MATRIX Course

Gauss Example

Go back

We will solve by ourselves the following system, using the GAUSS method. If you're wondering how you can do that in R, this was explained on this page.

\[ \begin{split}\left \{ \begin{array}{r c l} 0 - x_2 + 2x_3 + 13x_4 = 5 \\ x_1 - 2x_2 + 3x_3 + 17x_4 = 4 \\ -x_1 + 3x_2 - 3x_3 - 20x_4 = -1 \\ \end{array} \right .\end{split} \]

The corresponding matrix is

\[ \begin{split}\ \begin{pmatrix} \ 0 & -1 & 2 & 13 & 5 \\ \ 1 & -2 & 3 & 17 & 4 \\ \ -1 & 3 & -3 & -20 & -1 \end{pmatrix}\end{split} \]

Step 1

Operation: $L_1 \iff L_2$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 3 & 17 & 4 \\ \ 0 & -1 & 2 & 13 & 5 \\ \ -1 & 3 & -3 & -20 & -1 \end{pmatrix}\end{split} \]

Step 2

Operation: $L_3 = L_3 + L_1$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 3 & 17 & 4 \\ \ 0 & -1 & 2 & 13 & 5 \\ \ 0 & 1 & 0 & -3 & 3 \end{pmatrix}\end{split} \]

Step 3

Operation: $L_3 = L_3 + L_2$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 3 & 17 & 4 \\ \ 0 & -1 & 2 & 13 & 5 \\ \ 0 & 0 & 2 & 10 & 8 \end{pmatrix}\end{split} \]

At this point, the matrix is in the row echelon form. We could write back the system

\[ \begin{split}\left \{ \begin{array}{r c l} x_1 - 2x_2 + 3x_3 + 17x_4 = 4\\ - x_2 + 2x_3 + 13x_4 = 5\\ 2x_3 + 10x_4 = 8\\ \end{array} \right .\end{split} \]

But let's make the reduced row echelon form!


Step 4

Operation: $L_3 = L_3/2$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 3 & 17 & 4 \\ \ 0 & -1 & 2 & 13 & 5 \\ \ 0 & 0 & 1 & 5 & 4 \end{pmatrix}\end{split} \]

Step 5

Operation: $L_3 = L_1/-1$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 3 & 17 & 4 \\ \ 0 & 1 & -2 & -13 & -5 \\ \ 0 & 0 & 1 & 5 & 4 \end{pmatrix}\end{split} \]

Step 6

Operation: $L_1 = L_1 - 3 L_3$ and $L_2 = L_2 + 2 L_3$

\[ \begin{split}\ \begin{pmatrix} \ 1 & -2 & 0 & 2 & -8 \\ \ 0 & 1 & 0 & -3 & 3 \\ \ 0 & 0 & 1 & 5 & 4 \end{pmatrix}\end{split} \]

Step 7

Operation: $L_1 = L_1 + 2 L_2$

\[ \begin{split}\ \begin{pmatrix} \ 1 & 0 & 0 & -4 & -2 \\ \ 0 & 1 & 0 & -3 & 3 \\ \ 0 & 0 & 1 & 5 & 4 \end{pmatrix}\end{split} \]

Result

We are converting it back to a system

\[ \begin{split}\left \{ \begin{array}{r c l} x_1 - 4x_4 = -2\\ x_2 - 3x_4 = 3\\ x_3 + 5x_4 = 4\\ \end{array} \right .\end{split} \]

Giving us

\[ \begin{split}\left \{ \begin{array}{r c l} x_1 = -2 + 4x_4\\ x_2 = 3 + 3x_4\\ x_3 = 4 - 5x_4\\ \end{array} \right .\end{split} \]