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Question

If A=⎢ ⎢0tanα2tanα20⎥ ⎥ and I the identity matrix of order 2, show that I+A=(IA)[cosαsinαsinαcosα].

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Solution

A[0tanα/2tanα/20]
I+A=[1001]+[0tanα/2tanα/20]
=[1tanα/2tanα/21]
IA=[cosαsinαsinαcosα]
=[[1001][0tanα/2tanα/20]]×[cosαsinαsinαcosα]
=[1tanα/2tanα/21]×[cosαsinαsinαcosα]
=[cosα+tanα/2sinαsinα2+tanα/2cosαtanα/2cosα+sinαsinαtanα/2+cosα]
=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢1tanα2+tanα21+tan2α2×2tanα22tanα2+tanα21+tan2α2tan3α22tanα2tanα21+tan2α2tan3α21tanα2+tanα21+tan2α2×2tanα2⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
=[1tanα/2tanα/21]
=I+A

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