Show that the matrix $\displaystyle \begin{pmatrix}cos@ & sin@\\-sin@ & cos@\end{pmatrix}$ is nonsingular and compute its inverse.

(@ - theta)

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- Feb 9th 2009, 10:00 PMjennifer1004[SOLVED] Nonsingular matrix
Show that the matrix $\displaystyle \begin{pmatrix}cos@ & sin@\\-sin@ & cos@\end{pmatrix}$ is nonsingular and compute its inverse.

(@ - theta) - Feb 9th 2009, 10:51 PMjavax
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. - Feb 10th 2009, 06:09 AMjennifer1004Thank you!
Thank you. You helped me out so much and I love to learn how you can check your work. I'm taking an online class so my only instruction is the textbook, which isn't that great and lacks examples. Thanks again!