# 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.) Jan 21, 2014 at 21:39
• @DavidMitra you're right, it is $0=6$ Jan 21, 2014 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 $0=10$ (!). Jan 21, 2014 at 21:43
• @DavidMitra haha thanks! Jan 21, 2014 at 21:44
• So, long story short: carefully check your work. You seem to have the procedure down pat. Jan 21, 2014 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.