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Let F ( α ) = ⎡ ⎢ ⎣ Cos α − Sin α 0 Sin α Cos α 0 0 0 1 ⎤ ⎥ ⎦ and G ( β ) = ⎡ ⎢ ⎣ Cos β 0 Sin β 0 1 0 − Sin β 0 Cos β ⎤ ⎥ ⎦ Show that [ F ( α ) ] − 1 = F ( − α ) - Mathematics

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प्रश्न

Let
\[F \left( \alpha \right) = \begin{bmatrix}\cos \alpha & - \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\text{ and }G\left( \beta \right) = \begin{bmatrix}\cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ - \sin \beta & 0 & \cos \beta\end{bmatrix}\]

Show that

(i) \[\left[ F \left( \alpha \right) \right]^{- 1} = F \left( - \alpha \right)\]
(ii) \[\left[ G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)\]
(iii) \[\left[ F \left( \alpha \right)G \left( \beta \right) \right]^{- 1} = G \left( - \beta \right)F \left( - \alpha \right)\]
योग

उत्तर

(i) \[ F(\alpha) = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow F( - \alpha) = \begin{bmatrix}\cos\left( - \alpha \right) & - \sin\left( - \alpha \right) & 0 \\ \sin\left( - \alpha \right) & \cos\left( - \alpha \right) & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[Now, \]
\[ C_{11} = \begin{vmatrix}\cos\alpha & 0 \\ 0 & 1\end{vmatrix} = \cos\alpha, C_{12} = - \begin{vmatrix}\sin\alpha & 0 \\ 0 & 1\end{vmatrix} = - \sin\alpha\text{ and }C_{13} = \begin{vmatrix}\sin\alpha & \cos\alpha \\ 0 & 0\end{vmatrix} = 0\]
\[ C_{21} = - \begin{vmatrix}- \sin\alpha & 0 \\ 0 & 1\end{vmatrix} = \sin\alpha, C_{22} = \begin{vmatrix}\cos\alpha & 0 \\ 0 & 1\end{vmatrix} = \cos\alpha\text{ and }C_{23} = - \begin{vmatrix}\cos\alpha & - \sin\alpha \\ 0 & 0\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}- \sin\alpha & 0 \\ \cos\alpha & 0\end{vmatrix} = 0, C_{32} = - \begin{vmatrix}\cos\alpha & 0 \\ \sin\alpha & 0\end{vmatrix} = 0\text{ and }C_{33} = \begin{vmatrix}\cos\alpha & - \sin\alpha \\ \sin\alpha & \cos\alpha\end{vmatrix} = 1\]
\[ \Rightarrow adj\left\{ F(\alpha) \right\} = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}^T = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow \left| F(\alpha) \right| = 1\]
\[ \therefore \left[ F\left( \alpha \right) \right]^{- 1} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} . . . \left( 1 \right)\]
\[ \Rightarrow \left[ F\left( \alpha \right) \right]^{- 1} = F( - \alpha) \]
(ii) \[ G(\beta) = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[ \Rightarrow G( - \beta) = \begin{bmatrix}\cos\left( - \beta \right) & 0 & \sin\left( - \beta \right) \\ 0 & 1 & 0 \\ - \sin\left( - \beta \right) & 0 & \cos\left( - \beta \right)\end{bmatrix} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[Now, \]
\[ C_{11} = \begin{vmatrix}1 & 0 \\ 0 & \cos\beta\end{vmatrix} = \cos\beta, C_{12} = - \begin{vmatrix}0 & 0 \\ - \sin\beta & \cos\beta\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}0 & 1 \\ - \sin\beta & 0\end{vmatrix} = \sin\beta\]
\[ C_{21} = - \begin{vmatrix}0 & \sin\beta \\ 0 & \cos\beta\end{vmatrix} = 0, C_{22} = \begin{vmatrix}\cos\beta & \sin\beta \\ - \sin\beta & \cos\beta\end{vmatrix} = 1\text{ and }C_{23} = - \begin{vmatrix}\cos\beta & 0 \\ - \sin\beta & 0\end{vmatrix} = 0\]
\[ C_{31} = \begin{vmatrix}0 & \sin\beta \\ 1 & 0\end{vmatrix} = - \sin\beta, C_{32} = - \begin{vmatrix}\cos\beta & \sin\beta \\ 0 & 0\end{vmatrix} = 0\text{ and }C_{33} = \begin{vmatrix}\cos\beta & 0 \\ 0 & 1\end{vmatrix} = \cos\beta\]
\[adj\left\{ G(\beta) \right\} = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}^T = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[\left| G(\beta) \right| = 1\]
\[ \therefore G(\beta )^{- 1} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix} . . . \left( 2 \right) \]
\[ \Rightarrow G(\beta )^{- 1} = = G( - \beta) \]
(iii) \[ F(\alpha) = \begin{bmatrix}\cos\alpha & - \sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}\]
\[ \Rightarrow F( - \alpha) = \begin{bmatrix}\cos\left( - \alpha \right) & - \sin\left( - \alpha \right) & 0 \\ \sin\left( - \alpha \right) & \cos\left( - \alpha \right) & 0 \\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} . . . \left( 3 \right)\]
\[G(\beta) = \begin{bmatrix}\cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ - \sin\beta & 0 & \cos\beta\end{bmatrix}\]
\[ \Rightarrow G( - \beta) = \begin{bmatrix}\cos\left( - \beta \right) & 0 & \sin\left( - \beta \right) \\ 0 & 1 & 0 \\ - \sin\left( - \beta \right) & 0 & \cos\left( - \beta \right)\end{bmatrix} = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix} . . . \left( 4 \right)\]
\[ \left[ F(\alpha)G(\beta) \right]^{- 1} = \left[ G(\beta) \right]^{- 1} \left[ F(\alpha) \right]^{- 1} \]
\[ = \begin{bmatrix}\cos\beta & 0 & - \sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta\end{bmatrix}\begin{bmatrix}\cos\alpha & \sin\alpha & 0 \\ - \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix} \]               [Using equation (1) and (2)]
\[ = G( - \beta)F( - \alpha) \]                 [Using equatio (3) and (4)]

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अध्याय 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 16 | पृष्ठ २३

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