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cos⁡θ⋅[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ⋅[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \ \cosθ & \sinθ \end{bmatrix} is equal to:

  1. [1111]\begin{bmatrix} 1 & 1 \ 1 & 1 \end{bmatrix}

  2. [0110]\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}

  3. [1001]\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}

  4. none of these

Matrices

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Answer

Given,

cos⁡θ⋅[cos⁡θsin⁡θ−sin⁡θcos⁡θ]+sin⁡θ⋅[sin⁡θ−cos⁡θcos⁡θsin⁡θ]\cosθ \cdot \begin{bmatrix} \cosθ & \sinθ \ -\sinθ & \cosθ \end{bmatrix} + \sinθ \cdot \begin{bmatrix} \sinθ & -\cosθ \ \cosθ & \sinθ \end{bmatrix}

Solving:

⇒[cos⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θcos⁡2θ]+⋅[sin⁡2θ−sin⁡θcos⁡θsin⁡θcos⁡θsin⁡2θ]⇒[cos⁡2θ+sin⁡2θcos⁡θsin⁡θ−sin⁡θcos⁡θ−sin⁡θcos⁡θ+sin⁡θcos⁡θcos⁡2θ+sin⁡2θ]⇒[1001],\Rightarrow \begin{bmatrix} \cos^2θ & \cosθ\sinθ \ -\sinθ\cosθ & \cos^2θ \end{bmatrix} + \cdot \begin{bmatrix} \sin^2θ & -\sinθ\cosθ \ \sinθ\cosθ & \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} \cos^2θ + \sin^2θ & \cosθ\sinθ - \sinθ\cosθ\ -\sinθ\cosθ + \sinθ\cosθ & \cos^2θ + \sin^2θ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix},

Hence, option 3 is the correct option.

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