1. Tardigrade
  2. Question
  3. Mathematics
  4. Let F (α)= [ cos α - sin α 0 sin α cos α 0 0 0 1] where α ∈ R. Then [ F (α)]-1 is equal to

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Q. Let $F (\alpha)= \begin{bmatrix}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ where $\alpha \in R$. Then $[ F (\alpha)]^{-1}$ is equal to

Matrices

Solution:

$F (\alpha) \cdot F (-\alpha)= \begin{bmatrix}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\cos (-\alpha) & -\sin (-\alpha) & 0 \\ \sin (-\alpha) & \cos (-\alpha) & 0 \\ 0 & 0 & 1\end{bmatrix}$
$F (\alpha) \cdot F (-\alpha)= \begin{bmatrix}\cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix} \begin{bmatrix}\cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$
$= \begin{bmatrix}\cos ^{2} \alpha+\sin ^{2} \alpha+0 & \cos \alpha \sin \alpha-\cos \alpha \sin \alpha+0 & 0+0+0 \\ \sin \alpha \cos \alpha-\sin \alpha \cos \alpha+0 & \sin ^{2} \alpha+\cos ^{2} \alpha+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1\end{bmatrix}$
$= \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}= I$
$\left[\because \cos ^{2} \alpha+\sin ^{2} \alpha=1\right]$
$F (\alpha) \cdot F (-\alpha)=1 \therefore [ F (\alpha)]^{-1}= F (-\alpha)$