If E(θ )={bmatrix} cosθ & sinθ \\ -sinθ & cosθ {bmatrix} then E(α) E(β) is equal to

Correct Answer is : E(α+β)E(α + β)E(α+β)

Correct Answer is : E(α+β)E(α + β)E(α+β)

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UPSC NDA I 2015 Mathematics Paper

Question : 28 of 120

Marks: +1, -0

E(\theta )=\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}

then E(α) E(β) is equal to

Solution:

Concept:

Matrix multiplication of rotation matrices results in the addition of angles.

Explanation:

Given

E(\theta) = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}

Write

E(\alpha)

and

E(\beta)

and multiply them directly.

E(\alpha)E(\beta) = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} \begin{bmatrix} \cos\beta & \sin\beta \\ -\sin\beta & \cos\beta \end{bmatrix}

Compute each entry:

First row, first column:

\cos\alpha \cos\beta + \sin\alpha (-\sin\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha+\beta)

First row, second column:

\cos\alpha \sin\beta + \sin\alpha \cos\beta = \sin(\alpha+\beta)

Second row, first column:

-\sin\alpha \cos\beta + \cos\alpha (-\sin\beta) = -(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = -\sin(\alpha+\beta)

Second row, second column:

-\sin\alpha \sin\beta + \cos\alpha \cos\beta = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha+\beta)

Thus the product equals

\begin{bmatrix} \cos(\alpha+\beta) & \sin(\alpha+\beta) \\ -\sin(\alpha+\beta) & \cos(\alpha+\beta) \end{bmatrix} = E(\alpha+\beta)

Answer:

Option C:

E(\alpha + \beta)

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