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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?

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

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