JEE Main 2023 · 8 April · evening · Single correct · +4 / −1

If A=[1, 5; λ, 10], A⁻¹=α A+β I and α+β=-2, then 4 α²+β²+λ² is equal to :

  1. (a) 12
  2. (b) 10
  3. (c) 19
  4. (d) 14
Answer and solution

Answer (d) 14

Solution

Maths sir · 5 steps

Use Cayley–Hamilton to write A⁻¹ in terms of A and I, then compare with α A + β I.

  1. Cayley–Hamilton. For any 2 × 2 matrix, A² - (tr A)A + (det A)I = O. Here tr A = 1 + 10 = 11 and det A = 10 - 5λ.
  2. Multiply by inverse. Multiply A² - 11A + (10 - 5λ)I = O by A⁻¹ to get A - 11I + (10 - 5λ)A⁻¹ = O.

    A⁻¹ = -1/(10 - 5λ) A + 11/(10 - 5λ) I

  3. Find lambda. So α = -1/(10 - 5λ) and β = 11/(10 - 5λ), which gives α + β = 10/(10 - 5λ) = -2. Then 10 - 5λ = -5, so λ = 3.
  4. Find alpha, beta. With 10 - 5λ = -5, you get α = 1/5 and β = -11/5. Then 4α² + β² = 4/25 + 121/25 = 125/25 = 5.
  5. So. 4α² + β² + λ² = 5 + 9 = 14, which is option (d).

The idea to keep. Whenever A⁻¹ is written as a combination of A and I, reach for Cayley–Hamilton: A² - (tr A)A + (det A)I = O, then multiply through by A⁻¹.

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.