As an example, we will find the Smith normal form of the following matrix over the integers. \begin{equation*} \left(\begin{array}{ccc} 2 & 4 & 4 \\ -6 & 6 & 12 \\ 10 & -4 & -16 \end{array}\right) \end{equation*}The following matrices are the intermediate steps as the algorithm is applied to the above matrix.

\begin{equation*} \left(\begin{array}{ccc} 2 & 0 & 0 \\ -6 & 18 & 24 \\ 10 & -24& -36 \end{array}\right) \to \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 18 & 24 \\ 0 & -24& -36 \end{array}\right) \end{equation*} \begin{equation*} \to \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 18 & 24 \\ 0 & -6 & -12 \end{array}\right) \to \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 6 & 12 \\ 0 & 18 & 24 \end{array}\right) \end{equation*} \begin{equation*} \to \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 6 & 12 \\ 0 & 0 & -12 \end{array}\right) \to \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{array}\right) \end{equation*} So the Smith normal form is

\begin{equation*} \left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 12 \end{array}\right) \end{equation*} and the elementary divisors are $2$ $6$ and $12$