Elementary Row Operations (ERO)

In linear algebra, two matrices are row equivalent if one can be changed to the other by a sequence of elementary row operations. Alternatively, two $m \times n$matrices are row equivalent if and only if they have the same row space. The row space of a matrix is the set of all possible linear combinations of its row vectors. If the rows of the matrix represent a system of linear equations, then the row space consists of all linear equations that can be deduced algebraically from those in the system. Two matrices of the same size are row equivalent if and only if the corresponding homogeneous systems have the same set of solutions, or equivalently the matrices have the same null space. Because elementary row operations are reversible, row equivalence is an equivalence relation. It is commonly denoted by a tilde (~).  

An elementary row operation is any one of the following three moves:

1. Row switching (swap): Swap two rows of a matrix.
2. Row multiplication (scale): Multiply a row of a matrix by a nonzero constant.
3. Row addition (pivot): Add to one row of a matrix some multiple of another row.

Produce Equivalent Matrices Using Elementary Row Operations

Since the matrix is essentially the coefficients and constants of a linear system, the three row operations preserve the matrix. For example, swapping two rows simply means switching their position within the matrix. Also, when solving a system of linear equations by the elimination method, row multiplication would be the same as multiplying the whole equation by a number to obtain additive inverses so that a variable cancels. Finally, row addition is also the same as the elimination method, when one chooses to add or subtract the like terms of the equations to obtain the variable. Therefore, row operations preserve the matrix and can be used as an alternative method to solve a system of equations.

Example 1:  Show that these two matrices are row equivalent:

$\displaystyle A=\begin{pmatrix} 1 & -1 & 0 \\ 2 & 1 & 1 \end{pmatrix}\quad B=\begin{pmatrix} 3 & 0 & 1 \\ 0 & 3 & 1 \end{pmatrix}$

Start with $A$, add the second row to the first:

$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 2 & 1 & 1 \end{pmatrix}$

Then, multiply the second row by 3 and then subtract the first row from the second:

$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 3 & 3 & 2 \end{pmatrix}$

Finally, subtract the first row from the second:

$\displaystyle A=\begin{pmatrix} 3 & 0 & 1 \\ 0 & 3 & 1 \end{pmatrix}$

You can see that $A=B$, which we achieved through a series of elementary row operations.

Row Reduction:  Solving a System of Linear Equations

In row reduction, the linear system:

$\displaystyle x+3y-2z=5 \\ 3x+5y+6z=7 \\ 2x+4y+3z=8$   

Is represented as an augmented matrix:

$\displaystyle A=\begin{pmatrix} 1 & 3 & -2 & 5 \\ 3 & 5 & 6 & 7 \\ 2 & 4 & 3 & 8 \end{pmatrix}$

This matrix is then modified using elementary row operations until it reaches reduced row echelon form. 

Because these operations are reversible, the augmented matrix produced always represents a linear system that is equivalent to the original.

There are several specific algorithms to row-reduce an augmented matrix, the simplest of which are Gaussian elimination and Gauss-Jordan elimination. This computation can be done by hand (using the three types of ERO) or on the calculator under the matrix function 'rref' (reduced row echelon form). 

The final matrix is in reduced row echelon form, and represents the system $x=-15$, $y=8$$z=2$.

$\displaystyle A=\begin{pmatrix} 1 & 0 & 0 & -15 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & 2 \end{pmatrix}$