Question

Using elementary row transformation, find the inverse of $\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]$

Answer 1

$\left[\begin{array}{ccc}2& 0& -1\\ 5& 1& 0\\ 0& 1& 3\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

Apply $R_2\leftrightarrow R_1$

$\left[\begin{array}{ccc}5& 1& 0\\ 2& 0& -1\\ 0& 1& 3\end{array}\right]A^{-1}=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

$R_1\to R_1-2R_2$

$\left[\begin{array}{ccc}1& 1& 2\\ 2& 0& -1\\ 0& 1& 3\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-2& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right]$

$R_2\leftrightarrow R_3$

$\left[\begin{array}{ccc}1& 1& 2\\ 0& 1& 3\\ 2& 0& -1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-2& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$

$R_1\to R_1-R_2$

$\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 3\\ 2& 0& -1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-2& 1& -1\\ 0& 0& 1\\ 1& 0& 0\end{array}\right]$

$R_3\to R_3-2R_1$

$\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 3\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}-2& 1& -1\\ 0& 0& 1\\ 5& -2& 2\end{array}\right]$

$R_1\to R_1+R_3$, $R_2\to R_2-3R_3$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]$.