Let A=[aᵢⱼ]_(2 × 2), where aᵢⱼ ≠ 0 for all i, j and A²=I. Let a be the sum of all diagonal elements of A and b=|A|. Then 3 a²+4 b² is equal to :

Solution

Maths sir · 4 steps

Square A entry by entry: the nonzero off-diagonal entries force the trace to be zero.

1. Square A. Write A = (p, q; r, s) with all four entries nonzero. Then A² = (p² + qr, q(p+s); r(p+s), s² + qr).
2. Match with I. The off-diagonal entries of I are 0, so q(p+s) = 0. Since q ≠ 0, we get p + s = 0. So the trace is a = 0, which means s = -p.
3. Find the determinant. The diagonal entries give p² + qr = 1. Then b = ps - qr = -p² - qr = -(p² + qr) = -1.

   3a² + 4b² = 3(0)² + 4(-1)² = 4

4. So. With a = 0 and b = -1, 3a² + 4b² = 4, which is option (a).

The idea to keep. For a 2×2 matrix with A² = I that is not ± I, the trace must be 0 and the determinant must be -1. Nonzero off-diagonal entries rule out ± I.

Answer from the NTA final answer key. Solution by PrepWiser.