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Step-by-Step Solution
Step 1: Expressing A in terms of X and B
Given:
A = \begin{bmatrix}a_1 \\ a_2\end{bmatrix} ,
B = \begin{bmatrix}b_1 \\ b_2\end{bmatrix} ,
and
X = \frac{1}{\sqrt{3}}\begin{bmatrix}1 & -1 \\ 1 & k\end{bmatrix} .
It is stated that A = X\,B .
So,
A = \frac{1}{\sqrt{3}}
\begin{bmatrix}
1 & -1 \\
1 & k
\end{bmatrix}
\begin{bmatrix}
b_1\\
b_2
\end{bmatrix}
=
\frac{1}{\sqrt{3}}
\begin{bmatrix}
b_1 - b_2\\
b_1 + k\,b_2
\end{bmatrix}.
That means
a_1 = \frac{1}{\sqrt{3}}(b_1 - b_2)\quad \text{and}\quad a_2 = \frac{1}{\sqrt{3}}(b_1 + k\,b_2).
Step 2: Relating the squares of the entries
We are given that
a_1^2 + a_2^2 = \frac{2}{3} (b_1^2 + b_2^2).
Next, we use the expressions of a_1 and a_2 in terms of b_1 and b_2 to form equations.
Step 3: Substituting expressions for a_1^2 and a_2^2
Since a_1 = \frac{b_1 - b_2}{\sqrt{3}} and a_2 = \frac{b_1 + k\,b_2}{\sqrt{3}} , we have:
a_1^2 = \frac{(b_1 - b_2)^2}{3}, \quad a_2^2 = \frac{(b_1 + k\,b_2)^2}{3}.
Hence,
a_1^2 + a_2^2
= \frac{(b_1 - b_2)^2 + (b_1 + k\,b_2)^2}{3}.
But we are given:
\frac{(b_1 - b_2)^2 + (b_1 + k\,b_2)^2}{3} = \frac{2}{3}(b_1^2 + b_2^2).
Step 4: Simplifying the equation
Multiply both sides by 3:
(b_1 - b_2)^2 + (b_1 + k\,b_2)^2 = 2 (b_1^2 + b_2^2).
Expand each square:
(b_1 - b_2)^2 = b_1^2 + b_2^2 - 2b_1b_2 ,
(b_1 + k\,b_2)^2 = b_1^2 + k^2b_2^2 + 2k\,b_1b_2 .
So,
\bigl(b_1^2 + b_2^2 - 2b_1b_2\bigr) + \bigl(b_1^2 + k^2 b_2^2 + 2k\,b_1b_2\bigr)
= 2(b_1^2 + b_2^2).
Combine like terms:
b_1^2 + b_1^2 + b_2^2 + k^2 b_2^2 - 2b_1b_2 + 2k\,b_1b_2
= 2b_1^2 + 2b_2^2.
This simplifies to:
(2b_1^2 + (1 + k^2)b_2^2) + 2b_1b_2(k - 1) = 2b_1^2 + 2b_2^2.
Rearrange to compare terms involving b_2^2 and b_1b_2 :
(1 + k^2)b_2^2 + 2b_1b_2(k - 1) = 2b_2^2.
This gives two separate conditions (by comparing coefficients for b_2^2 and b_1b_2 ):
(1 + k^2) = 2 \quad \Longrightarrow \quad k^2 = 1 \quad \Longrightarrow \quad k = \pm 1.
2(k - 1) = 0 \quad \Longrightarrow \quad k - 1 = 0 \quad \Longrightarrow \quad k = 1.
Step 5: Concluding the value of k
From both conditions, we infer k = 1 (since k= -1 does not satisfy the second condition requiring k - 1 = 0 ). Thus, the value of k is
\boxed{1}.
