Solving a matrix with Gauss

Go back

The goal is to get a matrix with a strictly increasing number of zeros before the coefficients, called leading coefficients/pivots (pivots).

If the leading coefficients are all $1$, and on each coefficient's column we got only $0$, then the matrix is in a form called the reduced row echelon form (matrice échelonnée réduite). Otherwise, it's called row echelon form (matrice échelonnée).

The operations you can use are

  • Swapping two rows: $L_i \iff L_j$
  • Adding a row: $L_i = L_i + L_j$
  • Adding a row, multiplied by $c$: $L_i \leftarrow L_i + cL_j$
  • Subtracting a row: $L_i \leftarrow L_i + -1 * L_j$
  • Dividing a row by $c$: $L_i \leftarrow L_i + \frac{1}{c} * L_j$
  • Multiply a row by $c$: $L_i \leftarrow c * L_i$

And to summarize, you can

  • Swapping two rows: $L_i \iff L_j$
  • Use this formula: $L_i \leftarrow a * L_i + b * L_j$

To summarize, what you will have to do is to use the operations above, and extract a pivot for each line, while making sure that this column does not have a pivot yet. Then, you may sort the lines if needed to make a strictly increasing sequence of non-leading zero.