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1 Part 2: Systems of Equations Which Do Not Have A Unique Solution On the previous pages we learned how to solve systems of equations using Gaussian elimination. In each of the examples and exercises of part 1(except for exercise 1 parts d and e) the systems of equations had a unique solution. That is, a single value for each of the variables. In example 3 we found the solution to be  7 23 3,  . This means that the graphs of the two lines in example 3 intersect at this unique point. In 2-space, the xy-plane, we have the geometric bonus of being able to draw a picture of the solutions to a system of two equations two unknowns. Clearly, if we were asked to draw the graphs of two lines in the xy-plane we have 3 basic choices/cases: 1. Draw the two lines so they intersect. This point of intersection can only happen once for a given pair of lines. That is, the two lines intersect in a unique point. There is a unique common solution to the system of equations. Discussed in part 1. 2. Draw the two lines so that one is on "top of" the other. In this case there are an infinite number of common points, an infinite number of solutions to the given system. Discussed in part 2. 3. Draw two parallel lines. In this case there are no points common to both lines. There is no solution to the system of equations that describe the lines. Discussed in part 2. The 3 cases above apply to any system of equations. Theorem 1. For any system of m equations with n unknowns (m < n) one of the following cases applies: 1. There is a unique solution to the system. 2. There is an infinite number of solutions to the system. 3. There are no solutions to the system. Again, in this section of the notes we will illustrate cases 2 and 3. To solve systems of equations where these cases apply we use the matrix procedure developed previously. Example 6. Solve the system x + 2y = 1 2x + 4y = 2 2 It is probably already clear to the reader that the second equation is really the first in disguise. (Simply divide both sides of the second equation by 2 to obtain the first). So if we were to draw the graph of both we would obtain the same line, hence have an infinite number of points common to both lines, an infinite number of solutions. However it would be helpful in solving other systems where the solutions may not be so apparent to do the problem algebraically, using matrices. The matrix of the system with its simplification follows. Recall, we try to express the matrix 1 2 1 2 4 2       in the form 1 2 1 0 0 1 b b       from which we can read off the solution. However after one step we note that 1 2 1 2 4 2       1 22 R R  1 2 1 0 0 0       . It should be clear to the reader that no matter what further elementary row operations we perform on the matrix 1 2 1 0 0 0       we cannot change it to the form we hoped for, namel.

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