I want to find the Smith Normal form of the following matrix : $$\left(\begin{matrix}2 &-4&-1\\0&2&3\\-3&8&3\end{matrix}\right)$$ We have, $$\left(\begin{matrix}2 &-4&-1\\0&2&3\\-3&8&3\end{matrix}\right)\to \left(\begin{matrix}2 &0&0\\0&2&6\\-3&2&3\end{matrix}\right)\to \left(\begin{matrix}2 &0&0\\0&2&0\\-3&2&-3\end{matrix}\right)\to \left(\begin{matrix}2 &0&0\\0&2&0\\0&2&-3\end{matrix}\right) \to \left(\begin{matrix}2 &0&0\\0&2&0\\0&0&6\end{matrix}\right)$$
But determinant value mismatch. So somewhere the process is wrong ! I'm unable to find my mistake.
Edit:
We have, $$\left(\begin{matrix}2 &-4&-1\\0&2&3\\-3&8&3\end{matrix}\right)\to \left(\begin{matrix}-1 &0&0\\0&14&6\\0&20&3\end{matrix}\right)$$ Now, $$\left(\begin{matrix}14 & 6\\20 & 3\end{matrix} \right)\to \left(\begin{matrix}3 & 20\\6 & 14\end{matrix} \right)$$
From this step , how I can proceed ?
The Smith form for integer matrices is obtained by applying the following elemantary transformations to rows (columns) of the matrix:
1) Multiply the row (column) by $-1$
2) Interchange (permute) two rows (columns)
3) multiply a row (column) by an integer and add the result to another row (column)
The Smith form of a matrix $M$ of rank $\rho$ is a matrix with all elements equal to $0$ except the first diagonal emements $\rho$ $M_{1,1}, \cdots , M_{\rho,\rho}$ which, in addition satisfy the divisibility property $M_{i,i} | M_{i+1,i+1}$, $i=1,\cdots,\rho-1$.
Elemantary transformations of type 3) above correspond to pre (post) multiplication with triangular matrices with $1$s in the diagonal.
The mistake in the question is in the transformation which replaces the third column with $2$ times the third column plus the first column corresponds to multiplication with matrix
$\left[ \begin{matrix} 1&0&1\\ 0&1&0\\ 0&0&2 \end{matrix} \right]$
which clearly has determinant equal to $2$ and does not correspond to the allowed elemantary transformations leading to the Smith form.
The correct Smith form is
$ \left[ \begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&6 \end{matrix} \right]$
and is obtained by applying transformations of types 1),2) and 3) above as follows:
$\left[ \begin{matrix} 2&-4&-1\\ 0&2&3\\ -3&8&3 \end{matrix} \right] \sim \left[ \begin{matrix} 1&-4&2\\ -3&2&0\\ -3&8&-3 \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ -3&-10&6\\ -3&-4&3 \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&-10&6\\ 0&-4&3 \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&6&-4\\ 0&3&-1 \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&3&-1\\ 0&6&-4\\ \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&1&3\\ 0&4&6\\ \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&1&0\\ 0&4&-6\\ \end{matrix} \right] \sim \left[ \begin{matrix} 1&0&0\\ 0&1&0\\ 0&0&6 \end{matrix} \right]$