Question

Let $A=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$ and $B=\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]$ be two $2\times 1$ matrices with real entries such that $A=XB,$ where $X=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1& -1\\ 1& k\end{array}\right],$ and $k\in R.$ If $a_1^2+a_2^2=\frac{2}{3}(b_1^2+b_2^2)$ and $(k^2+1)b_2^2\ne -2b_1{b}_2,$ then the value of $k$ is

Answer 1

$XB=A$

$\Rightarrow \frac{1}{\sqrt{3}}\left[\begin{array}{c}{b}_1-b_2\\ b_1+k{b}_2\end{array}\right]=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]$

$b_1-b_2=\sqrt{3}{a}_1$
$\Rightarrow 3a_1^2=b_1^2+b_2^2-2b_1{b}_2$

and $b_1+k{b}_2=\sqrt{3}{a}_2$
$\Rightarrow 3a_2^2=b_1^2+k^2{b}_2^2+2k{b}_1{b}_2$

Adding both equations, we get $3(a_1^2+a_2^2)=2b_1^2+(k^2+1)b_2^2+2b_1{b}_2(k-1)$

$\Rightarrow 2b_1^2+2b_2^2=2b_1^2+(k^2+1)b_2^2+2b_1{b}_2(k-1)$

$\Rightarrow (k^2-1)b_2^2+2b_1{b}_2(k-1)=0$

$\Rightarrow (k-1)\left(\right(k+1)b_2^2+2b_1{b}_2)=0$

$\Rightarrow k=1$