Question

Find the inverse of each of the matrices(if it exists). \(\begin{bmatrix}1&0&0\\0& \cos\alpha& \sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}\)

Answer 1

Let A=\(\begin{bmatrix}1&0&0\\0& \cos\alpha& \sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}\)
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=\(\begin{bmatrix}-1&0&0\\0& -\cos\alpha& -\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}\)

therefore A^-1=\(\frac{1}{\mid A\mid}\).adj A=-\(\begin{bmatrix}-1&0&0\\0& -\cos\alpha& -\sin\alpha\\0&-\sin\alpha&\cos\alpha\end{bmatrix}\)

=\(\begin{bmatrix}1&0&0\\0& \cos\alpha& \sin\alpha\\0&\sin\alpha&-\cos\alpha\end{bmatrix}\)