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Question
Choose the correct answer in the following questions
If A = `[(alpha, beta),(y, - a)]` is such that A2 = I, then
Options
1 + α2 + βy = 0
1 – α2 + βy = 0
1 – α2 – βy = 0
1 + α2 – βy = 0
MCQ
Solution
1 – α2 – βy = 0
Explanation:
Given, A = `[(alpha, beta),(y, - alpha)]`
∴ A2 = A, A = `[(alpha, beta),(y, -alpha)][(alpha, beta),(y, -alpha)]`
= `[(alpha^2 + betay, alphabeta - alphabeta),(alphay - alphay, betay + alpha^2)] = [(alpha^2 + betay, 0),(0, betay + alpha^2)]`
Now, A2 = I
⇒ `[(alpha^2 + y, 0),(0, betay + alpha^2)] = [(1, 0),(0, 1)]`
We have equalting the corresponding elements.
`alpha^2 + betay` = 1
⇒ `alpha^2 + betay - 1` = 0
⇒ `1 - alpha^2 - betay` = 0
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