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# 06 Outcome of a linear system, and number of solutions.
Let us summarize what could happen when we solve a system of linear equations. There are three main outcomes.
Depending on the pivot structure of the echelon form (EF) you obtain after performing elementary row operations on the augmented matrix of a linear system, we have the following three situations:
- (1) **Inconsistent system.** This means **no solution**. We get this when there is a pivot in the augmented column of an EF of our augmented matrix.
- (2) **Consistent system.** This means there is at least one solution. So long as the pivots stay in the coefficient matrix part (not in the augmented column) of an EF of our augmented matrix, then we will have at least one solution. And further we have:
- (2A) **Consistent and with unique solution.** This is when we are consistent, and that each variable column has one pivot, and there is no free variables. In this case, we can uniquely solve for each variable, giving a set of unique solution.
- (2B) **Consistent and with multiple solutions (infinitely many if real scalars).** This is when we are consistent, and that there is at least one free variable column. In this case we get multiple solutions as the free variable is free to be any scalar. And if the scalar of choice is the real numbers, then we get infinitely many solutions, as there are infinitely many real numbers.
We can organize our analysis into a simple flow chart, again an algorithm:
- (1) Starting with a linear system, write down its augmented matrix.
- (2) Perform elementary row operations so it is in an echelon form (EF).
- (3) Identify where the pivots are, and ask, "Is there a pivot in the augmented column"?
- (3A) If the answer is "yes there is a pivot in the augmented column", then the system is inconsistent, with no solution. We are done.
- (3B) If the answer is "no, there is no pivot in the augmented column", then continue. Ask, "is there a free variable"?
- (3B1) If the answer is "yes there is a free variable", then we have a consistent system with many solutions. (And infinitely many solutions if we use scalars in $\mathbb{R}$ or $\mathbb{C}$.)
- (3B2) If the answer is "no, there is no free variable", then we have a consistent system with unique solution.
Here is a diagram describing above:
```tikz
\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd} [sep = .5 cm]
&\parbox{2cm}{\raggedright an augmented matrix} \arrow[d]& \\
& \parbox{2cm}{\raggedright EF of an augmented matrix}\arrow[dl] \arrow[dr]& \\
\parbox{2.2cm}{\raggedright pivot in the augmented column}\arrow[d] & & \parbox{3cm}{\raggedright no pivot in the augmented column} \arrow[dl] \arrow[d] \\
\parbox{2cm}{\raggedright no solution} & \parbox{2cm}{\raggedright yes, there are free variables} \arrow[d]& \parbox{2cm}{\raggedright no, there is no free variables} \arrow[d] \\
& \parbox{2.5cm}{\raggedright multiple solutions (and infinitely many if over R or C)} & \parbox{2cm}{\raggedright unique solution}
\end{tikzcd}
\end{document}
```
**AGAIN, IMPORTANT REMARK.** When we have free variables, then this means the system has a variable that can take on multiple different values of the scalars of choice. **If the scalars are the real numbers (or complex), then this further implies we have infinitely many solutions since there are infinitely many real numbers (as well as infinitely many complex numbers).** But in general, it depends on the scalars of choice! The text just says "infinitely many solutions", this is because it is assuming it will be using the reals as its scalars.
**Remark.** Quite often, algorithm becomes "you just follow it without thought". Now, this being math, you should understand why this is the case in addition to just "memorizing it"!