Question

Find the inverse of matrices by
elementary row transformation.
$A=\left|\begin{array}{ccc}1& 2& 1\\ -1& 0& 2\\ 2& 1& -3\end{array}\right|$
$B=\left|\begin{array}{ccc}cos\theta & -sin\theta & 0\\ sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right|$
$A^{-1}=\left(\begin{array}{ccc}cos\theta & sin\theta & 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right)$

Answer 1

$A\left[\begin{array}{ccc}1& 0& 0\\ -tan& 1& 0\\ 0& 0& 1\end{array}\right]=\left[\begin{array}{ccc}cos\theta & -sin\theta & 0\\ 0& \frac{1}{cos\theta }& 0\\ 0& 0& 1\end{array}\right]$
$R_2\to R_2.cos\theta$
$A\left|\begin{array}{ccc}1& 0& 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{ccc}cos\theta & -sin\theta & 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
$R_1\to R_1+sin\theta R_2$
$A\left|\begin{array}{ccc}1-sin& sin\theta & 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{ccc}cos\theta & 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$
$R_1\to R_1/cos\theta$
$A\left|\begin{array}{ccc}cos\theta & sin\theta & 0\\ -sin\theta & cos\theta & 0\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$