R = rref( A ) returns the reduced row echelon form of A using Gauss-Jordan elimination with partial pivoting.
R = rref( A , tol ) specifies a pivot tolerance that the algorithm uses to determine negligible columns.
[ R , p ] = rref( A ) also returns the nonzero pivots p .
Create a matrix and calculate the reduced row echelon form. In this form, the matrix has leading 1s in the pivot position of each column.
A = magic(3)
A = 3×3 8 1 6 3 5 7 4 9 2
RA = rref(A)
RA = 3×3 1 0 0 0 1 0 0 0 1
The 3-by-3 magic square matrix is full rank, so the reduced row echelon form is an identity matrix.
Now, calculate the reduced row echelon form of the 4-by-4 magic square matrix. Specify two outputs to return the nonzero pivot columns. Since this matrix is rank deficient, the result is not an identity matrix.
B = magic(4)
B = 4×4 16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1
[RB,p] = rref(B)
RB = 4×4 1 0 0 1 0 1 0 3 0 0 1 -3 0 0 0 0
p = 1×3 1 2 3
Use Gauss-Jordan elimination on augmented matrices to solve a linear system and calculate the matrix inverse. These techniques are mainly of academic interest, since there are more efficient and numerically stable ways to calculate these values.
Create a 3-by-3 magic square matrix. Add an additional column to the end of the matrix. This augmented matrix represents a linear system Ax = b , with the extra column corresponding to b .
A = magic(3); A(:,4) = [1; 1; 1]
A = 3×4 8 1 6 1 3 5 7 1 4 9 2 1
Calculate the reduced row echelon form of A . Index into R to extract the entries in the extra (augmented) column, which contains the solution to the linear system.
R = rref(A)
R = 3×4 1.0000 0 0 0.0667 0 1.0000 0 0.0667 0 0 1.0000 0.0667
x = R(:,end)
x = 3×1 0.0667 0.0667 0.0667
A more efficient way to solve this linear system is with the backslash operator, x = A\b .
Create a similar magic square matrix, but this time append an identity matrix of the same size to the end columns.
A = [magic(3) eye(3)]
A = 3×6 8 1 6 1 0 0 3 5 7 0 1 0 4 9 2 0 0 1
Calculate the reduced row echelon form of A . In this form the extra columns contain the inverse matrix for the 3-by-3 magic square matrix.
R = rref(A)
R = 3×6 1.0000 0 0 0.1472 -0.1444 0.0639 0 1.0000 0 -0.0611 0.0222 0.1056 0 0 1.0000 -0.0194 0.1889 -0.1028
inv_A = R(:,4:end)
inv_A = 3×3 0.1472 -0.1444 0.0639 -0.0611 0.0222 0.1056 -0.0194 0.1889 -0.1028
A more efficient way to calculate the inverse matrix is with inv(A) .
Consider a linear system of equations with four equations and three unknowns.
x 1 + x 2 + 5 x 3 = 6 2 x 1 + x 2 + 8 x 3 = 8 x 1 + 2 x 2 + 7 x 3 = 10 - x 1 + x 2 - x 3 = 2 .
Create an augmented matrix that represents the system of equations.
A = [1 1 5; 2 1 8; 1 2 7; -1 1 -1]; b = [6 8 10 2]'; M = [A b];
Use rref to express the system in reduced row echelon form.
R = rref(M)
R = 4×4 1 0 3 2 0 1 2 4 0 0 0 0 0 0 0 0
The first two rows of R contain equations that express x 1 and x 2 in terms of x 3 . The second two rows imply that there exists at least one solution that fits the right-hand side vector (otherwise one of the equations would read 1 = 0 ). The third column does not contain a pivot, so x 3 is an independent variable. Therefore, there are infinitely many solutions for x 1 and x 2 , and x 3 can be chosen freely.
x 1 = 2 - 3 x 3 x 2 = 4 - 2 x 3 .
For example, if x 3 = 1 , then x 1 = - 1 and x 2 = 2 .
From a numerical standpoint, a more efficient way to solve this system of equations is with x0 = A\b , which (for a rectangular matrix A ) calculates the least-squares solution. In that case, you can check the accuracy of the solution with norm(A*x0-b)/norm(b) and the uniqueness of the solution by checking if rank(A) is equal to the number of unknowns. If more than one solution exists, then they all have the form of x = x 0 + nt , where n is the null space null(A) and t can be chosen freely.