Answer
$A^{-1}=\left[\begin{array}{lll}
1/3 & 0 & 0\\
0 & 1/6 & 0\\
0 & 0 & 1/9
\end{array}\right]$
Work Step by Step
1. Form $[A|I] $
2. Row operations:
$\left[\begin{array}{lllllll}
3 & 0 & 0 & | & 1 & 0 & 0\\
0 & 6 & 0 & | & 0 & 1 & 0\\
0 & 0 & 9 & | & 0 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
\div 3\\
\div 6\\
\div 9
\end{array}\right.$
$\left[\begin{array}{lllllll}
1 & 0 & 0 & | & 1/3 & 0 & 0\\
0 & 1 & 0 & | & 0 & 1/6 & 0\\
0 & 0 & 1 & | & 0 & 0 & 1/9
\end{array}\right]$
3. We have obtained $[I|B], B=A^{-1}$
$A^{-1}=\left[\begin{array}{lll}
1/3 & 0 & 0\\
0 & 1/6 & 0\\
0 & 0 & 1/9
\end{array}\right]$
