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

If

A = [0 tan (\frac{θ}{2}) - tan (\frac{θ}{2}) 0]

and

(I_{2} + A) (I_{2} - A)^{- 1} = [a b - b a]

, then

13 (a^{2} + b^{2})

is equal to

Answer: 13

Solution

Let

t = tan (\frac{θ}{2})

. Then

A = [0 t - t 0]

.

I_{2} + A = [1 t - t 1]

and

I_{2} - A = [1 - t t 1]

.

(I_{2} - A)^{- 1} = \frac{1}{1 + t ^{2}} [1 t - t 1]

. Then

(I_{2} + A) (I_{2} - A)^{- 1} = \frac{1}{1 + t ^{2}} [1 t - t 1] [1 t - t 1] = \frac{1}{1 + t ^{2}} [1 - t^{2} 2 t - 2 t 1 - t^{2}]

. Using the identity

cos θ = \frac{1 - t ^{2}}{1 + t ^{2}}

and

sin θ = \frac{2 t}{1 + t ^{2}}

, we get

[cos θ sin θ - sin θ cos θ]

. So

a = cos θ

and

b = sin θ

. Then

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

, and

13 (a^{2} + b^{2}) = 13

.

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

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