[SOLVED] Nonsingular matrix

**jennifer1004** · February 9th 2009, 10:00 PM

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

(@ - theta)

**javax** · February 9th 2009, 10:51 PM

Hello.
A Matrix B is non-singular if $det(B) \neq 0.$
So in this case the determinant of your matrix is 1, that means it is non-singular.
I usually find inverse using Gauss method.
$\begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\-\sin x & \cos x & \vdots & 0 & 1\end{pmatrix}$
First to make $a_{21} = 0$ we do the transformation $Row2 +\frac{\sin x}{\cos x}*Row1$

after doing that transformation it will be

$\begin{pmatrix}\cos x & \sin x & \vdots & 1 & 0\\0 & \frac{1}{\cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}$

now a12 should be 0. So we do the transformation $Row1-\sin x*\cos x*Row2$
you'll have

$\begin{pmatrix}\cos x & 0 & \vdots & \cos^2 x & -\sin x \cos x \\0 & \frac{1}{cos x} & \vdots & \frac{\sin x}{\cos x} & 1\end{pmatrix}$

now you divide the first row by $\cos x$ and the second row with $\frac{1}{\cos x}$

and you'll have:

$\begin{pmatrix}1 & 0 & \vdots & \cos x & -\sin x \\ 0 & 1 & \vdots & \sin x & \cos x \end{pmatrix}$

So the inverse of that matrix is
$A^{-1}= \begin{pmatrix}\cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}$

You can check your results by multyplying $A x A^{-1}$ if the result is the Identity matrix $I$ , then it's done well.

**jennifer1004** · February 10th 2009, 06:09 AM

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!

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[SOLVED] Nonsingular matrix

Thank you!

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