JEE Main 2023 · 29 January · morning · Single correct · +4 / −1

Let α and β be real numbers. Consider a 3 × 3 matrix A such that A²=3A+α I. If A⁴=21A+β I, then

  1. (a) α=1
  2. (b) α=4
  3. (c) β=8
  4. (d) β=-8
Answer and solution

Answer (d) β=-8

Solution

Maths sir · 4 steps

Square the given relation to reach A⁴, then match the coefficient of A to find α first.

  1. Square both sides. A⁴ = (A²)² = (3A + α I)². Since A and I commute, this expands like ordinary algebra: 9A² + 6α A + α² I.
  2. Replace A squared again. Put A² = 3A + α I back in: 9(3A + α I) + 6α A + α² I. Collect the A terms and the I terms.

    A⁴ = (27 + 6α)A + (α² + 9α)I

  3. Compare coefficients. Match this with A⁴ = 21A + β I (this works because A is not a multiple of I). Then 27 + 6α = 21 gives α = -1. So β = α² + 9α = 1 - 9 = -8.
  4. So. α = -1 and β = -8. Of the options, only (d) β = -8 is true.

The idea to keep. When a matrix satisfies a quadratic like A² = pA + qI, every higher power reduces to the form aA + bI. Keep substituting for A² and compare coefficients.

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

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Asked about 110 times in JEE Main since 2020. This question is from 2023.