Smith normal form of an explicit three-by-three integer matrix

Swap the first two columns, replace the second row by $R_2-4R_1$ and the third by $R_3-2R_1$, and replace the second column by $C_2-2C_1$. The remaining lower block is $\left(\begin{array}{cc}-4& 6\\ -2& 6\end{array}\right)$. Swap its rows, replace the last row by the last row minus twice the preceding row, and clear the off-diagonal $6$ by adding three times the pivot column. Unit row scalings give $\mathrm{diag}(1,2,6)$.

The gcd of all entries is $1$; the minor from rows $1,3$ and columns $1,2$ is $2$, and every two-by-two minor is even, so their gcd is $2$; the determinant is $12$. Thus the determinantal divisors are $1,2,12$, whose successive quotients are $1,2,6$, confirming step 1.1 by [L1].

Quotienting the diagonal presentation gives $\mathbb{Z}\mathrm{/}(1)\oplus \mathbb{Z}\mathrm{/}(2)\oplus \mathbb{Z}\mathrm{/}(6)$; the first summand is zero, yielding the displayed cokernel with no free part because the determinant is nonzero.