# Row reduction to echelon form in ℤ/pℤ: rref

### 5.34.17 Row reduction to echelon form in ℤ/pℤ: rref

The rref command can find the reduced row echelon form of a matrix with elements in ℤ/pℤ. (See 5.56.3):

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

#### Example.

Input:

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

Output:

⎡
⎢
⎢
⎣

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

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 will returns a matrix [A1,B1] where A1 has 1s on its principal diagonal and zeros outside. The solutions in ℤ/pℤ of:

A1· X = B1

are the same as the solutions of:

A*X=B

A· X = B

#### Example.

Solve in ℤ/13ℤ

⎧
⎨
⎩

x + 2 · y | = | 9
3 · x +10 · y | = | 0

Input:

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

or:

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

Output:

⎡
⎢
⎣

1%13 | 0%13 | 3%13
0%13 | 1%13 | 3%13

⎤
⎥
⎦

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