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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix5 Marks
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$\big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$
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Answer

$\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$$\Rightarrow\ \text{F}(-\alpha)=\begin{bmatrix}\cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\-\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
Now, $\text{C}_{11}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha,\ \text{C}_{12}=-\begin{vmatrix}\sin\alpha & 0 \\ 0 & 1 \end{vmatrix}=-\sin\alpha$
$\text{and C}_{13}=\begin{vmatrix}\sin\alpha & \cos\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{21}=-\begin{vmatrix}-\sin\alpha & 0 \\0 & 1 \end{vmatrix}=\sin\alpha,\ \text{C}_{22}=\begin{vmatrix}\cos\alpha & 0 \\0 & 1 \end{vmatrix}=\cos\alpha$
$\text{and C}_{23}=-\begin{vmatrix}\cos\alpha & -\sin\alpha \\0 & 0 \end{vmatrix}=0$
$\text{C}_{31}=\begin{bmatrix} -\sin\alpha & 0 \\ \cos\alpha & 0 \end{bmatrix}=0, \text{C}_{32}=-\begin{bmatrix}\cos\alpha & 0 \\ \sin\alpha & 0 \end{bmatrix}=0$
$\text{and C}_{33}=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ 0 & 0 \end{bmatrix}=0$
$\Rightarrow\ \text{adj}\left\{\text{F}(\alpha)\right\}=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}^\text{T}$
$=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\Rightarrow\ |\text{F}(\alpha)|=1$
$\therefore\ \big[\text{F}(\alpha)\big]^{-1}=\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}\ .....(\text{i})$
$\Rightarrow\ \big[\text{F}(\alpha)\big]^{-1}=\text{F}(-\alpha)$

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