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Question

If A = [cosαsinαsinαcosα], find α satisfying 0<α<π2 when A+AT=2I2; where AT is transpose of A.

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Solution

We have A = [cosαsinαsinαcosα],

Now, AT=[cosαsinαsinαcosα]

But given that A+AT=2I2

[cosαsinαsinαcosα]+[cosαsinαsinαcosα]=2[1001]

[2cos α002cos α]=[2002]

By equality of matrices, we have

2cos α=2cos α=22

cosα=12cosα=cosπ4

α=π4 [ as 0<α<π2]

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