Show that the matrix $\displaystyle \begin{pmatrix}cos@ & sin@\\-sin@ & cos@\end{pmatrix}$ is nonsingular and compute its inverse.
(@ - theta)
Hello.
A Matrix B is non-singular if $\displaystyle det(B) \neq 0. $
So in this case the determinant of your matrix is 1, that means it is non-singular.
I usually find inverse using Gauss method.
$\displaystyle \begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\-\sin x & \cos x & \vdots & 0 & 1\end{pmatrix}$
First to make $\displaystyle a_{21} = 0$ we do the transformation $\displaystyle Row2 +\frac{\sin x}{\cos x}*Row1$
after doing that transformation it will be
$\displaystyle \begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\0 & \frac{1}{\cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}$
now a12 should be 0. So we do the transformation $\displaystyle Row1-\sin x*\cos x*Row2$
you'll have
$\displaystyle \begin{pmatrix}\cos x & 0 & \vdots & \cos^2 x & -\sin x \cos x \\0 & \frac{1}{cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}$
now you divide the first row by $\displaystyle \cos x$ and the second row with $\displaystyle \frac{1}{\cos x}$
and you'll have:
$\displaystyle \begin{pmatrix}1 & 0 & \vdots & \cos x & -\sin x \\ 0 & 1 & \vdots & \sin x & \cos x \end{pmatrix}$
So the inverse of that matrix is
$\displaystyle A^{-1}= \begin{pmatrix}\cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}$
You can check your results by multyplying $\displaystyle A x A^{-1}$ if the result is the Identity matrix $\displaystyle I$, then it's done well.