# Linear Algebra/Gauss' Method

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|previous=Solving Linear Systems

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--- we will see how to
describe the solution set of a system with many solutions.

## Exercises


## References

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