Question

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 = `A^2 + [(2, 2),(2, 2)]` and α₂ be the value of α which satisfies (A + B)² = B² . Then |α₁ – α₂| is equal to ______.

Options

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Answer 1

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 = `A^2 + [(2, 2),(2, 2)]` and α₂ be the value of α which satisfies (A + B)² = B² . Then |α₁ – α₂| is equal to 2.

Explanation:

Now A + B = `[(β + 1, 0),(3, α)]`

`\implies` (A + B)² = `[(β + 1, 0),(3, α)][(β + 1, 0),(3, α)]`

= `[((β + 1)^2, 0),(3(β + 1) + 3α, α^2)]`

Now A² = `[(1, -1),(2, α)][(1, -1),(2, α)] = [(-1, -1- α),(2 + 2α, α^2 - 2)]`

∴ `[(1, -α + 1),(2α + 4, α^2)] = [((β + 1)^2, 0),(3(α + β + 1), α^2)]`

`\implies` α = 1 = α₁

Now B² = `[(β, 1),(1, 0)][(β, 1),(1, 0)] = [(β^2 + 1, β),(β, 1)]`

= `[((β + 1)^2, 0),(3(β + 1) + 3α, α^2)]`

∴ β = 0, α = –1 = α₂

∴ |α₁ – α₂| = |1– (–1)| = 2