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 (integer)
1
Solution
$M=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right], M^{2}=\left[\begin{array}{cc}-\alpha^{2} & 0 \\ 0 & -\alpha^{2}\end{array}\right]=-\alpha^{2}$ I
<br/><br/>
$N=M^{2}+M^{4}+\ldots+M^{98}$
<br/><br/>
$=\left[-\alpha^{2}+\alpha^{4}-\alpha^{6}+\ldots\right] I$
<br/><br/>
$=\frac{-\alpha^{2}\left(1-\left(-\alpha^{2}\right)^{49}\right)}{1+\alpha^{2}} \cdot 1$
<br/><br/>
$I-M^{2}=\left(1+\alpha^{2}\right) I$
<br/><br/>
$\left(I-M^{2}\right) N=-\alpha^{2}\left(\alpha^{98}+1\right)=-2$
<br/><br/>
$\therefore \alpha=1$
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
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