Chapter 1
Section 1.3
Consistent Systems of Linear
Equations
Example
Find the augmented matrix for the 3 × 5
system of equations to the right and row
reduce it to reduced echelon form.
3 −6 9
0
−3 18
−1 2 −3 2
11 2
2 −4 6 −2 −12 4
1
−1
2
−2 3
2 −3
−4 6
1 −2
0 0
0 0
0
2
−2
−1 6
11 2
−12 4
3 0
−1 6
0 2
10 8
0 −2 −10 −8
⅓R1
R1+R2
-2R1+R3
½R2
1 −2
0 0
0 0
1
0
0
3 0
−1 6
0 1
5
4
0 −2 −10 −8
−2 3 0
0 0 1
0 0 0
−1 6
5 4
0 0
3𝑥1 − 6𝑥2 + 9𝑥3
− 3𝑥5 = 18
−𝑥1 + 2𝑥2 − 3𝑥3 + 2𝑥4 + 11𝑥5 = 2
2𝑥1 − 4𝑥2 + 6𝑥3 − 2𝑥4 − 12𝑥5 = 4
Rewriting the equivalent system:
𝑥1 − 2𝑥2 + 3𝑥3
− 𝑥5 = 6
𝑥4 + 5𝑥5 = 4
We can rewrite this so that the variables
whose coefficient is the leading 1 in row of
the matrix in reduced echelon form is on the
right side of the equation and all other
variables are on the left.
𝑥1 = 6 + 2𝑥2 − 3𝑥3 + 𝑥5
𝑥4 = 4 − 5𝑥5
General
Solution
Find the solution for: 𝑥2 = 1, 𝑥3 = 0, 𝑥5 = −2
2R2+R3
𝑥1 = 6
𝑥2 = 1
𝑥3 = 0
𝑥4 = 14
𝑥5 = −2
Particular Solution
Dependent vs Independent Variables
The reduced echelon form of an augmented matrix for a consistent system of equations
will never have a leading 1 in the last (or augmented) column. Each of the non-augmented
columns corresponds to a variable in the system. Each of the columns with a leading 1 the
corresponding variable appears on right side when the general solution is expressed.
The variables that correspond to the columns with the leading 1’s are the dependent
(constrained or determined) variables.
The variables that correspond to the columns without the leading 1’s are the independent
(unconstrained or free) variables.
1 −2 3 0
0 0 0 1
0 0 0 0
−1 6
5 4
0 0
𝑥1 = 6 + 2𝑥2 − 3𝑥3 + 𝑥5
𝑥4 = 4 − 5𝑥5
Dependent Variables: 𝑥1 , 𝑥4
Independent Variables: 𝑥2 , 𝑥3 , 𝑥5
Given values for the independent variables the values for the dependent variables are
automatically determined according to the equations in order to form a particular
solution. An independent variable can have any value between −∞ and ∞. So a
system with an independent variable will always have an infinite number of solutions.
Examples
Each of the matrices below is the augmented matrix for a system of equations. Given the
general solution for the system in terms of the dependent variables and tell which variables
are dependent and which are independent.
General Solution
Matrix
1
0
0
0
0 −5
1 3
0 0
0 0
0
0
0
0
1
0
0
0
1
0
0
0
−1
0
0
2
0
0
0
0
1
0
0
3
−2
0
0
1
0
0
0
3
0
0
0
0
1
0
0
−4 2 0
0 0 0
0 0 0
0
0
1
0
3
7
2
0
0 2
0 −5
0 0
0 0
𝒙𝟏 = 𝟑 + 𝟓𝒙𝟑
𝒙𝟐 = 𝟕 − 𝟑𝒙𝟑 + 𝒙𝟒
𝒙𝟓 = 𝟐
𝑥2 = 2 − 2𝑥3 − 3𝑥5
𝑥4 = −5 + 2𝑥5
Variables
Dependent: 𝑥1 , 𝑥2 , 𝑥5
Independent: 𝑥3 , 𝑥4
Dependent: 𝑥2 , 𝑥4
Independent: 𝑥1 , 𝑥3 , 𝑥5 , 𝑥6
No Solution the system
is not consistent
Dependent: 𝑥1
𝑥1 = 4𝑥2 − 2𝑥3
Independent: 𝑥2 , 𝑥3
Solutions to linear systems
The number of solutions to a linear system only
has three possible numbers; None, a unique (only
one) solution, or an infinite number of solutions.
1. The system has none when it is inconsistent.
2. The system has a unique solution when it is
consistent but has no independent variables
(i.e. all variables are dependent).
3. The system has infinitely many solutions when
it consistent but has at least one (maybe
more) independent variables.
1 0
0 1
0 0
−2 5
0 0
0 1
1 0
0 1
0 0
0 5
0 0
1 −2
1 0
0 1
0 0
−2
1
0
Homogeneous Systems of Equations
A homogeneous system of linear equations is a system
where all the constants are zero. In matrix form the
augmented (or last) column is all zeros. Because of this a
homogeneous system is always consistent. It always has
the solution of having all variables being zero. If there are
no independent variable this solution is unique (only one)
if there is an independent variable there is an infinite
number of solutions.
Inconsistent
No Solutions
No
Independent
Variable
Consistent
Infinite Solutions
𝑥1 = 5 + 𝑥3
𝑥2 = −𝑥3
5
0
0
𝑎11
⋮
𝑎𝑚1
One
Solution
𝑥1 = 5
𝑥2 = 0
𝑥3 = −2
⋯
⋯
𝑎1𝑛 0
⋮ ⋮
𝑎𝑚𝑛 0
Always has solution:
𝑥1 = 0
⋮
𝑥𝑛 = 0