JEE Main 2026 Apr 6 Shift 1, Mathematics Q21: Algebra Of Matrices

JEE Main2026Apr 6, Shift 1Mathematics

Let $A = - 11 - 1 101001$ satisfy

$A^{2} + α (adj (adj (A))) + β (adj (A) (adj (adj (A)))) = 2 - 22 - 20 - 1 00 - 1$

for some $α, β \in R$ . Then $(α - β)^{2}$ is equal to __________.

Your answer

Solution

Compute $∣ A ∣$ by expanding along the bottom row: $∣ A ∣ = 1 \cdot det (- 11 10) = - 1$ .

For a $3 \times 3$ matrix, $adj (adj (A)) = ∣ A ∣^{n - 2} A = ∣ A ∣ \cdot A = - A$ .

Also, $adj (A) \cdot A = ∣ A ∣ I = - I$ , so $adj (A) (adj (adj (A))) = adj (A) (- A) = - (- I) = I$ .

The equation becomes $A^{2} + α (- A) + β I = M$ , i.e. $A^{2} - α A + β I = M$ .

Compute $A^{2} = 2 - 11 - 110 001$ . Comparing entries with $M$ yields $α = 1$ and $β = - 1$ .

$∴ (α - β)^{2} = (1 - (- 1))^{2} = 4$ .

Concept behind this question

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