Let $$M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}$$, where $$\alpha$$ is a non-zero real number and $$N = \sum_{k=1}^{49} M^{2k}$$. If $$(I - M^2)N = -2I$$, then the positive integral value of $$\alpha$$ is ______.

We are given the matrices $$M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}$$ and $$N = \sum_{k=1}^{49} M^{2k}$$, together with the relation $$(I - M^2)N = -2I\,. $$

We first compute $$M^2$$ by multiplying the given matrix by itself: $$M^2 = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} = \begin{bmatrix} -\alpha^2 & 0 \\ 0 & -\alpha^2 \end{bmatrix} = -\alpha^2 I\,. $$ From this it follows that $$M^{2k} = (M^2)^k = (-\alpha^2)^k I = (-1)^k \alpha^{2k} I\,. $$

Consequently, the sum defining $$N$$ becomes $$N = \sum_{k=1}^{49}(-1)^k \alpha^{2k} I = \Bigl(\sum_{k=1}^{49}(-\alpha^2)^k\Bigr) I\,. $$ This is a geometric series with first term $$-\alpha^2$$ and common ratio $$-\alpha^2$$, so we use the formula $$\sum_{k=1}^{49}(-\alpha^2)^k = \frac{-\alpha^2\bigl(1-(-\alpha^2)^{49}\bigr)}{1-(-\alpha^2)} = \frac{-\alpha^2(1+\alpha^{98})}{1+\alpha^2}\,. $$

Next, since $$I - M^2 = I + \alpha^2 I = (1+\alpha^2)I$$, substituting into $$(I - M^2)N = -2I$$ gives $$(1+\alpha^2)\cdot \frac{-\alpha^2(1+\alpha^{98})}{1+\alpha^2} = -2\,, $$ which simplifies to $$-\alpha^2(1+\alpha^{98}) = -2\,. $$ Hence we obtain $$\alpha^2(1+\alpha^{98}) = 2 \quad\Longrightarrow\quad \alpha^2 + \alpha^{100} = 2\,. $$

Finally, testing positive integers shows that for $$\alpha = 1$$ we have $$1 + 1 = 2$$ ✔ whereas for $$\alpha = 2$$ we get $$4 + 2^{100} \gg 2$$ ✘. Therefore, the only positive integral solution is $$\alpha = \mathbf{1}\,. $$