# Solve a system of three equations by rewriting in row-echelon form

I tried solving this system of equations and I got what seems to be an inconsistent system. I wanted to post my results here to see if I'm correct.

Here is the original problem: \left\{ \begin{aligned} 3x_1-2x_2+4x_3&=1 \\ x_1+x_2-2x_3&=3 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right.

For the first step I interchanged equation one with equation two to give the following:

\left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ 3x_1-2x_2+4x_3&=1 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right.

I then added (-3) times equation one to equation two to give a new equation two:

\left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ 2x_1-3x_2+6x_3&=8 \end{aligned} \right.

Next I added (-2) times equation one to equation three to give a new equation three:

\left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ -5x_2+10x_3&=-2 \end{aligned} \right.

So it was at this point that I thought something was wrong, but I kept going. Next I added (-1) times equation two to equation three to give a new equation three:

\left\{ \begin{aligned} x_1+x_2-2x_3&=3 \\ -5x_2+10x_3&=-8 \\ 0=10 \end{aligned} \right.

Did I do something wrong, or is the properly solved? We haven't gotten to matrices yet, which I know will be tremendously easier, our teacher wants us to get comfortable with doing it this way first.

• Everything looks fine, except the last equation in your last system should be $0=6$. That doesn't change your conclusion, of course. (I'm horrible at arithmetic by the way, so you may want to check your arithmetic yourself. Your procedure is perfectly fine.) – David Mitra Jan 21 '14 at 21:39
• @DavidMitra you're right, it is $0=6$ – hax0r_n_code Jan 21 '14 at 21:40
• I see that in your third system, the last equation should be $-5x_2+10x_3=2$. So the last equation in the last system is $0=10$ (!). – David Mitra Jan 21 '14 at 21:43
• @DavidMitra haha thanks! – hax0r_n_code Jan 21 '14 at 21:44
• So, long story short: carefully check your work. You seem to have the procedure down pat. – David Mitra Jan 21 '14 at 21:44

$$\begin{bmatrix} 3 & -2 & 4 & 1 \\ 1 & 1 & -2 & 3 \\ 2 & -3 & 6 & 8 \\ \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 1 & 1 & -2 & 3 \\ 3 & -2 & 4 & 1 \\ 2 & -3 & 6 & 8 \\ \end{bmatrix} \xrightarrow{R_2 \gets R_2 -3 R_1} \begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 2 & -3 & 6 & 8 \\ \end{bmatrix} \xrightarrow{R_3 \gets R_3 -2 R_1} \begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 0 & -5 & 10 & 2 \\ \end{bmatrix} \xrightarrow{R_3 \gets R_3 -R_2} \begin{bmatrix} 1 & 1 & -2 & 3 \\ 0 & -5 & 10 & -8 \\ 0 & 0 & 0 & 10 \\ \end{bmatrix}$$ which matches your final answer.