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 $a_1^2+a_2^2=\frac{2}{3}\left(b_1^2+b_2^2\right)$ and $\left(k^2+1\right)b_2^2\ne -2b_1{b}_2$,then the value of $k$ is_______.

Answer 1

Answer: 1

$A=XB$
$\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1& -1\\ 1& k\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]$
$\left[\begin{array}{c}\sqrt{3}{a}_1\\ \sqrt{3}{a}_2\end{array}\right]=\left[\begin{array}{c}{b}_1-b_2\\ b_1+k{b}_2\end{array}\right]$
$b_1-b_2=\sqrt{3}{a}_1.......(1)$
$b_1+k{b}_2=\sqrt{3}{a}_2.......(2)$
Given, $a_1{}^2+a_2{}^2=\frac{2}{3}\left(b_1^2+b_2^2\right)$
$(1{)}^2+(2{)}^2$
${\left(b_1+b_2\right)}^2+{\left(b_1+k{b}_2\right)}^2=3\left(a_1^2+a_2^2\right)$
$a_1^2+a_2^2=\frac{2}{3}{b}_1^2+\frac{\left(1+k^2\right)}{3}{b}_2^2+\frac{2}{3}{b}_1{b}_2(k-1)$
Given, $a_1^2+a_2^2=\frac{2}{3}{b}_1^2+\frac{2}{3}{b}_2^2$
On comparing we get
$\frac{k^2+1}{3}=\frac{2}{3}\Rightarrow k^2+1=2$
$\Rightarrow k=\pm 1.....(3)$
$\&\frac{2}{3}(k-1)=0$
$\Rightarrow k=1.........(4)$
From both we get $k=1$