# Smith normal form of the following matrix

Let $$A = \begin{bmatrix} 66 & 30\\ 12 & 4 \end{bmatrix}$$ I've been trying to find the smith normal form of this matrix, and I keep getting the wrong answer. Here are my workings;

gcd of the entries of the first column is $$6$$, so we aim to get $$6$$ in the $$(1,1)$$ position via row operations. $$r_1 : r_1 - 5r_2$$ yields $$\begin{bmatrix} 6 & 10\\ 12 &4 \end{bmatrix}$$

Now, the gcd of the first row is $$2$$, so we aim to get this in the $$(1,1)$$ position. $$c_1 : 2c_1 - c_2$$ yields $$\begin{bmatrix} 2 & 10\\20 &4 \end{bmatrix}$$

Now, we aim to get zeroes in the $$(1,2)$$ and $$(2,1)$$ position. $$c_2 : 5c_1 - c_2$$ and $$r_2 : r_2 - 10r_1$$ yields the matrix $$\begin{bmatrix} 2 & 0\\ 0 & 96 \end{bmatrix}$$

But the answer is supposed to be $$\begin{bmatrix} 2 & 0\\0 & 48 \end{bmatrix}$$

But I can't see a mistake in my workings anywhere? Am I using the algorithm incorrectly?

$$c_1\leftarrow2c_1−c_2$$ is not an elementary operation.