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STD 12 - 3 and 4 . determinant and metrices — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 3 and 4 . determinant and metrices4 Marks
Question
Let $A=\left[\begin{array}{l}a_{1} \\ a_{2}\end{array}\right]$ and $B=\left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right]$ 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 $\left( k ^{2}+1\right) b _{2}^{2} \neq-2 b _{1} b _{2}$ then the value of $k$ is $....... .$
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Answer

$A=X B$$\left[\begin{array}{l} a _{1} \\ a _{2}\end{array}\right]$
$=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & k \end{array}\right]\left[\begin{array}{l} b _{1} \\ b _{2}\end{array}\right]$
$\left[\begin{array}{c}\sqrt{3} a_{1} \\ \sqrt{3} a_{2}\end{array}\right]=\left[\begin{array}{c}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{array}\right]$
$b_{1}-b_{2}=\sqrt{3} a_{1} ....(1)$
$b _{1}+ kb _{2}=\sqrt{3} a _{2} ....(2)$
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} $
$\Rightarrow k ^{2}+1=2$
$\Rightarrow k =\pm 1 ....(3)$
And $\frac{2}{3}( k -1)=0 $
$\Rightarrow k =1 ....(4)$
From both we get $k =1$

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