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Question

Question: Let E (a) =\(\begin{bmatrix} \cos^{2}\alpha & \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha & \sin^{2...

Let E (a) =[cos2αcosαsinαcosαsinαsin2α]\begin{bmatrix} \cos^{2}\alpha & \cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha & \sin^{2}\alpha \end{bmatrix}. If a and b differs by an odd multiple of p/2, then E(a) E(b) is a –

A

Null matrix

B

Unit matrix

C

Diagonal matrix

D

Orthogonal matrix

Answer

Null matrix

Explanation

Solution

We have

E(a)(b)=[cos2αcosαsinαcosαsinαsin2α]\left[ \begin{array} { c c } \cos ^ { 2 } \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin ^ { 2 } \alpha \end{array} \right] [cos2βcosβsinβcosβsinβsin2β]\begin{bmatrix} \cos^{2}\beta & \cos\beta\sin\beta \\ \cos\beta\sin\beta & \sin^{2}\beta \end{bmatrix}

= [cosαcosβcos(αβ)cosαsinβcos(αβ)sinαcosβcos(αβ)sinαsinβcos(αβ)]\begin{bmatrix} \cos\alpha\cos\beta\cos(\alpha - \beta) & \cos\alpha\sin\beta\cos(\alpha - \beta) \\ \sin\alpha\cos\beta\cos(\alpha - \beta) & \sin\alpha\sin\beta\cos(\alpha - \beta) \end{bmatrix}

As a and b differ by an odd multiple of p/2, a –b =

(2n 1) p/2 for some integer n. Thus, cos [(2n +1) p/2] = 0

\ E(a) E(b) = 0