Let \(\mathrm{M}=\left[\begin{array}{cc}0 & -\alpha \\ \alpha & 0\end{array}\right]\), where \(\alpha\) is a non-zero real number and \(N=\sum_{k=1}^{49} M^{2 k}\). If \(\left(I-M^{2}\right) N=-2 I\), then the positive integral value of \(\alpha\) is ______.

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Matrices

Let

M = [0 α - α 0]

, where

α

is a non-zero real number and

N = \sum_{k = 1}^{49} M^{2 k}

. If

(I - M^{2}) N = - 2 I

, then the positive integral value of

α

is ______.

Answer: 1

First compute

M^{2}

M^{2} = [0 α - α 0] [0 α - α 0] = [- α^{2} 0 0 - α^{2}] = - α^{2} I

Then:

N = M^{2} + M^{4} + \dots + M^{98} = - α^{2} I + α^{4} I - α^{6} I + \dots

This is a geometric series with first term

- α^{2}

and ratio

- α^{2}

N = - α^{2} I \cdot \frac{1 - ( - α ^{2} ) ^{49}}{1 + α ^{2}} = - α^{2} I \cdot \frac{1 + α ^{98}}{1 + α ^{2}}

Also,

I - M^{2} = (1 + α^{2}) I

.

Substituting into the given condition:

(1 + α^{2}) I \cdot (- α^{2} \frac{1 + α ^{98}}{1 + α ^{2}}) I = - 2 I

- α^{2} (1 + α^{98}) = - 2

α^{2} (1 + α^{98}) = 2

For

α = 1

1 (1 + 1) = 2

✓

Therefore,

α = 1

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer