Let F( )= & - & 0 & & 0 0 & 0 & 1 andG( )= & 0 & 0 & 1 & 0 - & 0 … - Vidyadip

Question

Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)$

✓

Answer

$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\Rightarrow\ |\text{G}(\beta)|=\cos^2\beta+\sin^2$$\text{C}_{11}=\cos\beta, \text{C}_{21}=0,\text{C}_{31}=\sin\beta$
$\text{C}_{12}=0,\text{C}_{22}=1,\text{C}_{32}=0$
$\text{C}_{13}=\sin\beta, \text{C}_{23}=0,\text{C}_{33}=\cos\beta$
$\big[\text{G}(\beta)\big]^{-1}=\frac{\text{adj}(\text{G}(\beta))}{|\text{G}(\beta)|}=\frac{1}{1}\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(i)}$
Now, $\text{G}(-\beta)=\begin{bmatrix} \cos(-\beta) & 0 & \sin(-\beta) \\ 0 & 1 & 0 \\ -\sin(-\beta) & 0 & \cos(-\beta) \end{bmatrix}$
$=\begin{bmatrix} \cos\beta & 0 & -\sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(ii)}$
From (i) & (ii)
$\big[\text{G}(\beta)\big]^{1}=\text{G}(-\beta)$

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