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Question
If A = `[(sin α, cos α), (-cos α, sin α)]`, then verify that A'A = I
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Solution
Given, A = `[(sin α, cos α),(-cos α, sin α)]`
So, A' = `[(sin α, -cos α),(cos α, sin α)]`
Now, A' A = `[(sin α, -cos α),(cos α, sin α)] xx [(sin α, cos α),(-cos α, sin α)]`
= `[(sin^2 α + cos^2 α, sin α cos α - cos α sin α),(cos α sin α - sin α cos α, cos^2 α + sin^2 α)]`
= `[(1, 0),(0, 1)]` = I ...[∵ sin2 α + cos2 α = 1]
Hence, it is proved that, A'A = I
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