Reduced Row Echelon Form (rref) Matrix in MATLAB

Last Updated : 14 May, 2021

Reduced Row Echelon Form of a matrix is used to find the rank of a matrix and further allows to solve a system of linear equations. A matrix is in Row Echelon form if

All rows consisting of only zeroes are at the bottom.
The first nonzero element of a nonzero row is always strictly to the right of the first nonzero element of the row above it.

Example :

$\left[ \begin{array}{ccccc} 1 & a_0 & a_1 & a_2 & a_3 \\ 0 & 0 & 2 & a_4 & a_5 \\ 0 & 0 & 0 & 1 & a_6\\ 0 & 0 & 0 & 0 & 0 \end{array} \right]$

A matrix can have several row echelon forms. A matrix is in Reduced Row Echelon Form if

It is in row echelon form.
The first nonzero element in each nonzero row is a 1.
Each column containing a nonzero as 1 has zeros in all its other entries.

Example:

$\left[{\begin{array}{ccccc}1&0&a_{1}&0&b_{1}\\0&1&a_{2}&0&b_{2}\\0&0&0&1&b_{3}\end{array}}\right]$

Where a1,a2,b1,b2,b3 are nonzero elements.

A matrix has a unique Reduced row echelon form. Matlab allows users to find Reduced Row Echelon Form using rref() method. Different syntax of rref() are:

R = rref(A)
[R,p] = rref(A)

Let us discuss the above syntaxes in detail:

rref(A)

It returns the Reduced Row Echelon Form of the matrix A using the Gauss-Jordan method.

Matlab

% creating a matrix using magic(n) 
% generates n*n matrix with values 
% from 1 to n^2 where every row sum 
% is equal to every column sum 
A = magic(4); 
disp("Matrix"); 
disp(A); 
  
% Reduced Row Echelon Form of A 
RA = rref(A); 
disp("rref :"); 
disp(RA);

Output :

rref(A)

It returns Reduced Row Echelon Form R and a vector of pivots p
p is a vector of row numbers that has a nonzero element in its Reduced Row Echelon Form.
The rank of matrix A is length(p).
R(1:length(p),1:length(p)) (First length(p) rows and length(p) columns in R) is an identity matrix.

Matlab

% creating a matrix using magic(n) 
% generates n*n matrix with values  
% from 1 to n^2 where every row sum 
% is equal to every column sum 
A = magic(5); 
disp("Matrix"); 
disp(A); 
  
% Reduced Row Echelon Form of A 
[RA,p] = rref(A); 
disp("rref :"); 
disp(RA); 
  
% Displaying pivot vector p 
disp("Pivot vector"); 
disp(p);

Output :

Finding solutions to a system of linear equations using Reduced Row Echelon Form:

The System of linear equations is $x+y+z=6\\ x+2y+3z=14\\ x+4y+7z=30$

Coefficient matrix A is $\left[ \begin{array}{ccccc} 1 & 1 & 1 \\ 1 & 2 & 3\\ 1 & 4 & 7 \end{array} \right]$

Constant matrix B is $\left[ \begin{array}{ccccc} 6 \\ 14\\ 30 \end{array} \right]$

Then Augmented matrix [AB] is $\left[ \begin{array}{ccccc} 1 & 1 & 1 & 6\\ 1 & 2 & 3 & 14\\ 1 & 4 & 7 & 30 \end{array} \right]$

Matlab

% Coefficient matrix 
A = [1  1  1; 
     1  2  3; 
     1  4  7]; 
       
% Constant matrix 
b = [6 ;14; 30]; 
  
% Augmented matrix 
M = [A b]; 
disp("Augmented matrix"); 
disp(M) 
  
% Reduced Row echelon form of  
% Augmented matrix 
R = rref(M); 
disp("rref"); 
disp(R)