Linear Algebra, 2020a

Section I. Solving Linear Systems 3
1.3 Example The ordered pair (−1, 5) is a solution of this system.
In contrast, (5, −1) is not a solution.
3x 1 + 2x 2 = 7
−x 1 + x 2 = 6
Finding the set of all solutions is solving the system. We don’t need guesswork
or good luck; there is an algorithm that always works. This algorithm is Gauss’s
Method (or Gaussian elimination or linear elimination).
1.4 Example To solve this system
3x 3 = 9
x 1 + 5x 2 − 2x 3 = 2
1
3 x 1 + 2x 2 = 3
we transform it, step by step, until it is in a form that we can easily solve.
The first transformation rewrites the system by interchanging the first and
third row.
swap row 1 with row 3
−→
1
3 x 1 + 2x 2 = 3
x 1 + 5x 2 − 2x 3 = 2
3x 3 = 9
The second transformation rescales the first row by a factor of 3.
multiply row 1 by 3
−→
x 1 + 6x 2 = 9
x 1 + 5x 2 − 2x 3 = 2
3x 3 = 9
The third transformation is the only nontrivial one in this example. We mentally
multiply both sides of the first row by −1, mentally add that to the second row,
and write the result in as the new second row.
add −1 times row 1 to row 2
−→
x 1 + 6x 2 = 9
−x 2 − 2x 3 =−7
3x 3 = 9
These steps have brought the system to a form where we can easily find the
value of each variable. The bottom equation shows that x 3 = 3. Substituting 3
for x 3 in the middle equation shows that x 2 = 1. Substituting those two into
the top equation gives that x 1 = 3. Thus the system has a unique solution; the
solution set is {(3, 1, 3)}.
We will use Gauss’s Method throughout the book. It is fast and easy. We
will now show that it is also safe: Gauss’s Method never loses solutions nor does
it ever pick up extraneous solutions, so that a tuple is a solution to the system
before we apply the method if and only if it is a solution after.