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If $$f(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ then the value of $\big(f(x)\big)^2$ is

  1. $f(x)$
  2. $f(2x)$
  3. $2f(x)$
  4. None of these
in Linear Algebra retagged by
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1 Answer

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$(f(x))^2$

$= f(x)*f(x)$

$=\begin{bmatrix} cos \ x & -sin\ x &0 \\ sin\ x& cos \ x &0 \\ 0& 0& 1 \end{bmatrix} * \begin{bmatrix} cos \ x & -sin\ x &0 \\ sin\ x& cos \ x &0 \\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} cos^2\ x - sin^2\ x & -cos\ x \ sin\ x - sin \ x\ cos \ x & 0 \\ sin\ x \ cos \ x + cos \ x \ sin\ x& -sin^2\ x + cos^2 \ x &0 \\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} cos^2\ x - sin^2\ x &  - 2sin \ x\ cos \ x & 0 \\ 2sin\ x \ cos \ x & cos^2\ x-sin^2\ x  &0 \\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} cos \ 2x & -sin\ 2x &0 \\ sin\ 2x& cos \ 2x &0 \\ 0& 0& 1 \end{bmatrix}$

$=f(2x)$

$\therefore$ Option $B.$ is correct.

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