Given matrix

$M=\begin{bmatrix}0&-\alpha\\ \alpha&0\end{bmatrix},\qquad \alpha\neq 0$

We need $N=\sum_{k=1}^{49} M^{2k}$ and it is given that $\left(I-M^2\right)N=-2I.$

Compute $M^2$

Find $M^{2k}$

Since $M^2=-\alpha^2 I$,

$M^{2k}=(M^2)^k=(-\alpha^2)^k I=(-1)^k\alpha^{2k}I$

Hence,

$N=\sum_{k=1}^{49} M^{2k}=\left(\sum_{k=1}^{49}(-\alpha^2)^k\right)I$

Let $r=-\alpha^2$ Then $N=(r+r^2+\cdots+r^{49})I$

Use the geometric series identity

$\left(I-M^2\right)N=(1-r)(r+r^2+\cdots+r^{49})I$

Since $r=-\alpha^2$ and $M^2=rI$, this is valid.

Now,

$(1-r)(r+r^2+\cdots+r^{49})=r-r^{50}$

Therefore,

$\left(I-M^2\right)N=(r-r^{50})I$

Given that this equals $-2I$, we get

$r-r^{50}=-2$

Substituting $r=-\alpha^2$,

$-\alpha^2-(-\alpha^2)^{50}=-2$

Since $50$ is even,

$-\alpha^2-\alpha^{100}=-2$

Rearranging,

$\alpha^{100}+\alpha^2=2$

Solve for positive integral $\alpha$

Let $x=\alpha^2$. Then $x$ is a positive integer and

$x^{50}+x=2$

Check positive integers:

• If $x=1$, then $1^{50}+1=2$ ✓
• If $x\ge 2$, then $x^{50}+x>2$ ✗

So,

$\alpha^2=1$

Since $\alpha$ is a positive integer,

$\alpha=1$

Final answer

$\boxed{1}$