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# 05a More examples of parametrized solutions.
Let us see more examples of how to write out the parametric solutions, if we encounter a consistent system with free variables.
Example.
Say we have a linear system where an EF of its augmented matrix is $$
\begin{array}{}
x \ \ \ \ \ y\ \ \ \ \ z \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
\colorbox{lightblue} 1 & -2 & 3 & \vdots & 3 \\
0 & 0 & \colorbox{lightblue}2 & \vdots & 14
\end{bmatrix}
\end{array}
$$
Then here $y$ is a free variable, while $x,y$ are pivot variables. If we adhere to our systematic approach: We solve all the pivot variables in terms of the free ones. So going backwards, we have $$
\begin{array}{ccl}
z & = & 7 \\
y &= & \text{free} \\
x & = & 3 - (-2y)-3z=3+2y-21=-18+2y
\end{array}
$$Hence the solutions are parametrized as $$
(x,y,z)=(-18+2y,y,7), \ \ \text{where } y \text{ is free.}
$$
Example 2.
Say we have a linear system where an EF of its augmented matrix is $$
\begin{array}{}
x \ \ \ \ \ y\ \ \ \ \ z \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
\colorbox{lightblue} 1 & -2 & 3 & \vdots & 3 \\
0 & 0 & 0 & \vdots & 0
\end{bmatrix}
\end{array}
$$
Then here **both** $y$ and $z$ are free variables, while $x$ is the only pivot variable. If we adhere to our systematic approach: We solve all the pivot variables in terms of the free ones. So going backwards, we have $$
\begin{array}{ccl}
z & = & \text{free} \\
y &= & \text{free} \\
x & = & 3 - (-2y)-3z=3 + 2y-3z
\end{array}
$$Hence the solutions are parametrized as $$
(x,y,z)=(3+2y-3z,\ y,\ z), \ \ \text{where } y,z \text{ are free.}
$$
Example 3.
Say we have a linear system where an EF of its augmented matrix is $$
\begin{array}{}
x \ \ \ \ \ y\ \ \ \ \ z \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
0 & \colorbox{lightblue} {$-2$} & 3 & \vdots & 3 \\
0 & 0 & 0 & \vdots & 0
\end{bmatrix}
\end{array}
$$
Then here **both** $x$ and $z$ are free variables, while $y$ is the only pivot variable. If we adhere to our systematic approach: We solve all the pivot variables in terms of the free ones. So going backwards, we have $$
\begin{array}{ccl}
z & = & \text{free} \\
-2y &= & 3-3z \implies y=-\frac{3}{2} + \frac{3}{2}z\\
x & = & \text{free}
\end{array}
$$Hence the solutions are parametrized as $$
(x,y,z)=\left( x,\ -\frac{3}{2} + \frac{3}{2}z,\ z \right), \ \ \text{where } x,z \text{ are free.}
$$Note that $y$ only depends on $z$ because of our pivot structure!
Example 4.
What if we have a linear system where an EF looks like this?$$
\begin{array}{}
x \ \ \ \ \ y\ \ \ \ \ z \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
0 & 0 &0 & \vdots & 0 \\
0 & 0 & 0 & \vdots & 0
\end{bmatrix}
\end{array}
$$In this case all variables are free! So the solutions are just $(x,y,z)$ where $x,y,z$ are all free. If our choice of scalars are the reals, this is all of $\mathbb{R}^{3}$.
Example 5.
What if we have
What if we have a linear system where an EF looks like this?$$
\begin{array}{}
x \ \ \ \ \ y\ \ \ \ \ z \ \ \ \ \ \ \ \ \ \ \ \\
\begin{bmatrix}
0 & 0 &0 & \vdots & \colorbox{lightblue}1 \\
0 & 0 & 0 & \vdots & 0
\end{bmatrix}
\end{array}
$$Note the pivot is in the augmented column, so we have a $0=1$ situation, which is **no solution**! Be careful.
**Remark.** It is important to note that our approach here is a **systematic way** of assigning the variables. It is possible to write a **different looking but equivalent** parametrization (generally there is always many ways to parametrize the same set). But doing it our way here is just an algorithmic way of approaching it, so we "don't have to think"....FOR NOW.