Week 5, Day 1

A System With No Solution:

When solving a system of linear equations, you should remember that it is possible for the system to have no solution. If, in the elimination process, you obtain a row with zeros except for the last entry, it is unnecessary to continue the elimination process. You can simply conclude that the system is inconsistent. The following is an example of a system with no solution:

x - y + 2y = 4

x + 0y + z = 6

2x - 3y + 5z = 4

3x + 2y - z = 1

Note that the third row of this matrix consists of zeros except for the last entry. This means that the original system of linear equations is inconsistent. You can see why this is true by converting back to a system of linear equations.

x - y + 2z = 4

0x + y - z = 2

0 = -2

0x + 5y - 7y = -11

Because the third equation is not possible, the system has no solution. 

A System With Infinite Solutions:

When solving a system of linear equations, it is also possible to produce an answer with infinitely many solutions. This occurs when there are two equations to solve for three variables. The following is an example of a system with an infinite number of solutions:

2x + 4y - 2z = 0

3x + 5y + 0z = 1

The corresponding system of equations is:

x + 5z = 2

y - 3z = -1

Solving for x and y in terms of z, you have x = -5z + 2 and y = 3z -1. Then, letting z = a, the solution set has the form: ( -5a + 2, 3a - 1, a) where a is a real number. Try substituting values for a to obtain a few solutions. Then check each solution in the original system of equations.