Let \(\mathrm{A}=\left(\begin{array}{cc}1+\mathrm{i} & 1 \\ -\mathrm{i} & 0\end{array}\right)\) where \(\mathrm{i}=\sqrt{-1}\). Then, the number of elements in the set \(\left\{\mathrm{n} \in\{1,2, \ldots, 100\}: \mathrm{A}^{\mathrm{n}}=\mathrm{A}\right\}\) is

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Matrices

Let

A = (1 + i - i 10)

where

i = - 1

.

Then, the number of elements in the set

{n \in {1, 2, \dots, 100} : A^{n} = A}

Answer: 25

Compute successive powers of

A

A^{2} = [1 + i - i 10] [1 + i - i 10] = [i 1 - i 1 + i - i]

A^{4} = (A^{2})^{2} = [1001] = I

.

Since

A^{4} = I

, we have

A^{4 k + 1} = A

for any integer

k \geq 0

.

The values of

n

{1, 2, \dots, 100}

of the form

4 k + 1

are:

1, 5, 9, 13, \dots, 97

.

This is an arithmetic progression with first term 1, common difference 4, and last term 97.
Number of terms =

\frac{97 - 1}{4} + 1 = 25

.

Answer: 25.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer