Question

Let $A=\begin{bmatrix}a_{1} \\ a_{2}\end{bmatrix}]$ and $B=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}]$ be two $2 \times 1$
matrices with real entries such that $A = XB$,
where $X=\frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k\end{bmatrix}]$, and $k \in R .$If $a _{1}^{2}+ a _{2}^{2}=\frac{2}{3}\left( b _{1}^{2}+ b _{2}^{2}\right)$ and $\left( k ^{2}+1\right) b _{2}^{2} \neq-2 b _{1} b _{2}$,then the value of $k$ is_______.

Answer 1

$A = XB$
$\begin{bmatrix} a _{1} \\ a _{2}\end{bmatrix}=\frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k \end{bmatrix} \begin{bmatrix} b _{1} \\ b _{2}\end{bmatrix}$
$\begin{bmatrix}\sqrt{3} a_{1} \\ \sqrt{3} a_{2}\end{bmatrix}=\begin{bmatrix}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{bmatrix}$
$b_{1}-b_{2}=\sqrt{3} a_{1}\,\,\,\, .......(1)$
$b _{1}+ kb _{2}=\sqrt{3} a _{2}\,\,\,\,\, .......(2)$
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 $k =1$