A linear system \(Ax=b\) can have only one of these three outcomes:
1) Unique solution (exactly one solution)
2) No solution (inconsistent system)
3) Infinitely many solutions (free variable(s))
In this lesson, we decide the solution type by converting the augmented matrix ([A\mid b]) into Reduced Row Echelon Form (RREF) using Gauss–Jordan elimination.
The key idea: look at the RREF of \([A \mid b]\)
1) No solution (inconsistent)
If you get a row like this:
$$
\left[\begin{array}{ccc|c}
0 & 0 & 0 & 1
\end{array}\right]
$$
That means \(0=1\), which is impossible \(\Rightarrow\) no solution.
2) Unique solution
You have a unique solution when:
- there is no inconsistent row, and
- every variable column has a pivot (so there are no free variables).
3) Infinitely many solutions
You have infinitely many solutions when:
- there is no inconsistent row, and
- at least one variable is free (a non-pivot column).
Worked examples
Example A: Unique solution
Suppose the RREF becomes:
$$
\left[\begin{array}{ccc|c}
1 & 0 & 0 & 2\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 4
\end{array}\right]
$$
Then the solution is:
$$
x=2,\quad y=-1,\quad z=4
$$
So the system has a unique solution.
Example B: No solution
If you get:
$$
\left[\begin{array}{ccc|c}
1 & 0 & 2 & 3\\
0 & 1 & -1 & 4\\
0 & 0 & 0 & 1
\end{array}\right]
$$
The last row means \(0=1\) \(\Rightarrow\) no solution.
Example C: Infinitely many solutions
If you get:
$$
\left[\begin{array}{ccc|c}
1 & 0 & 2 & 3\\
0 & 1 & -1 & 4\\
0 & 0 & 0 & 0
\end{array}\right]
$$
There is no contradiction, but there is at least one free variable.
Let \(z=t\) where \(t\in\mathbb{R}\). Then:
$$
x+2z=3 \Rightarrow x=3-2t
$$
$$
y-z=4 \Rightarrow y=4+t
$$
So the solution set is:
$$
(x,y,z)=(3-2t,\ 4+t,\ t),\quad t\in\mathbb{R}
$$
Therefore, the system has infinitely many solutions.
Quick checklist (in 10 seconds)
- If you see \([0\ \cdots\ 0\mid 1] \Rightarrow\) No solution
- Otherwise, if every variable column has a pivot \(\Rightarrow\) Unique solution
- Otherwise \(\Rightarrow\) Infinitely many solutions
Practice
1) Decide the solution type
$$
\left[\begin{array}{cc|c}
1 & 0 & 5\\
0 & 1 & -2
\end{array}\right]
$$
2) Decide the solution type
$$
\left[\begin{array}{cc|c}
1 & 3 & 7\\
0 & 0 & 2
\end{array}\right]
$$
3) Decide the solution type
$$
\left[\begin{array}{ccc|c}
1 & 2 & 0 & 1\\
0 & 0 & 1 & 4\\
0 & 0 & 0 & 0
\end{array}\right]
$$
Answers
1) Unique solution
2) No solution
3) Infinitely many solutions