Converting explicit solution to implicit solution of linear system

Question

I am trying to convert the explicit solution of a linear system back to its implicit form.

I am given the following system of equations:

$$ \begin{cases} x + y + z = 0 \\ x - y + 2z = 1 \end{cases} $$

We can solve the system using gaussian elemination to obtain the explicit solution:

$$ (x,y,z) = \left(\frac{1}{2}-\frac{3}{2}t,\ -\frac{1}{2}+\frac{1}{2}t,\ t\right) \text{ where } t \in \mathbb{R} $$

Now we can describe the solution set of this system in 2 ways:

Implicit solution:

$$ \{ \ \left(x,y,z\right) \ | \ x + y + z = 0 \text{ and } x - y + 2z = 1 \ \} $$

Explicit solution:

$$ (x,y,z) = \left(\frac{1}{2}-\frac{3}{2}t,\ -\frac{1}{2}+\frac{1}{2}t,\ t\right) \text{ where } t \in \mathbb{R} $$

From the above, we know how to convert from the implicit to the explicit solution. But I do not know how to convert from the explicit to implicit solution. Could somebody please advise me?

Answer 1

From the parametric equations

$$\begin{align} x &= \frac{1}{2} - \frac{3}{2} t \\ y &= -\frac{1}{2} + \frac{1}{2} t \\ z & = t \end{align}$$ derive $t=z$ and use it to eliminate $t$ from the first two equations. You get: $$\begin{align} x + \frac{3}{2} z &= \frac{1}{2} \\ \frac{1}{2} z &= y + \frac{1}{2} \enspace. \end{align}$$ This solves the "implicitization" problem. If you want to go back to the original equations, maybe as a verification step, write the first as

$$ x + \frac{1}{2}z + z = \frac{1}{2} \enspace, $$

and substitute the second to get $x + y + z = 0$. Also, from the second equation, get $-2y + z = 1$. Add $x + y + z = 0$ to it to get $x - y + 2z = 1$.