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Question
If A = `[(costheta, theta),(0, costheta)]`, B = `[(sintheta, 0),(0, sintheta)]` then show that A2 + B2 = I
Solution
L.H.S. = A2 + B2
A2 = `[(costheta, 0),(0, costheta)][(costheta, 0),(0, costheta)]`
= `[(cos^2theta, 0),(0, cos^2theta)]`
B2 = `[(sintheta, 0),(0, sintheta)] [(sintheta, 0),(0, sintheta)]`
= `[(sin^2theta, 0),(0, sin^2theta)]`
A2 + B2 = `[(cos^2theta, 0),(0, cos^2theta)] + [(sin^2theta, 0),(0, sin^2theta)]`
= `[(sin^2theta + cos^2theta, 0),(0, sin^2theta + cos^2theta)]`
= `[(1, 0),(0, 1)]`
= I
= R.H.S.
Hence proved.
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