This problem combines matrix multiplication with algebraic manipulation of quadratic expressions. The core idea is to express the elements of matrix $A$ in terms of the elements of matrix $B$ and the unknown $k$ using the given matrix equation. Then, these expressions are substituted into the quadratic relationship between the elements to form an equation in terms of $b_1, b_2,$ and $k$. Finally, careful simplification and the use of the given inequality condition allow us to determine the unique value of $k$.

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1. Expressing $a_1$ and $a_2$ in terms of $b_1, b_2,$ and $k$

We are given the matrix equation $A = XB$. Let's perform the matrix multiplication to find $a_1$ and $a_2$: $\begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$ To multiply these matrices, we take the dot product of each row of $X$ with the column of $B$: $a_1 = \frac{1}{\sqrt{3}}(1 \cdot b_1 + (-1) \cdot b_2) = \frac{1}{\sqrt{3}}(b_1 - b_2)$ $a_2 = \frac{1}{\sqrt{3}}(1 \cdot b_1 + k \cdot b_2) = \frac{1}{\sqrt{3}}(b_1 + kb_2)$ These expressions relate the elements of $A$ to $B$ and $k$.

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2. Substituting into the Given Quadratic Relation

Next, we use the given condition $a_1^2 + a_2^2 = \frac{2}{3}(b_1^2 + b_2^2)$. We substitute the expressions for $a_1$ and $a_2$ derived in the previous step: $\left(\frac{1}{\sqrt{3}}(b_1 - b_2)\right)^2 + \left(\frac{1}{\sqrt{3}}(b_1 + kb_2)\right)^2 = \frac{2}{3}(b_1^2 + b_2^2)$ Squaring the terms: $\frac{1}{3}(b_1 - b_2)^2 + \frac{1}{3}(b_1 + kb_2)^2 = \frac{2}{3}(b_1^2 + b_2^2)$ To simplify, we can multiply the entire equation by 3: $(b_1 - b_2)^2 + (b_1 + kb_2)^2 = 2(b_1^2 + b_2^2)$ This step eliminates the fractional coefficient and sets up for further algebraic expansion.

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3. Expanding and Simplifying the Equation

Now, we expand the squared terms and simplify the equation: $(b_1^2 - 2b_1b_2 + b_2^2) + (b_1^2 + 2kb_1b_2 + k^2b_2^2) = 2b_1^2 + 2b_2^2$ Combine like terms on the left side: $(b_1^2 + b_1^2) + (-2b_1b_2 + 2kb_1b_2) + (b_2^2 + k^2b_2^2) = 2b_1^2 + 2b_2^2$ $2b_1^2 + 2b_1b_2(k - 1) + b_2^2(1 + k^2) = 2b_1^2 + 2b_2^2$ Subtract $2b_1^2$ from both sides: $2b_1b_2(k - 1) + b_2^2(1 + k^2) = 2b_2^2$ Rearrange the terms to one side to form a quadratic-like equation in terms of $b_1, b_2,$ and $k$: $2b_1b_2(k - 1) + b_2^2(1 + k^2) - 2b_2^2 = 0$ $2b_1b_2(k - 1) + b_2^2(1 + k^2 - 2) = 0$ $2b_1b_2(k - 1) + b_2^2(k^2 - 1) = 0$ We can factor $k^2 - 1$ as $(k-1)(k+1)$: $2b_1b_2(k - 1) + b_2^2(k - 1)(k + 1) = 0$ Now, factor out the common term $(k-1)$: $(k - 1)[2b_1b_2 + b_2^2(k + 1)] = 0$ This equation gives us two possibilities: $k-1=0$ or $2b_1b_2 + b_2^2(k+1) = 0$.

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4. Analyzing the Given Inequality Condition

We are given the condition $(k^2 + 1)b_2^2 \ne -2b_1b_2$. This can be rewritten as: $(k^2 + 1)b_2^2 + 2b_1b_2 \ne 0$ This condition is crucial for determining the unique value of $k$. It implies that the matrix $X$ is well-behaved or that we avoid trivial solutions.

First, let's consider if $b_2 = 0$. If $b_2=0$, the condition becomes $(k^2+1)(0)^2 + 2b_1(0) \ne 0$, which simplifies to $0 \ne 0$. This is a contradiction. Therefore, $b_2$ cannot be zero.

Since $b_2 \ne 0$, we can divide the expression inside the bracket in our simplified equation by $b_2$: $(k - 1)b_2[2b_1 + b_2(k + 1)] = 0$ Since $b_2 \ne 0$, we must have: $(k - 1)[2b_1 + b_2(k + 1)] = 0$ This leads to two cases:

Case 1: $k - 1 = 0 \implies k=1$. Let's check if $k=1$ satisfies the inequality condition $(k^2 + 1)b_2^2 + 2b_1b_2 \ne 0$. Substitute $k=1$: $(1^2 + 1)b_2^2 + 2b_1b_2 \ne 0$ $2b_2^2 + 2b_1b_2 \ne 0$ $2b_2(b_2 + b_1) \ne 0$ Since we established $b_2 \ne 0$, this condition further implies $b_1 + b_2 \ne 0$. So, $k=1$ is a valid solution if $b_1 + b_2 \ne 0$.

Case 2: $2b_1 + b_2(k + 1) = 0$. This implies $2b_1 = -b_2(k+1)$. Let's substitute this into the inequality condition $(k^2 + 1)b_2^2 + 2b_1b_2 \ne 0$: $(k^2 + 1)b_2^2 + (-b_2(k+1))b_2 \ne 0$ $(k^2 + 1)b_2^2 - (k+1)b_2^2 \ne 0$ Factor out $b_2^2$: $b_2^2[(k^2 + 1) - (k+1)] \ne 0$ $b_2^2[k^2 + 1 - k - 1] \ne 0$ $b_2^2[k^2 - k] \ne 0$ $b_2^2 k(k-1) \ne 0$ Since $b_2 \ne 0$, this implies $k \ne 0$ and $k \ne 1$. So, if $2b_1 + b_2(k+1) = 0$, then for this to be a valid path to a solution for $k$, we must have $k \ne 0$ and $k \ne 1$.

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5. Determining the Unique Value of $k$

We have two potential scenarios for $k$:

1. $k=1$, which is valid if $b_1+b_2 \ne 0$.
2. $2b_1 + b_2(k+1) = 0$, which is valid if $k \ne 0$ and $k \ne 1$. (This means $k = -1 - \frac{2b_1}{b_2}$)

The problem asks for "the value of k", implying a unique numerical value, independent of $b_1$ and $b_2$ (as long as they satisfy the conditions). Let's investigate if $b_1+b_2=0$ is possible.

Assume for contradiction that $b_1+b_2=0$. This means $b_1 = -b_2$. If $b_1 = -b_2$, then $k=1$ is not a valid solution (from Case 1, as $b_1+b_2 \ne 0$ would be violated). So, if $b_1=-b_2$, we must fall into Case 2: $2b_1 + b_2(k+1) = 0$. Substitute $b_1 = -b_2$ into the equation for Case 2: $2(-b_2) + b_2(k+1) = 0$ $-2b_2 + b_2(k+1) = 0$ Factor out $b_2$: $b_2(-2 + k + 1) = 0$ $b_2(k - 1) = 0$ Since we established $b_2 \ne 0$, this implies $k-1=0$, so $k=1$. However, Case 2 explicitly requires $k \ne 1$. This creates a contradiction.

The assumption that $b_1+b_2=0$ leads to a contradiction. Therefore, $b_1+b_2$ cannot be zero. Since $b_1+b_2 \ne 0$, Case 1 ($k=1$) is always a valid solution. Furthermore, because $b_1+b_2 \ne 0$, the scenario where $2b_1 + b_2(k+1) = 0$ (which led to $k=1$ when $b_1+b_2=0$) is inconsistent with the condition $k \ne 1$ that must hold for Case 2. Specifically, if $2b_1 + b_2(k+1) = 0$ and $k=1$, it forces $b_1+b_2=0$, which is forbidden. Thus, the only value of $k$ that consistently satisfies all conditions is $k=1$.

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Tips for Success & Common Mistakes

• Systematic Approach: Break down the problem into smaller, manageable steps like matrix multiplication, substitution, and algebraic simplification.
• Don't Forget All Conditions: The inequality condition is often overlooked but is crucial for eliminating extraneous solutions and pinpointing the unique answer, especially in "find the value of..." problems.
• Factorization: Look for opportunities to factor expressions, as this often reveals the possible values of the unknown variable.
• Handle Special Cases (like $b_2=0$): Always consider if denominators or common factors could be zero, as this might lead to contradictions or additional constraints. Here, $b_2 \ne 0$ was a critical deduction.
• "The value of k": This phrase strongly suggests a unique numerical answer, implying that any solution dependent on $b_1, b_2$ (like $k = -1 - \frac{2b_1}{b_2}$) must be ruled out by the given conditions.

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Summary

By performing matrix multiplication to relate $a_1, a_2$ to $b_1, b_2,$ and $k$, and then substituting these into the given quadratic relationship, we obtained an equation $(k-1)[2b_1b_2 + b_2^2(k+1)] = 0$. Analyzing this equation in conjunction with the inequality condition $(k^2+1)b_2^2 \ne -2b_1b_2$ (which simplified to $2b_2(b_1+b_2) \ne 0$ for $k=1$ and $b_2^2k(k-1) \ne 0$ for other $k$ values), we systematically eliminated all possibilities except $k=1$. The key was demonstrating that the assumption $b_1+b_2=0$ leads to a contradiction, thus ensuring $b_1+b_2 \ne 0$, which validates $k=1$ as the unique solution.

The final answer is $\boxed{1}$.