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Let $\mathrm{M=\left[\begin{array}{cc} 0 & -\alpha \\ \alpha & 0 \end{array}\right]}$ , where $\alpha$ is a non-zero real number and $\mathrm{N=\sum_{k=1}^{49} M^{2 k}}$ . If $\mathrm{\left(I-M^{2}\right) N=-2 I}$ , then the positive integral value of $\alpha$ is _______.
Option: 1
1

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{M=\left[\begin{array}{cc} 0 & -\alpha \\ \alpha & 0 \end{array}\right]}$

$\mathrm{M^{2}=\left[\begin{array}{cc} 0 & -\alpha \\ \alpha & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -\alpha \\ \alpha & 0 \end{array}\right]}$

$\mathrm{=\left[\begin{array}{cc} -\alpha^{2} & 0 \\ 0 & -\alpha^{2} \end{array}\right] }\\$

$\mathrm{=-\alpha^{2} \cdot I}$

$\mathrm{\therefore M^{4} =M^{2} \cdot M^{2}=\alpha^{4} I } \\$

$\mathrm{M^{6} =M^{4} \cdot M^{2}=-\alpha^{6} I}$

......

......

$\mathrm{\therefore N =M^{2}+M^{4}+M^{6}+--+M^{98} }\\$

$\mathrm{=-\alpha^{2} I+\alpha^{4} I-\alpha^{6} I \ldots . . .49 \text { terms }} \\$

$\mathrm{=\left(-\alpha^{2}+\alpha^{4}-\alpha^{6}-\ldots\right) I}$

$\mathrm{=-\alpha^{2}\left(\frac{\left(-\alpha^{2}\right)^{49}-1}{-\alpha^{2}-1}\right) I}\\$

$\mathrm{Given \left(I-M^{2}\right) N=-2 I}\\$

$\mathrm{\Rightarrow\left(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} -\alpha^{2} & 0 \\ 0 & -\alpha^{2} \end{array}\right]\right)\left(\frac{\alpha^{2}\left(-\alpha^{9 8}-1\right)}{\left(\alpha^{2}+1\right)}\right) I=-2I }$

$\mathrm{\Rightarrow\left(1+\alpha^{2}\right) I \cdot \frac{\alpha^{2}\left(-\alpha^{98}-1\right)}{\left(\alpha^{2}+1\right)}=-2 I }. \\$

$\mathrm{\Rightarrow \quad \alpha^{2}\left(-\alpha^{98}-1\right)=-2}$

Only $\mathrm{\alpha =1}$ satisfies it

Hence answer is $\mathrm{1}$

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Kshitij

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