{~b}_{2}^{2}+\frac{2}{3} \mathrm{~b}_{1} \mathrm{~b}_{2}(\mathrm{k}-1)$
Given, $a_{1}^{2}+a_{2}^{2}=\frac{2}{3} b_{1}^{2}+\frac{2}{3} b_{2}^{2}$
On comparing we get
$\frac{\mathrm{k}^{2}+1}{3}=\frac{2}{3} \Rightarrow \mathrm{k}^{2}+1=2$
$\Rightarrow \mathrm{k}=\pm 1$ .............(3)
$\& \frac{2}{3}(\mathrm{k}-1)=0 \Rightarrow \mathrm{k}=1$...........(4)
From both we get $\mathrm{k}=1$