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(D) Given $M = \begin{bmatrix} 0 & -\alpha \ \alpha & 0 \end{bmatrix}$.Calculating $M^2$: $M^2 = \begin{bmatrix} 0 & -\alpha \ \alpha & 0 \end{bmatr

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$1$

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(D) Given $M = \begin{bmatrix} 0 & -\alpha \ \alpha & 0 \end{bmatrix}$.Calculating $M^2$: $M^2 = \begin{bmatrix} 0 & -\alpha \ \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & -\alpha \ \alpha & 0 \end{bmatrix} = \begin{bmatrix} -\alpha^2 & 0 \ 0 & -\alpha^2 \end{bmatrix} = -\alpha^2 I$.Now,$N = \sum_{k=1}^{49} M^{2k} = M^2 + M^4 + \dots + M^{98}$.Since $M^2 = -\alpha^2 I$,we have $M^{2k} = (M^2)^k = (-\alpha^2)^k I$.Thus,$N = \sum_{k=1}^{49} (-\alpha^2)^k I = I \sum_{k=1}^{49} (-\alpha^2)^k$.This is a geometric progression with first term $a = -\alpha^2$ and common ratio $r = -\alpha^2$ for $49$ terms.$N = I \left( \frac{-\alpha^2(1 - (-\alpha^2)^{49})}{1 - (-\alpha^2)} ight) = I \left( \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} ight)$.Given $(I - M^2)N = -2I$.Since $M^2 = -\alpha^2 I$,$I - M^2 = I - (-\alpha^2 I) = (1 + \alpha^2)I$.Substituting this into the equation: $(1 + \alpha^2)I \cdot \left( \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} ight) I = -2I$.$(1 + \alpha^2) \cdot \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} = -2$.$-\alpha^2(1 + \alpha^{98}) = -2$.$\alpha^2(1 + \alpha^{98}) = 2$.If $\alpha = 1$,then $1^2(1 + 1^{98}) = 1(1 + 1) = 2$. This satisfies the equation.Thus,the positive integral value of $\alpha$ is $1$.

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

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