$A=XB$

$\left[\begin{array}{l}{a}_1\\ a_2\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1& -1\\ 1& k\end{array}\right]\left[\begin{array}{l}{b}_1\\ { b}_2\end{array}\right]$

$\left[\begin{array}{l}\sqrt{3}{a}_1\\ \sqrt{3}{a}_2\end{array}\right]=\left[\begin{array}{l}{b}_1-b_2\\ { b}_1+k{b}_2\end{array}\right]$

$b_1-b_2=\sqrt{3}{a}_1 \dots \left(1\right)$

$b_1+k{b}_2=\sqrt{3}{a}_2 \dots \left(2\right)$

Given, $a_1^2+a_2^2=\frac{2}{3}\left(b_1^2+b_2^2\right)$

$(1{)}^2+(2{)}^2$

${\left(b_1+b_2\right)}^2+{\left(b_1+k{b}_2\right)}^2=3\left(a_1^2+a_2^2\right)$

$a_1^2+a_2^2=\frac{2}{3}{b}_1^2+\frac{\left(1+k^2\right)}{3}{b}_2^2+\frac{2}{3}{b}_1{b}_2\left(k-1\right)$

Given, $a_1^2+a_2^2=\frac{2}{3}{b}_1^2+\frac{2}{3}{b}_2^2$

On comparing we get

$\frac{k^2+1}{3}=\frac{2}{3}\Rightarrow k^2+1=2$

$\Rightarrow k=\pm 1 \dots \left(3\right)$

$\frac{2}{3}(k-1)=0\Rightarrow k=1 \dots \left(4\right)$

From both we get $k=1$