As part of another question I asked here, the answer was to invert a matrix $$ \begin{bmatrix} \cos\varphi &\sin\varphi \\ -\rho\sin\varphi &\sin\varphi \end{bmatrix} $$ I tried inversing the matrix by the formula $$ \frac{1}{\text{det}} \cdot \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}, $$ but with my determinant computed to $1/(\cos\varphi\sin\varphi + \rho\sin²\varphi)$ I get this matrix: $$ \dfrac{1}{\cos\varphi\sin\varphi + \rho\sin²\varphi} \cdot \begin{bmatrix} \sin\varphi & -\sin\varphi \\ \rho\sin\varphi & \cos\varphi \end{bmatrix} $$ And I've looked at trig identities if there is some trick I can do, but I really don't get it. How do you get the inverse matrix to be $$ \frac{1}{\rho}\cdot \begin{bmatrix} \rho\cos\varphi & -\sin\varphi\\ \rho\sin\varphi & \cos\varphi \end{bmatrix}? $$ Which I was told it should be.
edit (solved)
I had made a mistake (pointed out in comments) when I wrote the matrix, it should have been $$ \begin{bmatrix} \cos\varphi &\sin\varphi\\ -\rho\sin\varphi &\rho\cos\varphi \end{bmatrix} $$ so that makes the determinant $\rho$ which makes it a lot simpler.
It's simple:$$\begin{bmatrix}\cos\varphi&\sin\varphi\\-\rho\sin\varphi&\sin\varphi\end{bmatrix}^{-1}\neq\frac1\rho\begin{bmatrix}\rho\cos\varphi & -\sin\varphi\\\rho\sin\varphi & \cos\varphi\end{bmatrix}.$$For instance,$$\det\begin{bmatrix}\cos\varphi&\sin\varphi\\-\rho\sin\varphi&\sin\varphi\end{bmatrix}=\cos\varphi\sin\varphi+\rho\sin^2\varphi,$$whereas$$\det\left(\frac1\rho\begin{bmatrix}\rho\cos\varphi & -\sin\varphi\\\rho\sin\varphi & \cos\varphi\end{bmatrix}\right)=\frac1\rho.$$In order to be the inverse of each other, the determinant of the first matrix should be the inverse of the determinant of the second one.