Let \(\mathrm{S}=\left\{\mathrm{n} \in \mathrm{N} \left\lvert\,\left(\begin{array}{ll}0 & \mathrm{i} \\ 1 & 0\end{array}\right)^{\mathrm{n}}\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right) \forall \mathrm{a}\right., \mathrm{b}, \mathrm{c}, \mathrm{d} \in \mathrm{R}\right\}\), where \(\mathrm{i}=\sqrt{-1}\). Then the number of 2-digit numbers in the set S is

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Matrices

Let

S = {n \in N (01 i 0)^{n} (a c b d) \forall a, b, c, d \in R}

, where

i = - 1

. Then the number of 2-digit numbers in the set S is

Answer: 11

Writing

A = (01 i 0)

A^{2} = (i 0 0 i) = i I, A^{4} = (i I)^{2} = - I, A^{8} = I

The product

A^{n} M

returns

M

for every real matrix

M

exactly when

A^{n} = I

. For even

n = 2 m

A^{n} = (i I)^{m} = i^{m} I

, which equals

I

only when

m

is a multiple of

4

, that is when

n

is a multiple of

8

. For odd

n

A^{n} = i^{(n - 1) /2} A

has both diagonal entries zero and can never be

I

.
So

S

consists of the multiples of

8

, and its two-digit members are

16, 24, 32, \dots, 96

, which number

\frac{96 - 16}{8} + 1 = 11

The set

S

contains

11

two-digit numbers.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer