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Question
if `A = [(0, -tan alpha/2), (tan alpha/2, 0)]` and I is the identity matrix of order 2, show that I + A = `(I -A)[(cos alpha, -sin alpha),(sin alpha, cos alpha)]`
Solution
`A = [(0, -tan alpha/2), (tan alpha/2, 0)], I = [(1,0),(0,1)]`
`I + A = [(1,0),(0,1)] + [(0, -tan alpha/2), (tan alpha/2, 0)]`
`= [(1, -tan alpha/2), (tan alpha/2, 1)]`
`(I - A) [(cos alpha, -sin alpha),(sin alpha, cos alpha)] = ([(1,0),(0,1)] - [(0, -tan alpha/2), (tan alpha/2, 0)]) [(cos alpha, -sin alpha),(sin alpha, cos alpha)]`
`= [(1, tan alpha/2), (-tan alpha/2, 1)] [(cos alpha, -sin alpha),(sin alpha, cos alpha)]`
`= [(1, tan alpha/2), (-tan alpha/2, 1)]` `[((1 - tan^2 alpha/2)/(1 + tan^2 alpha/2)(-2 tan alpha/2)/(1+ tan^2 alpha/2)),((-2 tan alpha/2)/(1+ tan^2 alpha/2)(1 - tan^2 alpha/2)/(1 + tan^2 alpha/2))]`
`= [((1 + tan^2 alpha/2)/(1 + tan^2 alpha/2)(-tan alpha/2 - tan^3 alpha/2)/(1+ tan^2 alpha/2)),((tan alpha/2 + tan^3 alpha/2)/(1+ tan^2 alpha/2)(1 + tan^2 alpha/2)/(1 + tan^2 alpha/2))]`
`= [(1, -tan alpha/2),(tan alpha/2, 1)]`
Hence, `I + A = (I - A) [(cos alpha, -sin alpha),(sin alpha, cos alpha)]`
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