If $A=\left[\begin{array}{cc}\cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta\end{array}\right],\left(\theta=\frac{\pi}{24}\right)$ and $A^{5}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, where $i=\sqrt{-1}$, then, which one of the following is not true?

(B) $\mathrm{a}^{2}-\mathrm{b}^{2}=\frac{1}{2}$

(C) $0 \leq$ a $^{2}+$ b $^{2} \leq 1$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Matrices

A = [cos θ i sin θ i sin θ cos θ], (θ = \frac{π}{24})

and

A^{5} = [a c b d]

, where

i = - 1

, then, which one of the following is not true?

Answer: (B)

By induction

A^{n} = [cos n θ i sin n θ i sin n θ cos n θ]

, so

A^{5}

has

a = d = cos 5 θ

and

b = c = i sin 5 θ

with

θ = \frac{π}{24}

, giving

5 θ = \frac{5 π}{24} = 7 5^{\circ} /2

. Then

a^{2} - c^{2} = cos^{2} 5 θ - (i sin 5 θ)^{2} = cos^{2} 5 θ + sin^{2} 5 θ = 1

, and likewise

a^{2} - b^{2} = 1

, not

\frac{1}{2}

; also

a^{2} + b^{2} = cos^{2} 5 θ - sin^{2} 5 θ = cos 10 θ = cos 7 5^{\circ}

, which lies between

0

and

1

, and

a^{2} - d^{2} = 0

. The relation that fails is

a^{2} - b^{2} = \frac{1}{2}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option