Verified Solution • 2025–26 Session moderate

QJEEM22Ma79aa: JEE Main 2022 Mathematics - Class 12 JEEM22Ma79aa Solution: Let M = [ 0 − α α 0 ] , where is a non-zero real number an . If …

Let M=[0−αα0], where α\alpha is a non-zero real number an N=∑k=149M2kN = \sum\limits_{k = 1}^{49} {{M^{2k}}} . If (I−M2)N=−2I(I - {M^2})N = - 2I , then the positive integral value of α\alpha is ____________.

Step-by-Step Verified Solution

M=[0amp;−ααamp;0],M2=[−α2amp;00amp;−α2]=−α2M=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right], M^{2}=\left[\begin{array}{cc}-\alpha^{2} & 0 \\ 0 & -\alpha^{2}\end{array}\right]=-\alpha^{2} I

N=M2+M4+…+M98N=M^{2}+M^{4}+\ldots+M^{98}

=[−α2+α4−α6+…]I=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I

=−α2(1−(−α2)49)1+α2⋅1=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1

I−M2=(1+α2)II-M^{2}=\left(1+\alpha^{2}\right) I

(I−M2)N=−α2(α98+1)=−2\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2

∴α=1\therefore \alpha=1

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