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