Mathematics - Matrices Question with Solution | TestHub

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

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

Also given, $N=\sum _{\mathrm{k}=1}^{49}{M}^{2k}$

$\Rightarrow N=M^2+M^4+\dots \dots .+M^{98}=\left[-{\alpha }^2+{\alpha }^4-{\alpha }^6+\dots ..\right]I$

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

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

$\Rightarrow N\left(I-M^2\right)=-{\alpha }^2\left({\alpha }^{98}+1\right)$

Now comparing with $\left(I-M^2\right)N=-2I$

We get, $\alpha =1$.