Consider the following system of linear equations: \begin{aligned} 4x − 2y &= 1 \\ 16x − 8y &= 4 \end{aligned} We can eliminate $x$ from the second equation by adding $(−4) \times$ the first equation to the second. This yields the simplified system \begin{aligned} 4x − 2y &= 1 \\ 0 &= 0 \\ \end{aligned} The second equation does not impose any restrictions on $x$ and $y$ and hence can be omitted. Thus, the solutions of the system are those values of $x$ and $y$ that satisfy the single equation $$ 4x − 2y = 1 $$ One way to describe the solution set is to solve this equation for $x$ in terms of $y$ to obtain $x = \frac{1}{4} + \frac{1}{2}y$ and then assign an arbitrary value $t$ (called a parameter) to $y$. This allows us to express the solution by the pair of equations (called parametric equations) $$ x = \frac{1}{4} + \frac{1}{2}t, \qquad y = t $$ My Question is: What do we gain by representing the 2nd-last equation in a parametric way? Are there any obvious advantages that I am not seeing ? I have some understanding of parametric equations but I am unable to relate the solution of linear system to the concept of parametric equations ? or stated otherwise how do they relate to each other ?



$$x(y) = \frac{1}{4} + \frac{1}{2}y$$


$$x(t) = \frac{1}{4} + \frac{1}{2}t, \qquad y(t) = t$$

In the first you represent one of the original variables in terms of the other, describing your solution set with the mapping

$$y \to (x(y),y)$$

In the second it's not much different other than using an arbitrary variable, $ \ t$:

$$t \to (x(t),y(t))=\left(\frac{1}{2}t+\frac{1}{4},t\right)$$

Which only will yield the same set of $(x,y)$'s just like above. But here you can make the substitution

$$\quad t\to 2t-\frac{1}{2} \quad $$

for example, leading you to another description of your solution set:

$$t\to (x_2(t),y_2(t))=\left(t,2t-\frac{1}{2}\right)$$

Which again yields that very same set of $(x,y)$'s.

Using an arbitrary parameter is a degree of freedom that, say, the first solution set description doesn't have. It gives you a way of describing your solution set in more than one way, which will come into play later in your linear algebraic studies, and in higher dimensions.


The following is more of a philosophical nature. It does not refer to your specific set of equations.

An equation $$\Phi(x)=0,\quad x\in X,\tag{1}$$ which might also be a system of equations, implicitly defines a solution set $$S:=\bigl\{x\in X\bigm|\Phi(x)=0\bigr\} \ \subset X\ .$$ For any trial point $x\in X$ it is easy to check whether it belongs to $S$ or not, but we don't have a birds eye view over the set $S$. Solving the equation means producing an explicit representation of $S$ in the form of a "list". This list may be empty (if there are no solutions), or consist of exactly one entry, which is then called the solution of $(1)$. As long as $S$ is a finite set we can present it in the form $S=\{a, b,\ldots, p\}$. If the set $S$ is infinite we have to resort to a "production scheme" consisting of an index set $I$ and a function $f:\>I\to X$ that outputs for each $\iota\in I$ an element $x_\iota\in S$ in an essentially bijective way. Such an $f$ is then called a parametric representation of $S$.

In your example we can write $S$ in the form $S=f({\mathbb R})$ with $$f:\quad{\mathbb R}\to{\mathbb R}^2,\qquad t\mapsto\left({1\over4}+{t\over2},\>t\right)\ .\tag{1}$$ Here $X={\mathbb R}^2$, the index set is the $t$-axis ${\mathbb R}$, and the $f$ in $(1)$is the "production scheme" for $S$.

  • $\begingroup$ Hi Christian, thanks for the quick reply. I was not being able to understand you completely. It would be helpful if you could provide an example. $\endgroup$ May 22 '18 at 13:48

In your example the solution of the linear system is a line.

Here are some representations of a line in 2D I use: $$ y = a x + b \\ a x + b y + c = 0 \\ n \cdot (x, y)^\top = d \\ (x, y)^\top = (p_x, p_y)^\top + t\, (d_x, d_y)^\top \\ (1-\lambda) u_1 + \lambda u_2 $$ It depends on the problem, what form I use. E.g. the parametric case is interesting for physical problems, where we are interested in the movement of a point on the line.

If we move to three dimensions, I would need to update that list, e.g. using the intersection of linear equations / affine planes, I would need two of them, and they need to be linear independent: $$ n_1 \cdot (x, y, z)^\top = d_1 \\ n_2 \cdot (x, y, z)^\top = d_2 \\ $$ But I could keep the parametric forms more or less: $$ (x, y, z)^\top = (p_x, p_y, p_z)^\top + t \, (d_x, d_y, d_z)^\top \\ (1-\lambda) u_1 + \lambda u_2 $$


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