linear-guest

G.2 Systems of Linear Equations 371
Solution set in set notation
Here is an augmented matrix, let’s think about what the solution set looks
like (
1 0 3
)
2
0 1 0 1
This looks like the system
1 · x 1 + 3x 3 = 2
1 · x 2 = 1
Notice that ( when ) the system is written this way the copy of the 2 × 2 identity
1 0
matrix
makes it easy to write a solution in terms of the variables
0 1
( 3
x 1 and x 2 . We will call x 1 and x 2 the pivot variables. The third column
0)
does not look like part of an identity matrix, and there is no 3 × 3 identity
in the augmented matrix. Notice there are more variables than equations and
that this means we will have to write the solutions for the system in terms of
the variable x 3 . We’ll call x 3 the free variable.
Let x 3 = µ. (We could also just add a ‘‘dummy’’ equation x 3 = x 3 .) Then we
can rewrite the first equation in our system
x 1 + 3x 3 = 2
x 1 + 3µ = 2
x 1 = 2 − 3µ.
Then since the second equation doesn’t depend on µ we can keep the equation
and for a third equation we can write
x 2 = 1,
x 3 = µ
so that we get the system
⎛ ⎞
x 1
⎝x 2
⎠ =
x 3
=
=
⎛ ⎞
2 − 3µ
⎝ 1⎠
µ
⎛ ⎞ ⎛ ⎞
2 −3µ
⎝1⎠ + ⎝ 0⎠
0 µ
⎛ ⎞ ⎛ ⎞
2 −3
⎝1⎠ + µ ⎝ 0⎠ .
0 1
371