Question

Find the inverse of A=$\left[\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right]$ by
elementary column transformations.

Answer 1

$|A|=\left|\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right|$
$=\cos \theta (\cos \theta -0)+\sin \theta (\sin \theta -0)+0$
$={\cos }^2\theta +{\sin }^2\theta =1\ne 0$
$\therefore A^{-1}$ exists.
Consider $A{A}^{-1}=I$
$\therefore \left[\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
By $\cos \theta \times C_1$, we get,
$\left[\begin{array}{ccc}{\cos }^2\theta & -\sin \theta & 0\\ \sin \theta \cos \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}\cos \theta & 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
By $C_1-\sin \theta \times C_2$ we get,
$\left[\begin{array}{ccc}1& -\sin \theta & 0\\ 0& \cos \theta & 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}\cos \theta & 0& 0\\ -\sin \theta & 1& 0\\ 0& 0& 1\end{array}\right]$
By $C_2+\sin \theta \times C_1$, we get,
$\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}\cos \theta & \sin \theta & 0\\ -\sin \theta & {\cos }^2\theta & 0\\ 0& 0& 1\end{array}\right]$
By $\left(\frac{1}{\cos \theta }\right)\times C_2$ we get,
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}\cos \theta & \cos \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$
$\therefore A^{-1}=\left[\begin{array}{ccc}\cos \theta & \cos \theta & 0\\ -\sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$