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Question
If A = `[(α, β),(γ, -α)]` is such that A2 = I, then ______.
Options
1 + α2 + βγ = 0
1 – α2 + βγ = 0
1 – α2 – βγ = 0
1 + α2 – βγ = 0
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Solution
If A = `[(α, β),(γ, -α)]` is such that A2 = I, then 1 – α2 – βγ = 0.
Explanation:
A = `[(α, β), (γ, -α)]`
`A^2 = A * A [(α, β), (γ, -α)][(α, β), (γ, -α)]`
= `[(α^2 + βγ, αβ - αβ), (αγ - αγ, βγ + α^2)] = [(1, 0), (0, 1)]`
Now, A2 = I
⇒ `[(α^2 + βγ,0), (0, βγ + α^2)] = [(1, 0), (0, 1)]`
α2 + βγ = 1 or 1 – α2 – βγ = 0
Accordingly, option (1 – α2 – βγ = 0) is correct.
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