Question

Find the inverse of $A=\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]$ by column transformation.

Answer 1

Take $A=AI$

$\left[\begin{array}{ccc}1& 3& 3\\ 1& 4& 3\\ 1& 3& 4\end{array}\right]=A\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$c_2\to c_2-3c_1$ and $c_3\to c_3-3c_1$

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

$c_1\to c_1-c_2$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]=A\left[\begin{array}{ccc}4& -3& -3\\ -1& 1& 0\\ 0& 0& 1\end{array}\right]$

$c_1\to c_1-c_3$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=A\left[\begin{array}{ccc}7& -3& -3\\ -1& 0& 0\\ -1& 0& 1\end{array}\right]$

$\therefore A^{-1}=\left[\begin{array}{ccc}7& -3& -3\\ -1& 0& 0\\ -1& 0& 1\end{array}\right]$