Question

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

Then, the value of $k$ is

Answer 1

Step 1: Applying matrix operations

Given the matrices are $A=\left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]\&B=\left[\begin{array}{c}{b}_1\\ b_2\end{array}\right]$

And also given

$\begin{array}{cl}A=& XB\end{array}\&X=\frac{1}{\sqrt{3}}\left[\begin{array}{cc}1& -1\\ 1& k\end{array}\right],\forall k\in R.$

$$\begin{gathered} \Rightarrow \left[\begin{array}{c}{a}_1\\ a_2\end{array}\right]=\begin{array}{cl}\frac{1}{\sqrt{3}}& \left[\begin{array}{cc}{b}_1& -b_2\\ b_1& k{b}_2\end{array}\right]\end{array} \\ \begin{array}{lcl}\Rightarrow b_1–b_2& =& \sqrt{3}{a}_1\&\end{array}\begin{array}{lcl}{b}_1+k{b}_2& =& \sqrt{3}{a_2}^2\end{array} \\ \Rightarrow \begin{array}{lcl}3{a_1}^2& =& {b_1}^2+{b_2}^2–2b_1{b}_2...\left(i\right)\&\begin{array}{lcl}3{a_2}^2& =& {b_1}^2+k^2{b_2}^2+2k{b}_1{b}_2.....\left(ii\right)\end{array}\end{array} \end{gathered}$$

Adding both the equations, we have

$\Rightarrow \begin{array}{lcl}3{a_1}^2+3{a_2}^2& =& \left({b_1}^2+{b_2}^2–2b_1{b}_2\right)+\left({b_1}^2+k^2{b_2}^2+2k{b}_1{b}_2\right)\end{array}$

$\Rightarrow \begin{array}{lcl}3{\left({a_1}^2+a_2\right)}^2& =& {2b_1}^2+\left(1+k\right){b_2}^2–2(1+k)b_1{b}_2\end{array}$

Step 2: Applying the given information

$\because ({a_1}^2+{a_2}^2)=\left(\frac{2}{3}\right)({b_1}^2+{b_2}^2)\text{\&}(k^2+1){b_2}^2\ne -2b_1{b}_2$ is given,

$\begin{array}{rcl}\therefore (k^2–1){b_2}^2+2b_1{b}_2(k–1)& =& 0\\ & \Rightarrow & (k–1)\left[\right(k+1){b_2}^2+2b_1{b}_2]=0\\ (k-1)& =& 0\\ k& =& 1\end{array}$

Thus, value of $k$ is $1$.

Hence, option (A) is the correct answer.