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Question
If A = `[(sin alpha, cos alpha), (-cos alpha, sin alpha)]` then verify that A'A = I
Solution
Given, A = `[(sin alpha, cos alpha),(-cos alpha, sin alpha)]`
So, A' = `[(sin alpha, -cos alpha),(cos alpha, sin alpha)]`
Now, A' A = `[(sin alpha, -cos alpha),(cos alpha, sin alpha)] xx [(sin alpha, cos alpha),(-cos alpha, sin alpha)]`
`= [(sin^2 alpha+ cos^2 alpha, sin alpha cos alpha - cos alpha sin alpha),(cos alpha sin alpha - sin alpha cos alpha, cos^2 alpha + sin^2 alpha)]`
`= [(1, 0),(0,1)] = I` ... [Because `sin^2 alpha + cos^2 alpha = 1`]
Hence, it is proved that, A'A = I
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