RREF Calculator

In linear algebra, the "ref" stands for row echelon form, and "rref" stands for "reduced row echelon form. These methods are used to solve systems of linear equations but they have slight differences.

Row Echelon Form (REF):

A matrix is said to be in row echelon form if it satisfies the conditions such as all zero rows are at the bottom of the matrix.

The leading entry of each nonzero row is in a column to the right of the leading entry of the previous row. The leading entry in each nonzero row is 1.

Reduced Row Echelon Form (RREF):

A matrix is said to be in reduced row echelon form if it satisfies the conditions of row echelon form, and in addition:

The leading entry in each nonzero row is the only nonzero entry in its column. The leading term should be 1 in its column.

Therefore, the reduced row echelon form is a more simplified version of the row echelon form. The reduced echelon form is more often used because it's more structured and easier, especially when solving systems of linear equations.

The Reduced Row Echelon Form (RREF) and the Gauss-Jordan Elimination Method are related methods used in linear algebra for solving systems of linear equations. The differences are:

RREF:

This method is used to obtain the reduced row echelon form to find a simplified form of a matrix that makes it easier to calculate the rank, linear independence, and solutions to systems of linear equations. The resulting matrix is in reduced row echelon form, where the leading entity is 1 and zeros from above and below the diagonal value inside the matrix

Gauss-Jordan Elimination:

The Gauss-Jordan elimination method is used to solve systems of linear equations after converting it into the augmented matrix and then using the row operation to change it into reduced row echelon form. Its result is also a matrix in reduced row echelon form, but its focus is on changing the augmented matrix to the value of the variable during the process.

Therefore, both the reduced row echelon form and the Gauss-Jordan elimination method have led to the same result, but their emphasis, application, and the specific steps involved have variations.

To find the determinant of a matrix using its reduced row echelon form (RREF), you can use the following steps:

• Use the row operations to convert the original matrix into its reduced row echelon form.

Suppose we have the following 3×3 matrix:

$$ A \;=\; \left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \\ \end{matrix} \right] $$

Performing row operations on A, we get:

$$ RREF\; (A) \;=\; \left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ \end{matrix} \right] $$

• Check the number of times you swapped “s” rows during the row operations to obtain the RREF.
• Once you have the matrix in RREF, for the determinant of a given matrix, three cases are generated that show:
• If the RREF has a row of zeros, then the determinant is 0.
• If the RREF is a diagonal matrix, then the determinant is the product of the diagonal elements.
• If the RREF has row swaps “s”, then the determinant is the product of the diagonal elements multiplied by (-1)^s.

Since no row swaps in the above example when you do a row operation to get the RREF, s = 0.

So, the determinant of A is 0.

After getting the reduced row echelon form (RREF) of a matrix result that contains the vectors in question, we can find the linear independence from the structure of the RREF. Here's how you can find out that a set of vectors is linearly independent with the help of an example.

Suppose, we have the vectors v1, v2, and v3.

$$ v_1 \;=\; \left[ \begin{matrix} 1 \\ 2 \\ 3 \\ \end{matrix} \right],\; v_2 \;=\; \left[ \begin{matrix} 2 \\ 4 \\ 6 \\ \end{matrix} \right] \;and\; v_3 \;=\; \left[ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right] $$

Convert the given vector into an augmented matrix,

$$ \left[ \begin{matrix} 1 & 2 & 1 \\ 2 & 4 & 1 \\ 3 & 6 & 1 \\ \end{matrix} \right] $$

Change the given matrix into reduced row echelon form,

$$ RREF\; (A) \;=\; \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \\ \end{matrix} \right] $$

In the RREF, there are two pivot columns (v1 and v3), so there are two pivot variables. Since there are three vectors, the vectors are linearly dependent. So, in this example, the vectors v1, v2, and v3 are linearly dependent.

To find the inverse of a matrix into the reduced row echelon form (RREF), use Gaussian elimination augmented with the identity matrix such as, suppose we want to find the inverse of the matrix:

$$ A \;=\; \left( \begin{matrix} 2 & 1 \\ 5 & 3 \\ \end{matrix} \right) $$

• Take the matrix whose inverse you want to find A and augment it with the identity matrix of the same size. This results in a new matrix of the form [A | I], where I is the identity matrix.

$$ [A | I] \;=\; \left[ \begin{matrix} 1 & 2 & | & 1 & 0 \\ 3 & 4 & | & 0 & 1 \\ \end{matrix} \right] $$

Use the gaussian elimination method to transform the left side of the matrix into the identity matrix. The goal is to get the left side to I and the right side will then be the inverse of A. The augmented matrix becomes [A | I] :

$$ \left( \begin{matrix} 2 & 1 & | & 1 & 0 \\ 5 & 3 & | & 0 & 1 \\ \end{matrix} \right) $$

Divide the first row R1 by 2 (to make the leading coefficient 1),

$$ \left( \begin{matrix} 1 & 0.5 & | & 0.5 & 0 \\ 5 & 3 & | & 0 & 1 \\ \end{matrix} \right) $$

• Subtract 5 times the first row R1 from the second row R2

$$ \left( \begin{matrix} 1 & 0.5 & | & 0.5 & 0 \\ 0 & 0.5 & | & -2.5 & 1 \\ \end{matrix} \right) $$

• Divide the second row R2 by 0.5 (to make the leading coefficient 1),

$$ \left( \begin{matrix} 1 & 0.5 & | & 0.5 & 0 \\ 0 & 1 & | & -5 & 2 \\ \end{matrix} \right) $$

• Subtract 0.5 times the second row R2 from the first row R1,

$$ \left( \begin{matrix} 1 & 0 & | & 3 & -1 \\ 0 & 1 & | & -5 & 2 \\ \end{matrix} \right) $$

• The resulting matrix is the inverse of reduced row echelon form [I | A^-1] which is denoted with A^-1.

$$ A^{-1} \;=\; \left( \begin{matrix} 3 & -1 \\ -5 & 2 \\ \end{matrix} \right) $$