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