Let $A = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$ and $B = \left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right]$ be two 2 $\times$ 1 matrices with real entries such that A = XB, where

$$X = {1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & { - 1} \cr 1 & k \cr } } \right]$$, and k$\in$R.

If $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$) and (k² + 1) b$_2^2$ $\ne$ $-$2b₁b₂, then the value of k is __________.

$XB = A$ <br><br>$\Rightarrow$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 &amp; { - 1} \cr 1 &amp; k \cr } } \right]\left[ {\matrix{ {{b_1}} \cr {{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$ <br><br>$\Rightarrow$ $${1 \over {\sqrt 3 }}\left[ {\matrix{ {{b_1} - {b_2}} \cr {{b_1} + k{b_2}} \cr } } \right] = \left[ {\matrix{ {{a_1}} \cr {{a_2}} \cr } } \right]$$<br><br>${b_1} - {b_2} = \sqrt 3 {a_1} \Rightarrow 3a_1^2 = b_1^2 + b_2^2 - 2{b_1}{b_2}$<br><br>$${b_1} + k{b_2} = \sqrt 3 {a_2} \Rightarrow 3a_2^2 = b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}$$<br><br>$$3\left( {a_1^2 + a_2^2} \right) = 2b_1^2 + \left( {{k^2} + 1} \right)b_2^2 + 2{b_1}{b_2}(k - 1)$$ <br><br>$\Rightarrow$ ${a_1^2 + a_2^2}$ = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$ <br><br>Given $a_1^2$ + $a_2^2$ = ${2 \over 3}$(b$_1^2$ + b$_2^2$) <br><br>$\therefore$ ${2 \over 3}$(b$_1^2$ + b$_2^2$) = $${2 \over 3}b_1^2 + {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$ <br><br>$\Rightarrow$ $${2 \over 3}b_2^2 = {{\left( {{k^2} + 1} \right)} \over 3}b_2^2 + {2 \over 3}{b_1}{b_2}\left( {k - 1} \right)$$ <br><br>Comparing both sides, We get <br><br>${{\left( {{k^2} + 1} \right)} \over 3} = {2 \over 3}$ <br><br>$\Rightarrow$ k² = 1 <br><br>$\Rightarrow$ k = $\pm$ 1 ......(1) <br><br>and ${2 \over 3}\left( {k - 1} \right) = 0$ $\Rightarrow$ k = 1 ....(2) <br><br>From (1) and (2), <br><br>k = 1