Question

Obtain the inverse of the following matrix using elementary operations:
$A=\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$

Answer 1

We know that

$A{A}^{-1}=I$

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

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

$\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 2\\ 0& -5& -8\end{array}\right]A^{-1}=\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& -3& 1\end{array}\right]$

$R_3\to R_3+5R_2,R_1\to R_1-2R_2$

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

$R_3\to \frac{1}{2}{R}_3$

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

$R_2\to R_2-2R_3,R_1\to R_1+R_3$

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]A^{-1}=\left[\begin{array}{ccc}1/2& -1/2& 1/2\\ -4& 3& -1\\ 5/2& -3/2& 1/2\end{array}\right]$

$\therefore I{A}^{-1}=A^{-1}=\left[\begin{array}{ccc}1/2& -1/2& 1/2\\ -4& 3& -1\\ 5/2& -3/2& 1/2\end{array}\right]$