Solve the system using Gaussian elimination with back-substitution or Gauss-Jordan elimination

Here is the system:

\left\{ \begin{aligned} x_1-3x_3&=-2 \\ 3x_1+x_2-2x_3&=5 \\ 2x_1+2x_2+x_3&=4 \end{aligned} \right.

This is my very first problem actually using a matrix so here is my attempt

First I setup the augmented matrix:

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 3&1&-2&5 \\ 2&2&1&4 \end{array} \right]$$

Then I did $(-3)R_1+R_2->R_2$ which produced:

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 0&1&7&11 \\ 2&2&1&4 \end{array} \right]$$

And then I did $(-2)R_1+R_3->R_3$:

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 0&1&7&11 \\ 0&2&7&8 \end{array} \right]$$

And then $(1/2)R_3->R_3$:

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 0&1&7&11 \\ 0&1&\frac{7}{2}&4 \end{array} \right]$$

And then $(-1)R_2+R_3->R_3$

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 0&1&7&11 \\ 0&0&\frac{-7}{2}&7 \end{array} \right]$$

And finally $\left(\frac{-1}{7}\right)R_3->R_3$

$$\left[ \begin{array}{ccc|c} 1&0&-3&-2\\ 0&1&7&11 \\ 0&0&1&-1 \end{array} \right]$$

But this is not correct according to the solution. Can someone tell me where I went wrong? Clearly I did something really wrong here.

• Also, in the last step, after you fix the "augmented" column, you want to multiply the third row through by $\ - \frac{2}{7} \ ...$ (A warning about solution of systems of equations, by matrices or otherwise: arithmetic must be handled very carefully, because there is not much of anything that will cause your calculations to hang up, so you can get all the way to an "end" and only discover a problem when you check your answers in the original system of equations.) – colormegone Jan 23 '14 at 4:22

2 Answers

From your step: And then $(-1)R_2+R_3->R_3$

If you look at where you have $−11+4$.

• A second error is in the last step, but yes, that's the first mistake. – user127.0.0.1 Jan 23 '14 at 4:21

Your mistake is at row: $(-1)R_2 + R_3 \to R_3$.