A First Course in Linear Algebra, 2017a

1.2. Systems of Equations, Algebraic Procedures 29
we used in the previous section. First, we construct the augmented matrix, given by
[ 2 1 −1
] 0
1 2 −2 0
Then, we carry this matrix to its reduced row-echelon form, given below.
[ 1 0 0
] 0
0 1 −1 0
The corresponding system of equations is
x = 0
y − z = 0
Since z is not restrained by any equation, we know that this variable will become our parameter. Let z = t
where t is any number. Therefore, our solution has the form
x = 0
y = z = t
z = t
Hence this system has infinitely many solutions, with one parameter t.
♠
Suppose we were to write the solution to the previous example in another form. Specifically,
can be written as
⎡
⎣
x
y
z
⎤
⎦ = ⎣
x = 0
y = 0 +t
z = 0 +t
⎡
0
0
0
⎤
⎡
⎦ +t ⎣
Notice that we have constructed a column from the constants in the solution (all equal to 0), as well as a
column corresponding to the coefficients on t in each equation. While we will discuss this form of solution
more in further ⎡chapters, ⎤ for now consider the column of coefficients of the parameter t. In this case, this
0
is the column ⎣ 1 ⎦.
1
There is a special name for this column, which is basic solution. The basic solutions of a system are
columns constructed from the coefficients on parameters in the solution. We often denote basic solutions
by X 1 ⎡,X 2 etc., ⎤ depending on how many solutions occur. Therefore, Example 1.30 has the basic solution
0
X 1 = ⎣ 1 ⎦.
1
We explore this further in the following example.
0
1
1
⎤
⎦