Find the Inverse of the Following Matrix. ⎡ ⎢ ⎣ 1 0 0 0 Cos α Sin α 0 Sin α − Cos α ⎤ ⎥ ⎦ - Mathematics

Question

Find the inverse of the following matrix.

\[\begin{bmatrix}1 & 0 & 0 \\ 0 & \cos \alpha & \sin \alpha \\ 0 & \sin \alpha & - \cos \alpha\end{bmatrix}\]

Solution

\[G = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & - \cos\alpha\end{bmatrix}\]
Now,
\[ C_{11} = \begin{vmatrix}\cos\alpha & \sin\alpha \\ \sin\alpha & - \cos\alpha\end{vmatrix} = - 1, C_{12} = - \begin{vmatrix}0 & \sin\alpha \\ 0 & - \cos\alpha\end{vmatrix} = 0\text{ and }C_{13} = \begin{vmatrix}0 & \cos\alpha \\ 0 & \sin\alpha\end{vmatrix} = 0\]
\[ C_{21} = - \begin{vmatrix}0 & 0 \\ \sin\alpha & - \cos\alpha\end{vmatrix} = 0, C_{22} = \begin{vmatrix}1 & 0 \\ 0 & - \cos\alpha\end{vmatrix} = - \cos\alpha\text{ and }C_{23} = - \begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix} = - \sin\alpha\]
\[ C_{31} = \begin{vmatrix}0 & 0 \\ \cos\alpha & \sin\alpha\end{vmatrix} = 0, C_{32} = - \begin{vmatrix}1 & 0 \\ 0 & \sin\alpha\end{vmatrix} = - \sin\alpha\text{ and }C_{33} = \begin{vmatrix}1 & 0 \\ 0 & \cos\alpha\end{vmatrix} = \cos\alpha\]
\[adjF = \begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix}^T = \begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix}\]
\[and \left| F \right| = - 1\]
\[ \therefore F^{- 1} = - 1\begin{bmatrix}- 1 & 0 & 0 \\ 0 & - \cos\alpha & - \sin\alpha \\ 0 & - \sin\alpha & \cos\alpha\end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & - \cos\alpha\end{bmatrix}\]

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Inverse of Matrix - Inverse of a Square Matrix by the Adjoint Method

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Q 8.7 Q 8.6 Q 9.1

Chapter 7: Adjoint and Inverse of a Matrix - Exercise 7.1 [Page 23]

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RD Sharma Mathematics [English] Class 12

Chapter 7 Adjoint and Inverse of a Matrix
Exercise 7.1 | Q 8.7 | Page 23

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