Linear Algebra/Gauss' Method

Finding the set of all solutions is solving the system. No guesswork or good fortune is needed to solve a linear system. There is an algorithm that always works. The next example introduces that algorithm, called Gauss' method. It transforms the system, step by step, into one with a form that is easily solved.

Most of this subsection and the next one consists of examples of solving linear systems by Gauss' method. We will use it throughout this book. It is fast and easy. But, before we get to those examples, we will first show that this method is also safe in that it never loses solutions or picks up extraneous solutions.

Each of those three operations has a restriction. Multiplying a row by $$0$$ is not allowed because obviously that can change the solution set of the system. Similarly, adding a multiple of a row to itself is not allowed because adding $$-1$$ times the row to itself has the effect of multiplying the row by $$0$$. Finally, swapping a row with itself is disallowed to make some results in the fourth chapter easier to state and remember (and besides, self-swapping doesn't accomplish anything).

When writing out the calculations, we will abbreviate "row $$i$$" by "$$\rho_i$$". For instance, we will denote a pivot operation by $$k\rho_i+\rho_j$$, with the row that is changed written second. We will also, to save writing, often list pivot steps together when they use the same $$\rho_i$$.

As these examples illustrate, Gauss' method uses the elementary reduction operations to set up back-substitution.

Strictly speaking, the operation of rescaling rows is not needed to solve linear systems. We have included it because we will use it later in this chapter as part of a variation on Gauss' method, the Gauss-Jordan method.

All of the systems seen so far have the same number of equations as unknowns. All of them have a solution, and for all of them there is only one solution. We finish this subsection by seeing for contrast some other things that can happen.

That example's system has more equations than variables. Gauss' method is also useful on systems with more variables than equations. Many examples are in the next subsection.

Another way that linear systems can differ from the examples shown earlier is that some linear systems do not have a unique solution. This can happen in two ways.

The first is that it can fail to have any solution at all.

The other way that a linear system can fail to have a unique solution is to have many solutions.

Don't be fooled by the "$$ 0=0 $$" equation in that example. It is not the signal that a system has many solutions.

We will finish this subsection with a summary of what we've seen so far about Gauss' method.

Gauss' method uses the three row operations to set a system up for back substitution. If any step shows a contradictory equation then we can stop with the conclusion that the system has no solutions. If we reach echelon form without a contradictory equation, and each variable is a leading variable in its row, then the system has a unique solution and we find it by back substitution. Finally, if we reach echelon form without a contradictory equation, and there is not a unique solution (at least one variable is not a leading variable) then the system has many solutions.

The next subsection deals with the third case&mdash; we will see how to describe the solution set of a system with many solutions.

Exercises
/Solutions/