JEE MAIN - Mathematics (2022 - 29th June Evening Shift - No. 19)
Let $$M = \left[ {\matrix{
0 & { - \alpha } \cr
\alpha & 0 \cr
} } \right]$$, where $$\alpha$$ is a non-zero real number an $$N = \sum\limits_{k = 1}^{49} {{M^{2k}}} $$. If $$(I - {M^2})N = - 2I$$, then the positive integral value of $$\alpha$$ is ____________.
Answer
1
Explanation
$M=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right], M^{2}=\left[\begin{array}{cc}-\alpha^{2} & 0 \\ 0 & -\alpha^{2}\end{array}\right]=-\alpha^{2}$ I
$N=M^{2}+M^{4}+\ldots+M^{98}$
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
$=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1$
$I-M^{2}=\left(1+\alpha^{2}\right) I$
$\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2$
$\therefore \alpha=1$
$N=M^{2}+M^{4}+\ldots+M^{98}$
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
$=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1$
$I-M^{2}=\left(1+\alpha^{2}\right) I$
$\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2$
$\therefore \alpha=1$
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