Let \[A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right]\] be two \[2 \times 1\] matrices with real entries such that \[A = XB\], where \[X = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&k
\end{array}} \right]\] , and \[k \in R\]. If \[a_1^2 + a_2^2 = (\dfrac{2}{3})(b_1^2 + b_2^2)\] and \[({k^2} + 1)b_2^2 \ne - 2{b_1}{b_2}\], then the value of \[k\]is ________.

Hint : We are given two \[2 \times 1\] matrices \[A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right]\] where \[A = XB\], and \[X = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&k
\end{array}} \right]\]. Also, it is given that \[a_1^2 + a_2^2 = (\dfrac{2}{3})(b_1^2 + b_2^2)\] and \[({k^2} + 1)b_2^2 \ne - 2{b_1}{b_2}\]. We have to find the value of \[k\] . We will be first multiplying the matrices \[B = \left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right]\] and \[X = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&k
\end{array}} \right]\]. Then equating this with \[A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right]\]. Then we will find the values of \[{a_1},{a_2}\] in terms of \[{b_1},{b_2}\], we will square \[{a_1},{a_2}\] and add them. Then we will compare the resultant answer with \[(\dfrac{2}{3})(b_1^2 + b_2^2)\] to find the value of \[k\].

Formula used : Following formulae will be useful in solving the given question
\[
  \left[ {{a_1}} \right] = \left[ {\begin{array}{*{20}{c}}
  {{a_2}}&{{a_3}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right] \\
   \Rightarrow {a_1} = {a_2}{b_1} + {a_3}{b_2} \\
 \]
\[{(a \pm b)^2} = {a^2} \pm 2ab + {b^2}\]

Complete Step-by-step Solution:
We are given \[A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right]\] , \[B = \left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right]\], \[X = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&k
\end{array}} \right]\]
On putting this value in \[A = XB\], we get
\[
  A = XB \\
  \;\left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right] = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  1&k
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {{b_1}} \\
  {{b_2}}
\end{array}} \right] \\
  \;\left[ {\begin{array}{*{20}{c}}
  {{a_1}} \\
  {{a_2}}
\end{array}} \right] = \dfrac{1}{{\sqrt 3 }}\left[ {\begin{array}{*{20}{c}}
  {{b_1} - {b_2}} \\
  {{b_1} + k{b_2}}
\end{array}} \right] \\
 \]
From this we can write
\[
  \;{a_1} = \dfrac{1}{{\sqrt 3 }}({b_1} - {b_2}) \\
  {a_2} = \dfrac{1}{{\sqrt 3 }}({b_1} + k{b_2}) \\
 \]
On squaring these two values, we get
\[
  {\left( {\;{a_1}} \right)^2} = {\left[ {\dfrac{1}{{\sqrt 3 }}({b_1} - {b_2})} \right]^2} \\
   = \dfrac{1}{3}(b_1^2 + b_2^2 - 2{b_1}{b_2}) \\
  {\left( {{a_2}} \right)^2} = {\left[ {\dfrac{1}{{\sqrt 3 }}({b_1} + k{b_2})} \right]^2} \\
   = \dfrac{1}{3}(b_1^2 + {k^2}b_2^2 - 2k{b_1}{b_2}) \\
 \]
Now, on summing these above values, we get
\[
  {\left( {\;{a_1}} \right)^2} + {\left( {{a_2}} \right)^2} = \dfrac{1}{3}(b_1^2 + b_2^2 - 2{b_1}{b_2}) + \dfrac{1}{3}(b_1^2 + {k^2}b_2^2 - 2k{b_1}{b_2}) \\
   = \dfrac{1}{3}(b_1^2 + b_2^2 - 2{b_1}{b_2} + b_1^2 + {k^2}b_2^2 + 2k{b_1}{b_2}) \\
   = \dfrac{1}{3}(2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1)) \\
 \]
It is also given that \[a_1^2 + a_2^2 = (\dfrac{2}{3})(b_1^2 + b_2^2)\]. So, we can say that
\[
  (\dfrac{2}{3})(b_1^2 + b_2^2) = \dfrac{1}{3}(2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1)) \\
  \dfrac{2}{3}(b_1^2 + b_2^2) = \dfrac{1}{3}(2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1)) \\
 \]
On multiplying both sides with \[3\]and further simplifying, we get
\[
  3 \times \dfrac{2}{3}(b_1^2 + b_2^2) = 3 \times \dfrac{1}{3}(2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1)) \\
  2(b_1^2 + b_2^2) = 2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1) \\
  2b_1^2 + 2b_2^2 = 2b_1^2 + ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1) \\
  2b_2^2 = ({k^2} + 1)b_2^2 - 2{b_1}{b_2}(k - 1) \\
 \]
On comparing L.H.S and R.H.S, we get
\[
  2 = {k^2} + 1 \\
  {k^2} = 1 \\
  k = \pm 1 \\
 \]
And
\[
   - 2{b_1}{b_2}(k - 1) = 0 \\
  k - 1 = 0 \\
  k = 1 \\
 \]
Thus value of \[k\] is \[1\]

Note : Here, students generally make mistakes while doing matrix multiplication. They generally multiply the first row of the first matrix with the first row of the second matrix, which is wrong. They should multiply the first row of the first matrix with the first column of the second matrix to obtain the first element in matrix multiplication.