Reduced Row-Echelon Form

Reduced Row-Echelon Form (RREF) is a special form of a matrix in which each leading entry is 1 and is the only non-zero entry in its column. It is commonly used to solve systems of linear equations and simplify matrix calculations.

Row Echelon Form

In linear algebra, a matrix is in row echelon form if it is obtained as the result of Gaussian Elimination on the rows of that matrix. This form of the matrix must satisfy some conditions which are discussed below:

• The leading non-zero element or pivot of every row is at the right of the leading element of the first row.
• Any row consisting entirely of zeros comes at the bottom of the matrix.

Example:

\begin{bmatrix}1 &6 &9 \\0& 0& 8 \\0& 0& 0 \\\end{bmatrix}

Here, matrix C is in row-echelon form. The first and second rows are non-zero rows. The third row is a zero row.

In reduced row echelon form, a matrix must satisfy the following properties:

• The leading non-zero entry is 1
• Leading 1 is the only nonzero entry in its column
• The leading 1 in the rows is always right of the leading 1 of the first row
• If there are any rows containing all zeros, they should be at the bottom of the matrix.

Example 1:

\begin{bmatrix}1 &0& 0 \\0& 1& 0 \\0& 0 &0 \\\end{bmatrix}

The above matrix is in reduced row-echelon form. The third row is the zero row which is at the bottom of the non-zero rows. The first and second row are non-zero rows. A₁₁ & A₂₃ are non-zero entries and equal to 1.

Example 2:

\begin{bmatrix}1 0 0 & & \\0 1 0 & & \\0 0 1 & & \\\end{bmatrix}

The above matrix is in reduced row-echelon form. All rows are non-zero rows.

Solved Example

Solve the following system of equations

• x + y + z = 2
• x + 2y + 3z = 7
• 2x + 3y + 4z = 13

Solution:

The given linear equations are

x + y + z = 2

x + 2y + 3z = 7

2x + 3y + 4z = 13

The augmented matrix of these linear equations is

1 1 1 : 2

1 2 3 : 7

2 3 4 : 13

Now subtracting R₁ from R₂, i.e., (R₂=R₂-R₁)

1 1 1 : 2

0 1 2 : 5

2 3 4 : 13

Now subtracting 2R₁ from R₃, i.e., (R₃=R₃-2R₁), we get,

1 1 1 : 2

0 1 2 : 5

0 1 2 : 9

Now we will subtract R₂ from R₁, after that we will get new elements of R₁

1 0 -1 : -3

0 1 2 : 5

0 1 2 : 9

Lastly we will subtract R₂ from R₃ to get new elements of R₃

1 0 -1 : -3

0 1 2 : 5

0 0 0 : 4

Now putting the values of x, y and z in the equations, we get,

x - y = -3

y + 2z = 5

0 = 4 , which is impossible.

∴The above system of equations have no solutions.

Steps to Transformation of a Matrix to Reduced Row-Echelon Form

The step-by-step process to convert a matrix to reduced row echelon form is discussed below:

Let us consider a matrix

\begin{bmatrix} 1 & 2 & -1 & -4 \\ 2& 3& -1& -11 \\-2 &0 & -3& 22 & \\\end{bmatrix}

Now, we will discuss the steps one by one,

1. Make the leading entry of the first row = 1

If there is a 1 in the leading entry of the first row then, no change is required in this step. We have 1 in the leading entry in the given matrix A.

2. Eliminate the other entries in the first column

• First, we will make changes in the second row of the matrix: R2 = R2 - 2R1
• Then, for the third row: R3 = R3 + 2R1

1 2 -1 -4

A = 0 -1 1 -3

0 4 -5 14

3. Make the leading entry of the second row = 1

After the first row and column, we have to make the leading entry of the second row = 1: R2 = - R2

1 2 -1 -4

A = 0 1 -1 3

0 4 -5 14

4. Eliminate the other entries in the second column

• First, we will make changes in the first row: R1 = R1 - 2R2
• For the third row: R3 = R3 - 4R2

1 0 1 -10

A = 0 1 -1 3

0 0 -1 2

5. Make the leading entry of the third row = 1

Now we have to make the leading entry of the third row = 1.

Therefore, R3 = -R3

1 0 1 -10

A = 0 1 -1 3

0 0 1 -2

6. Eliminate the other entries in the third column

• First, we will make changes in the first row: R1 = R1 - R3
• then, for the second row: R2 = R2 + R3

1 0 0 -8

A = 0 1 0 1

0 0 1 -2

Finally, we get the reduced row echelon form of the matrix A

1 0 0 -8

A = 0 1 0 1

0 0 1 -2

Solved Problems

Find the reduced row echelon form of the given matrix.

3 −2 −3 3

2 3 3 2

Solution:

The given matrix is

3 −2 −3 3

2 3 3 2

Now, R1 - R2 → R1

1 −5 −6 1

2 3 3 2

Now, R2 - 2R1 → R2

1 −5 −6 1

0 13 15 0

Now, 1/13 R2 → R2

1 −5 −6 1

0 1 15/13 0

At the end, R1 + 5R2→ R1

1 0 -3/13 1

0 1 15/13 0

∴ The row reduced echelon form of the given matrix is

1 0 -3/13 1

0 1 15/13 0

Practice Questions

Q1. Solve the following system of equations using Gaussian Elimination.

• 2x - 2y = -6
• x - y + z = 1
• 3y - 2z = -7

Q2. Solve this system of equations using Gaussian Elimination.

• x + y = 5
• 3x - 2y = 6

Q3. Solve this system of equations using Gaussian Elimination.

• -x + 4y - 2z = -15
• -4x + 6y + z = - 5
• -6x - 6y - 2z = - 10

Q4. Is the following matrix in reduced row echelon form?

5 0 0 7

0 1 0 5

0 0 1 4

Q5. Solve the following system of equations.

x + y = 2

3x + 4y = 5

4x + 5y = 9