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Mathematics Question on Determinants

Find the inverse of each of the matrices(if it exists). [100\0cos⁡αsin⁡α\0sin⁡α−cos⁡α]\begin{bmatrix}1&0&0\\\0& \cos\alpha& \sin\alpha\\\0&\sin\alpha&-\cos\alpha\end{bmatrix}

Answer

Let A=[100\0cos⁡αsin⁡α\0sin⁡α−cos⁡α]\begin{bmatrix}1&0&0\\\0& \cos\alpha& \sin\alpha\\\0&\sin\alpha&-\cos\alpha\end{bmatrix}
We have,
IAI=1(-cos2α-sin2α)=-(cos2α+sin2α)
Now A11=-cos2α-sin2α=-1, A12=0, A13=0
A21=0, A22=-cosα, A23=-sinα
A31=0, A32=-sinα, A33=cosα

therefore adj A=[−100\0−cos⁡α−sin⁡α\0−sin⁡αcos⁡α]\begin{bmatrix}-1&0&0\\\0& -\cos\alpha& -\sin\alpha\\\0&-\sin\alpha&\cos\alpha\end{bmatrix}

therefore A-1=1∣A∣\frac{1}{\mid A\mid}.adj A=-[−100\0−cos⁡α−sin⁡α\0−sin⁡αcos⁡α]\begin{bmatrix}-1&0&0\\\0& -\cos\alpha& -\sin\alpha\\\0&-\sin\alpha&\cos\alpha\end{bmatrix}

=[100\0cos⁡αsin⁡α\0sin⁡α−cos⁡α]\begin{bmatrix}1&0&0\\\0& \cos\alpha& \sin\alpha\\\0&\sin\alpha&-\cos\alpha\end{bmatrix}