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Question
If A = `[(alpha, beta),(gamma, -alpha)]` is such that A2 = I then ______.
Options
1 + α² + βγ = 0
1 – α² + βγ = 0
1 – α² – βγ = 0
1 + α² – βγ = 0
Solution
If A = `[(alpha, beta),(gamma, -alpha)]` is such that A2 = I then 1 – α² – βγ = 0.
Explanation:
A = `[(alpha, beta), (ϒ, -alpha)]`
`"A"^2 = "A" * "A"[(alpha, beta), (ϒ, -alpha)][(alpha, beta), (ϒ, -alpha)]`
= `[(alpha^2 + betaϒ, alphabeta - alphabeta), (alphaϒ - alphaϒ, betaϒ + alpha^2)] = [(1, 0), (0, 1)]`
Now, A2 = I
⇒ `[(alpha^2 + betaϒ,0), (0, betaϒ + alpha^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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