Let $$A = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$$ and $$B = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$$ be two $$2 \times 1$$ matrices with real entries such that $$A = XB$$, where $$X = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix}$$, and $$k \in R$$. If $$a_1^2 + a_2^2 = \frac{2}{3}(b_1^2 + b_2^2)$$ and $$(k^2 + 1)b_2^2 \neq -2b_1 b_2$$, then the value of $$k$$ is ________.

We have $$A = XB$$ where $$X = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix}$$, giving us $$a_1 = \frac{b_1 - b_2}{\sqrt{3}}$$ and $$a_2 = \frac{b_1 + kb_2}{\sqrt{3}}$$.

Computing $$a_1^2 + a_2^2$$: $$a_1^2 = \frac{(b_1-b_2)^2}{3} = \frac{b_1^2 - 2b_1b_2 + b_2^2}{3}$$ and $$a_2^2 = \frac{(b_1+kb_2)^2}{3} = \frac{b_1^2 + 2kb_1b_2 + k^2b_2^2}{3}$$.

Adding: $$a_1^2 + a_2^2 = \frac{2b_1^2 + (1+k^2)b_2^2 + 2(k-1)b_1b_2}{3}$$.

Setting equal to $$\frac{2}{3}(b_1^2 + b_2^2) = \frac{2b_1^2 + 2b_2^2}{3}$$, and clearing the denominator of 3: $$2b_1^2 + (1+k^2)b_2^2 + 2(k-1)b_1b_2 = 2b_1^2 + 2b_2^2$$.

Simplifying: $$(k^2 - 1)b_2^2 + 2(k-1)b_1b_2 = 0$$, which factors as $$(k-1)\bigl[(k+1)b_2^2 + 2b_1b_2\bigr] = 0$$.

So either $$k = 1$$ or $$(k+1)b_2^2 + 2b_1b_2 = 0$$. We are given the condition $$(k^2+1)b_2^2 \neq -2b_1b_2$$, i.e., $$(k^2+1)b_2^2 + 2b_1b_2 \neq 0$$. Suppose the second factor holds: $$(k+1)b_2^2 + 2b_1b_2 = 0$$, meaning $$2b_1b_2 = -(k+1)b_2^2$$. Substituting into $$(k^2+1)b_2^2 + 2b_1b_2$$: $$(k^2+1)b_2^2 - (k+1)b_2^2 = (k^2 - k)b_2^2 = k(k-1)b_2^2$$. For this to be non-zero (as given), we need $$k \neq 0$$, $$k \neq 1$$, and $$b_2 \neq 0$$. So in principle the second factor could hold without contradicting the given condition when $$k \neq 0, 1$$. However, the second factor $$(k+1)b_2^2 + 2b_1b_2 = 0$$ imposes a specific relationship between $$b_1$$ and $$b_2$$, namely $$b_1 = -\frac{(k+1)b_2}{2}$$. The problem states the condition must hold for the given $$b_1, b_2$$ without any such restriction, so the equation $$(k-1)\bigl[(k+1)b_2^2 + 2b_1b_2\bigr] = 0$$ must be satisfied universally. The only way to guarantee this for arbitrary $$b_1, b_2$$ satisfying the constraint is $$k = 1$$.

When $$k = 1$$, we can verify: $$X = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$$, so $$a_1^2+a_2^2 = \frac{(b_1-b_2)^2+(b_1+b_2)^2}{3} = \frac{2b_1^2+2b_2^2}{3} = \frac{2}{3}(b_1^2+b_2^2)$$, which matches the given condition for all $$b_1, b_2$$.

The answer is $$1$$.