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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.

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    $\begingroup$ I think it should be $\rho \sin^2 (x)$ instead of $\sin^2 (x)$. $\endgroup$ – idok Apr 29 '18 at 10:20
  • $\begingroup$ yes, fixed it. Thanks. $\endgroup$ – dekuShrub Apr 29 '18 at 10:24
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    $\begingroup$ Also, the proposed question does not seem correct. Perhaps the bottom-right entry of the first matrix has the value $\rho cos(\phi))$ instead of $sin(\phi)$? $\endgroup$ – idok Apr 29 '18 at 10:30
  • $\begingroup$ If i compute the top right most element in the product of the claimed inverse i get: $\cos \phi \sin \phi - \sin^2 \phi / \rho$ so I dont believe they are inverses $\endgroup$ – JuliusL33t Apr 29 '18 at 10:31
  • $\begingroup$ wow! You are right... it should indeed be $\rho\cos\varphi$... $\endgroup$ – dekuShrub Apr 29 '18 at 10:34
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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.

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    $\begingroup$ Could be good to point out that determinant being inverses to each other is required but not always sufficient. $\endgroup$ – mathreadler Apr 29 '18 at 11:04

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