Let M = [ cases 0 & - α α & 0 cases ] , where α is a non-zero real number and N = ∑ _ k = 1 ^ 49 M ^ 2k . If (I - M ^ 2 )N = - 2I , then the positive integral value of α is ______.

Correct answer: 0 Given, M = [ cases 0 & - α α & 0 cases ] Now, M ^ 2 = [ cases - α ^ 2 & 0 0 & - α ^ 2 cases ] = - α ^ 2 I Also given, N = ∑ _ k = 1 ^ 49 M ^ 2k ⇒ N = M ^ 2 + M ^ 4 + . + M ^ 98 = [- α ^ 2 + α ^ 4 - α ^ 6 + . .]I = - α ^ 2 (1 - (- α ^ 2 ) ^ 49 ) 1 + α ^ 2 . I ⇒ N(1 + α ^ 2 ) = - α ^ 2 ( α ^ 98 + 1) ⇒ N(I - M ^ 2 ) = - α ^ 2 ( α ^ 98 + 1) Now comparing with (I - M ^ 2 )N = - 2I We get, α = 1 .

JEE Main 202229 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Matrices

Let M = [ cases 0 & - α α & 0 cases ] , where α is a non-zero real...

Let

M = \left[

\begin{cases}0 & - \alpha \\ \alpha & 0\end{cases}

\right]

, where

\alpha

is a non-zero real number and

N = \mathop{∑}_{k = 1}^{49}{M}^{2k}

. If \left(I - {M}^{2}\right)N = - 2I

, then the positive integral value of

\alpha is ______.

Correct answer

Step-by-step solution

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

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

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

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

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

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

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

Now comparing with \left(I - {M}^{2}\right)N = - 2I $</p><p>We get,$ \alpha = 1$.

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