This example problem finds the solution to a system of equations containing two equations and two unknowns. Two different techniques are used to find the solution, substitution and graphing.This video also describes the meaning of a consistent systems of equations and inconsistent system of equations.

Systems of linear equations can be solved by using elementary row operations to manipulate the augmented matrix into a row reduced echelon form or a simplified form where the equations can easily be solved using back substitution.In this first example, the system of equations has a unique solution and this solution is found by performing a sequence of row operations. In the subsequent videos a system of equations with no solutions is examined, as well as a system of equations with an infinite number of equations.

In this second example, the system of equations does not have a solution. During the row operations we encounter a mathematical contradiction which tells us the system has no solution.

In this final example, the system of equations investigated has an infinite number of solutions as the variable x3 turns out to be a free variable.

This problem also investigates the solution of a system of linear equations. However, in this problem one coefficient within the system of equations is a variable. Using the augmented matrix of the system we determine for what values of this unknown the system has a solution (i.e. it's consistent), and for what values it does not (i.e. the system of equations is inconsistent).

This example performs row operations on a matrix to obtain a row reduced echelon form matrix. This matrix form has the following structure:1) The first non-zero entry in each row is a one. The column with this one is called a pivot column.2) A pivot column has zeros in every other row (except where the one is located)3) Any rows consisting of all zeros are placed at the bottom of the matrix.

We've considered representing linear systems of equations in an augmented matrix form in previous videos and examined how to solve them using simple row operations to manipulate them into a row reduced echelon form.This video provides 3 additional examples of finding the solution of a system of linear equations after the augmented matrix has been manipulated into row reduced echelon form. We don't work through the row operation details, but instead focus on identifying basic variables, free variables, and constructing the system of equation solution.

We’ve looked at the equation Ax = b several times in previous videos. In this video we investigate the same equation for the special case where b = 0 (i.e. the all-zero vector). When b = 0, we call the equation Ax = 0 a homogeneous system of equations. A homogeneous system of equations will always have x = 0 as a trivial solution to the system of equations. We’re often interested in finding if the system has a non-trivial solution. This can be determined by using our typical strategy for solving systems of equations, namely:Construct the augmented matrix, row reduce, identify basic and free variables, write the parametric solution.