# matrices with real entries such that A=XB,

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 $\mathrm{A}=\mathrm{XB}$,

where $X=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & \mathrm{k}\end{array}\right]$, and $\mathrm{k} \in \mathrm{R}$

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 $\mathrm{k}$ is___________.

Solution:

$\mathrm{A}=\mathrm{XB}$

$\left[\begin{array}{l}\mathrm{a}_{1} \\ \mathrm{a}_{2}\end{array}\right]=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1 & -1 \\ 1 & \mathrm{k}\end{array}\right]\left[\begin{array}{l}\mathrm{b}_{1} \\ \mathrm{~b}_{2}\end{array}\right]$

$\left[\begin{array}{l}\sqrt{3} a_{1} \\ \sqrt{3} a_{2}\end{array}\right]=\left[\begin{array}{l}b_{1}-b_{2} \\ b_{1}+k b_{2}\end{array}\right]$

$\mathrm{b}_{1}-\mathrm{b}_{2}=\sqrt{3} \mathrm{a}_{1} \ldots .(1)$

$\mathrm{b}_{1}+\mathrm{kb}_{2}=\sqrt{3} \mathrm{a}_{2} \ldots . .(2)$

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

$\frac{k^{2}+1}{3}=\frac{2}{3} \Rightarrow k^{2}+1=2$

$\Rightarrow k=\pm 1$

$\& \frac{2}{3}(\mathrm{k}-1)=0 \Rightarrow \mathrm{k}=1$

From both we get $\mathrm{k}=1$
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