Let A=[1, -1; 2, α] and B=[β, 1; 1, 0], α, β ∈ R. Let α₁ be the value of α which satisfies (A+B)²=A²+[2, 2; 2, 2] and α₂ be the value of α which satisfies (A+B)²=B². Then |α₁-α₂| is equal to ___________.

Solution

Maths sir · 4 steps

Square A+B once, then match entries; the zero and off-diagonal entries pin down α and β.

1. Add and square. A + B = (1+β, 0; 3, α). The top-right entry is 0, so the square stays simple.

   (A+B)² = ((1+β)², 0; 3(1+β) + 3α, α²)

2. First condition. A² = (-1, -1-α; 2+2α, α²-2), so A² + (2, 2; 2, 2) = (1, 1-α; 4+2α, α²). The top-right entries give 1 - α = 0, so α = 1. The bottom-left entries then give 3 + 3β + 3 = 6, so β = 0, and (1+β)² = 1 checks out. So α₁ = 1.
3. Second condition. B² = (β²+1, β; β, 1). The top-right entries give β = 0. The bottom-left entries then give 3 + 3α = 0, so α = -1, and α² = 1 checks out. So α₂ = -1.
4. So. |α₁ - α₂| = |1 - (-1)| = 2.

The common slip: 0. Reading α from the α² = 1 entry alone and taking α = 1 in both cases gives 0. In the second case the bottom-left entry 3 + 3α = 0 forces α₂ = -1.

The idea to keep. For matrix equations with unknowns, compute both sides fully and equate entry by entry. Start with the simplest entries, then check the rest. Never expand (A+B)² as A² + 2AB + B².

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