JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Matrices

Let

A = [a_{ij}]_{2 \times 2}

where

a_{ij} \neq = 0

for all

i, j

and

A^{2} = I

. Let a be the sum of all diagonal elements of A and

b = ∣ A ∣

, then

3 a^{2} + 4 b^{2}

is equal to

Answer: (D)

Solution

Let

A = (p r q s)

. From

A^{2} = I

p^{2} + q r = 1

s^{2} + q r = 1

pq + q s = 0

p r + r s = 0

. From the last two equations:

q (p + s) = 0

and

r (p + s) = 0

. Since

q, r \neq = 0

, we have

p + s = 0

. From the first two equations:

p^{2} + q r = s^{2} + q r

, so

p^{2} = s^{2}

, confirming

p = - s

. The trace

a = p + s = 0

. The determinant

b = p s - q r = - p^{2} - q r = - (p^{2} + q r) = - 1

. Thus

3 a^{2} + 4 b^{2} = 3 \cdot 0 + 4 \cdot 1 = 4

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option