Let A = begin{bmatrix} a_1 a_2 end{bmatrix} | vedclass

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(B) Given $A = XB$,we have $\begin{bmatrix} a_1 \ a_2 \end{bmatrix} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & -1 \ 1 & k \end{bmatrix} \begin{bmatrix

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$1$

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(B) Given $A = XB$,we have $\begin{bmatrix} a_1 \ a_2 \end{bmatrix} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & -1 \ 1 & k \end{bmatrix} \begin{bmatrix} b_1 \ b_2 \end{bmatrix}$.Multiplying,we get $\sqrt{3} a_1 = b_1 - b_2$ and $\sqrt{3} a_2 = b_1 + k b_2$.Squaring and adding these equations:$3(a_1^2 + a_2^2) = (b_1 - b_2)^2 + (b_1 + k b_2)^2$$3(a_1^2 + a_2^2) = (b_1^2 + b_2^2 - 2b_1b_2) + (b_1^2 + k^2b_2^2 + 2kb_1b_2)$$3(a_1^2 + a_2^2) = 2b_1^2 + (1 + k^2)b_2^2 + 2b_1b_2(k - 1)$.We are given $a_1^2 + a_2^2 = \frac{2}{3}(b_1^2 + b_2^2)$,so $3(a_1^2 + a_2^2) = 2b_1^2 + 2b_2^2$.Comparing the two expressions for $3(a_1^2 + a_2^2)$:$2b_1^2 + (1 + k^2)b_2^2 + 2b_1b_2(k - 1) = 2b_1^2 + 2b_2^2$.This simplifies to $(k^2 + 1)b_2^2 + 2b_1b_2(k - 1) = 2b_2^2$.$(k^2 - 1)b_2^2 + 2b_1b_2(k - 1) = 0$.$(k - 1)[(k + 1)b_2^2 + 2b_1b_2] = 0$.Since $(k^2 + 1)b_2^2 
eq -2b_1b_2$,the term in the bracket cannot be zero for arbitrary $b_1, b_2$ unless $k=1$. Thus,$k = 1$.

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