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# 05 Pivots, free variables, unique solutions, parametrized solutions, no solutions.
Let us look at **THREE** scenarios of what could happen when solving a linear system and an echelon form (EF) we obtain at the end.
## Scenario 1. Unique solution.
Let us consider the linear system $$
\left\{
\begin{array}{}
x & - & 2y & = & 3 \\
2x & + & y & =&1
\end{array}
\right.
$$We can write down its augmented matrix and apply elementary row operations to obtain an echelon form: $$
\begin{array}{}
x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ & &x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ \ \
\\
\begin{bmatrix}
1 & -2 & \vdots & 3 \\
2 & 1 & \vdots & 1
\end{bmatrix} & \stackrel{\text{row}}\sim & \begin{bmatrix}
\colorbox{lightblue} 1 & -2 & \vdots & 3 \\
0 & \colorbox{lightblue} 5 & \vdots & -5
\end{bmatrix}
\end{array}
$$(Recall that $\stackrel{\text{row}}\sim$ means these matrices are related by some sequence of elementary row operations, that they are row equivalent. Also note that this is just an EF, and not REF nor RREF.)
Here we color all the $\colorbox{lightblue}{pivots}$ in blue for emphasis. Observe also that each pivot is in each of the variable columns (the $x$ and the $y$ column). Notice this final EF is the augmented matrix of the linear system $$
\left\{
\begin{array}{}
x & - & 2y & = & 3 \\
& & 5y & =& -5,
\end{array}
\right.
$$ which we can solve by performing backwards and back-substitution, giving us $$
y = -1,\ \ \ x= 1.
$$This shows this system has a **unique solution**, only $(x,y)=(1,-1)$ works.
## Scenario 2. Multiple solutions -- parameterized solutions.
Let us consider the linear system $$
\left\{
\begin{array}{}
x & - & 2y & = & 3 \\
2x & - & 4y & =&6
\end{array}
\right.
$$ Writing out its augmented matrix and performing elementary row operations to obtain an EF, we get:$$
\begin{array}{}
x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ & &x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ \ \
\\
\begin{bmatrix}
1 & -2 & \vdots & 3 \\
2 & -4 & \vdots & 6
\end{bmatrix} & \stackrel{\text{row}}\sim & \begin{bmatrix}
\colorbox{lightblue} 1 & -2 & \vdots & 3 \\
0 & 0 & \vdots & 0
\end{bmatrix}
\end{array}
$$ This final EF corresponds to the system $\left\{\begin{array}{}x-2y & =3\\0 & =0\end{array}\right.$.
Again we highlight all the $\colorbox{lightblue}{pivots}$ as blue. Something different happens here. Only the $x$ column has a pivot, while the $y$ column does not. In this situation we actually have some degree of freedom in the solutions, namely, not unique solution. But how do we write them out?
To help us to be systematic, we declare $y$ is **free variable** (without pivot), and $x$ is a **pivoted variable**. And we **solve the pivot variables all in terms of the free variables**. So we get $$
y=\text{free}, \ \ \ x=3+2y
$$This gives us **multiple** solutions, which we can write **parametrically** as follows: $$
(x,y)=(3+3y,y), \ \ \text{where } y \text{ is free.}
$$Here free means it can be **any** scalars. If the choice of scalars are real numbers, then we have infinitely many.
If you like, you can also use another variable to parametrize this situation, say using $t$ to be free, so we get $$
(x,y)=(3+3t,t),\ \ \text{where } t \text{ is free.}
$$This is unnecessary, though still correct, however.
## Scenario 3. Inconsistent system -- no solution.
Let us consider the linear system $$
\left\{
\begin{array}{}
x & - & 2y & = & 3 \\
3x & - & 6y & =&6
\end{array}
\right.
$$Writing out its augmented matrix and performing elementary row operations to obtain an EF, we get:$$
\begin{array}{}
x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ & &x\ \ \ \ \ y \ \ \ \ \ \ \ \ \ \ \ \ \ \
\\
\begin{bmatrix}
1 & -2 & \vdots & 3 \\
3 & -6 & \vdots & 6
\end{bmatrix} & \stackrel{\text{row}}\sim & \begin{bmatrix}
\colorbox{lightblue} 1 & -2 & \vdots & 3 \\
0 & 0 & \vdots & \colorbox{lightblue}{$-3$}
\end{bmatrix}
\end{array}
$$Again, highlight the $\colorbox{lightblue}{pivots}$ as blue (or you just box it to identify it). Now here we have two pivots, but something interesting happens -- **a pivot shows up in the augmented column**! What does this mean? Let us write what linear system this augmented matrix is representing: $$
\left\{
\begin{array}{}
x & - & 2y & = & 3 \\
0x & - & 0y & =& -3
\end{array}
\right.
$$
Or in other words, $0=-3$. Well, this is not possible, no matter how you choose $x,y$. So in this case we have **no solution**!
Hence, if you have an EF for an augmented matrix where a pivot shows up in the augmented column, then there is no solution what-so-ever.
Some terminology. If the linear system has at least one solution (whether unique or many), we say the linear system is **consistent**. And if the linear system has no solution what-so-ever, then it is said to be **inconsistent.**