Find the inverse of the matrix (if it exists) | vedclass

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(A) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos a & \sin a \ 0 & \sin a & -\cos a\end{array}ight]$.First,we calculate the determinant $|

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$\left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos a & \sin a \ 0 & \sin a & -\cos a\end{array}ight]$

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(A) Let $A = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos a & \sin a \ 0 & \sin a & -\cos a\end{array}ight]$.First,we calculate the determinant $|A|$:$|A| = 1(\cos a(-\cos a) - \sin a(\sin a)) = -\cos^2 a - \sin^2 a = -(\cos^2 a + \sin^2 a) = -1$.Since $|A| 
eq 0$,the inverse $A^{-1}$ exists.Next,we find the cofactor matrix $C$ where $C_{ij}$ is the cofactor of element $a_{ij}$:$C_{11} = +(-\cos^2 a - \sin^2 a) = -1$,$C_{12} = -(0) = 0$,$C_{13} = +(0) = 0$.$C_{21} = -(0) = 0$,$C_{22} = +(-\cos a) = -\cos a$,$C_{23} = -(\sin a) = -\sin a$.$C_{31} = +(0) = 0$,$C_{32} = -(\sin a) = -\sin a$,$C_{33} = +(\cos a) = \cos a$.The adjoint matrix $\operatorname{adj} A$ is the transpose of the cofactor matrix:$\operatorname{adj} A = \left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos a & -\sin a \ 0 & -\sin a & \cos a\end{array}ight]$.Finally,$A^{-1} = \frac{1}{|A|} \operatorname{adj} A = \frac{1}{-1} \left[\begin{array}{ccc}-1 & 0 & 0 \ 0 & -\cos a & -\sin a \ 0 & -\sin a & \cos a\end{array}ight] = \left[\begin{array}{ccc}1 & 0 & 0 \ 0 & \cos a & \sin a \ 0 & \sin a & -\cos a\end{array}ight]$.

Find the inverse of the matrix (if it exists): $\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & \sin a & -\cos a\end{array}\right]$

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