Row reduction to echelon form in ℤ/pℤ

### 11.8.17 Row reduction to echelon form in ℤ/pℤ

The rref command finds the reduced row echelon form of a matrix with elements in ℤ/pℤ (see 15.7.3).

- rref takes A, a matrix in ℤ/pℤ.
- rref(A) returns the echelon form of A.

#### Example

rref([[0,2,9]%15,[1,10,1]%15,[2,3,4]%15])

⎡ ⎢ ⎢ ⎣ |
1%15 | 0%15 | 0%15
0%15 | 1%15 | 0%15
0%15 | 0%15 | 1%15
| ⎤ ⎥ ⎥ ⎦

This can be used to solve a linear system of equations with coefficients in ℤ/pℤ by rewriting it in matrix form

A X=B. (1)

rref can then take as argument the augmented matrix of the system (the matrix obtained by augmenting matrix A to the right with the column vector B).

rref returns a matrix [A1,B1] where A1 has ones on its principal diagonal and zeros outside. The solutions in ℤ/pℤ of:

A1 X=B1

are the same as the solutions of (1).

#### Example

Solve in ℤ/13ℤ:

| ⎧ ⎨ ⎩ |
x | + | 2y | = | 9,
3x | + | 10y | = | 0.

rref([[1,2,9]%13,[3,10,0]%13])

or:

rref([[1,2,9],[3,10,0]])%13

⎡ ⎢ ⎣ |
1%13 | 0%13 | 3%13
0%13 | 1%13 | 3%13
| ⎤ ⎥ ⎦

hence x=3%13 and y=3%13.