Elementary Row Operations

Recall that an equation such as:
7(x-4)=14,
may be solved for x by applying the following operations:
  • dividing both sides of the equation by the same value, namely 7, to yield x-4=2,
  • then adding the same quantity to both sides, namely 4, to yield x=6.
The solution is x=6, as can be verified by substituting it back into the original equation and finding the identity 14=14.

Similarly, the solution of a system of equations is any set of values of all of the variables that satisfies all of the equations simultaneously. For example the system:
has the solution:

{x = 7, y = 5, z = 3}.
This can be verified by substituting these values into all three of the equations and producing three identities.

A system of equations can be solved by generalizing the two operations described above and observing that the solution of a system of equations is not changed by:
  • dividing an both sides of an equation by a constant, or
  • subtracting a multiple of one equation from another equation.
These same operations can be applied to the rows of an augmented matrix, since each row just represents an equation. They are then called Elementary Row Operations.


The Elementary Row Operations (E.R.O.'s) are:

  • E.R.O.#1: Choose a row of the augmented matrix and divide (every element of) the row by a constant.
Example:
The notation means to divide the first row of the augmented matrix by 2 to produce the new augmented matrix.
  • E.R.O.#2: Choose any row of the augmented matrix and subtract a multiple of any other row from it (element by element).
Example:
The notation means to take row 2 and subtract 3 times row 1 from it to produce the new augmented matrix.


We will apply the E.R.O.'s in a certain sequence (the Gaussian elimination method, described below) to transform the augmented matrix into triangular echelon form. In this form the augmented matrix has 1's on the diagonal, 0's below the diagonal and any numbers above the diagonal. For example, the augmented matrix:
transformed into the triangular echelon form is:
This new augmented matrix represents the system of equations:
It is solved by back-substitution. Substituting z = 3 from the third equation into the second equation gives y = 5, and substituting z = 3 and y = 5 into the first equation gives x = 7. Thus the complete solution is:

{x = 7, y = 5, z = 3}.


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