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Question: The product of the matrices A = \(\begin{bmatrix} \cos^{2}\theta & \cos\theta\sin\theta \\ \cos\thet...

The product of the matrices A = [cos⁡2θcos⁡θsin⁡θcos⁡θsin⁡θsin⁡2θ]\begin{bmatrix} \cos^{2}\theta & \cos\theta\sin\theta \\ \cos\theta\sin\theta & \sin^{2}\theta \end{bmatrix} and

B =[cos⁡2φcos⁡2φsin⁡φcos⁡φsin⁡φsin⁡2φ]\begin{bmatrix} \cos^{2}\varphi & \cos 2\varphi\sin\varphi \\ \cos\varphi\sin\varphi & \sin^{2}\varphi \end{bmatrix} is a null matrix if q – f =

A

(2n + 1)π2\frac{\pi}{2}

B

np

C

2np

D

nπ2\frac{n\pi}{2}

Answer

(2n + 1)π2\frac{\pi}{2}

Explanation

Solution

AB =[cos⁡2θcos⁡θsin⁡θcos⁡θsin⁡θsin⁡2θ]\begin{bmatrix} \cos^{2}\theta & \cos\theta\sin\theta \\ \cos\theta\sin\theta & \sin^{2}\theta \end{bmatrix} [cos⁡2φcos⁡φsin⁡φcos⁡φsin⁡φsin⁡2φ]\begin{bmatrix} \cos^{2}\varphi & \cos\varphi\sin\varphi \\ \cos\varphi\sin\varphi & \sin^{2}\varphi \end{bmatrix}

AB = [cos⁡θcos⁡φcos⁡(θ−φ)cos⁡θsin⁡φcos⁡(θ−φ)sin⁡θcos⁡φcas(θ−φ)sin⁡θsin⁡φcos⁡(θ−φ)]\begin{bmatrix} \cos\theta\cos\varphi\cos(\theta - \varphi) & \cos\theta\sin\varphi\cos(\theta - \varphi) \\ \sin\theta\cos\varphi cas(\theta - \varphi) & \sin\theta\sin\varphi\cos(\theta - \varphi) \end{bmatrix}

Now AB is null matrix if cos (q – f) = 0It means q – f is odd

multiple of π2\frac{\pi}{2}.