Let A = [ array ll a _ 1 a _ 2 array ] and B = [ array ll b _ 1 b _...

$A = XB$

\left[$$\begin{array}{ll}{a}_{1} \\ {a}_{2}\end{array}$$\right] = \dfrac{1}{\sqrt{3}}\left[$\begin{cases}1 & - 1 \\ 1 & k\end{cases}$\right]\left[$$\begin{array}{ll}{b}_{1} \\ {b}_{2}\end{array}$$\right]</p><p>\left[$\begin{array}{ll}\sqrt{3}{a}_{1} \\ \sqrt{3}{a}_{2}\end{array}$\right] = \left[$\begin{array}{ll}{b}_{1} - {b}_{2} \\ {b}_{1} + k{b}_{2}\end{array}$\right]

${b}_{1} - {b}_{2} = \sqrt{3}{a}_{1} \ldots \left(1\right)$

${b}_{1} + k{b}_{2} = \sqrt{3}{a}_{2} \ldots \left(2\right)$

Given, ${a}_{1}^{2} + {a}_{2}^{2} = \dfrac{2}{3}\left({b}_{1}^{2} + {b}_{2}^{2}\right)$

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

${\left({b}_{1} + {b}_{2}\right)}^{2} + {\left({b}_{1} + k{b}_{2}\right)}^{2} = 3\left({a}_{1}^{2} + {a}_{2}^{2}\right)$

${a}_{1}^{2} + {a}_{2}^{2} = \dfrac{2}{3}{b}_{1}^{2} + \dfrac{\left(1 + {k}^{2}\right)}{3}{b}_{2}^{2} + \dfrac{2}{3}{b}_{1}{b}_{2}\left(k - 1\right)$

Given, ${a}_{1}^{2} + {a}_{2}^{2} = \dfrac{2}{3}{b}_{1}^{2} + \dfrac{2}{3}{b}_{2}^{2}$

On comparing we get

$\dfrac{{k}^{2} + 1}{3} = \dfrac{2}{3} ⇒ {k}^{2} + 1 = 2$

$⇒ k = \pm 1 \ldots \left(3\right)$

$\dfrac{2}{3} ( k - 1 ) = 0 ⇒ k = 1 \ldots \left(4\right)$

From both we get $k = 1$