linear algebra - Reducing the Matrix to Reduced Row Echelon Form

Reduce the matrix $\begin{bmatrix}1&-1&-6\\4&-1&-15\\-2&2&12\end{bmatrix}$ to reduced row-echelon form

How is my answer incorrect?

I performed the row operations:

1) $R_2 = 4R_1 - R_2$

2) $R_3 = 2R_1 + R_3$

3) $R_2 = R_2 / -3$;

4) $R_3 = R_3/18$

5) $R_2 = R_2 + 7R_3$

6) $R_1 = R_1 + -6R_3$

7) $R_1 = R_1 + R_2$

Which gives me the RREF of the matrix

$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

So how in the world is my solution incorrect?

Answer

After Step 1:

$$\left[\begin{matrix}1 & -1 & -6 \\
0 & -3 & -9 \\
-2 & 2 & 12\end{matrix}\right]$$

After Step 2:

$$\left[\begin{matrix}1 & -1 & -6 \\
0 & -3 & -9 \\

0 & 0 & 0\end{matrix}\right]$$

After Step 3:

$$\left[\begin{matrix}1 & -1 & -6 \\
0 & 1 & 3 \\
0 & 0 & 0\end{matrix}\right]$$

All of the steps with $R_3$ are unnecessary since $R_3$ is all zeroes. Skip to Step 7:

$$\left[\begin{matrix}1 & 0 & -3 \\
0 & 1 & 3 \\
0 & 0 & 0\end{matrix}\right]$$