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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?

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  • $\begingroup$ According to Wolfram Alpha, this has a zero determinant, so the RREF can not be the identity matrix. I don't know exactly where you went wrong, though. $\endgroup$ Jun 23, 2016 at 20:30
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    $\begingroup$ $-2R_1=R_3$ so there's no way this reduces to the identity. $\endgroup$
    – user137731
    Jun 23, 2016 at 20:30
  • $\begingroup$ I don't either... $\endgroup$
    – Shammy
    Jun 23, 2016 at 20:30
  • $\begingroup$ @Shammy After step 2, you should get that the third row is completely zero. $\endgroup$ Jun 23, 2016 at 20:32
  • $\begingroup$ $R_3 = -2 R_1$. After step 2, Row 3 should be a zero vector. Any steps involving row 3 that follow must be incorrect. $\endgroup$
    – Doug M
    Jun 23, 2016 at 20:32

1 Answer 1

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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:

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

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  • $\begingroup$ -9 in first step?? what??????? how???????? $\endgroup$
    – Shammy
    Jun 23, 2016 at 20:39
  • $\begingroup$ @Shammy $$4\cdot -6-(-15)=-24+15=-9$$ $\endgroup$ Jun 23, 2016 at 20:39
  • $\begingroup$ I'm an idiot. On my paper I had a positive 6 in the third column when I wrote it down instead of a -6 -_-. I hate matrices $\endgroup$
    – Shammy
    Jun 23, 2016 at 20:41
  • $\begingroup$ That's why I like to use Wolfram Alpha to check my work. For example, here's Wolfram Alpha's answer to the RREF of this matrix. $\endgroup$ Jun 23, 2016 at 20:44
  • $\begingroup$ I finally got it. Thank you @noble you truly are, Noble $\endgroup$
    – Shammy
    Jun 23, 2016 at 20:50

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