Eliminating variables

1. Oct 8, 2009

BobbyBear

I'm a bit stuck with the idea of eliminating variables from a set of simultaneous equations . . . for example, suppose you have two equations (with more than 3 variables), could you, in principle, reduce it to one equation with one variable less?

And if you had three equations with lets say 4 variables, could you, in principle, reduce them to 2 equations with 3 variables, and them in turn to one equation with 2 variables?

So in general, if you have m equations and n variables, with n>m, then can you, in principle, eliminate m-1 variables, so that you end up with a single equation with n-(m-1) variables? Is this correct? (I say in principle because I suppose you need to assume all the equations are independent and stuff like that:P).

2. Oct 8, 2009

HallsofIvy

Staff Emeritus
Yes, in general, if you have m equations with n variables, n> m, you can reduce to m-1 equations with n-1 variables and, repeating, eventually arrive at one equation with n-(m-1)= n-m+1 variable. You could then write one of those variables as a function of the other n-m variables and, working back, eventually write all variables as functions of those n-m "free variables".

Here is an example. Find a basis for the subspace of all (x, y, z) in R3 satisfying x+ y- z= 0 and x- y+ 2z= 0 (two equations in 3 variables). Adding those equations together I get 2x+ z= 0 (one equation in two variabes) so z= -2x. Putting that into the first equation, x+ y-(-2x)= 3x+ y= 0 so y= -3x. Any such vector is of the form (x, -3x, -2x)= x(1, -3, -2). The single vector (1, -3, -2) forms a basis for that (one dimensional) subspace.

3. Oct 8, 2009

BobbyBear

Thank you Ivy:)

so basically what you have done in that example is write a set of two implicit functions as explicit functions of x, that is, you have solved the system for y and z (can I put it like that?). It's like, you've considered the 'free' variable to be x, ie., as if it were a parameter instead of a variable, and you've solved the system of two equations with two unknown variables y and z.

So eliminating variables is the first step to actually solving an implicit system of equations to end up with explicit equations, or parametric equations, yes? But it's also obtaining implicit equations from explicit ones, no?

Say for example, I'm given a curve in space by its parametric equations:

$$x=f(s)$$
$$y=g(s)$$
$$z=h(s)$$

I could eliminate s to obtain two implicit functions by doing the following:

$$x=f(s) \rightarrow s=f^{-1}(x)$$
$$y=g(s) \rightarrow s=g^{-1}(y)$$
$$z=h(s) \rightarrow s=h^{-1}(z)$$

from which:

$$f^{-1}(x)=g^{-1}(y)$$
$$g^{-1}(y)=h^{-1}(z)$$

ie.
$$F(x,y)=0$$
$$G(y,z)=0$$

which would be the two surfaces whose intersection is the curve.

But now suppose I'm given the parametric equations of a surface:

$$x=f(u,v)$$
$$y=g(u,v)$$
$$z=h(u,v)$$

In theory, I could eliminate u and v and I'd be left with a single equation in the variables x,y and z, which would be the implicit equation of the surface. But, say if from the first two equations I solve for u and v, and I do the same using the second pair of equations, I'd have:

$$u=f_1(x,y)$$
$$v=f_2(x,y)$$
and
$$u=g_1(y,z)$$
$$v=g_2(y,z)$$

and eliminating u and v from these four equations, I have:

$$f_1(x,y)=g_1(y,z)$$
$$f_2(x,y)=g_2(y,z)$$

ie. I end up with two equations in the variables x,y and z instead of just one!

$$f_1(x,y)=g_1(y,z) \rightarrow F(x,y,z)=0$$
$$f_2(x,y)=g_2(y,z) \rightarrow G(x,y,z)=0$$

which define a curve, not a surface. Can you tell me where I am going wrong?

4. Oct 9, 2009

HallsofIvy

Staff Emeritus
The two equations are not independent. For example, suppose your equations are x= u+v, y= u- v, z= 2u+ v (describing a plane). From the first two equations you get u= (x+y)/2 and then v= u- y= (x- y)/2. From the last two equations, u= (y+z)/3 and then v= u- y= (z- 2y)/3. Now combine those: u= (x+y)/2= (y+z)/3 give 3x+ 3y= 2y+ 2z or 3x+ y- 2z= 0. v= (x- y)/2= (z- 2y)/3 or 3x- 3y= 2z- 4y: 3x+ y- 2z= 0 again!

5. Oct 10, 2009

BobbyBear

Ivy, I love you!

Okay but, I assume this will always be the case . . . but how can one see this if it's not through a particular example?

6. Oct 10, 2009

BobbyBear

Okay wait, in the case of linear equations it's easy to show in general that the equations are dependent:

$$x=f(u,v)=Au+Bv$$
$$y=g(u,v)=Cu+Dv$$
$$z=h(u,v)=Eu+Fv$$

Applying Cramer's rule, from the frist two we get:

$$\begin{array}{cc} x=Au+Bv\\ y=Cu+Dv \end{array} \right\} \rightarrow}$$

$$u=\frac{ det \left( \begin{array}{cc} x & B\\ y & D \end{array} \right)}{det \left( \begin{array}{cc} A & B\\ C & D \end{array} \right)} = \frac{xD-yB}{AD-CB}; (1)$$

$$v=\frac{ det \left( \begin{array}{cc} A & x\\ C & y \end{array} \right)}{det \left( \begin{array}{cc} A & B\\ C & D \end{array} \right)} = \frac{yA-xC}{AD-CB}; (2)$$

and from the second two:

$$\begin{array}{cc} y=Cu+Dv\\ z=Eu+Fv \end{array} \right\} \rightarrow}$$

$$u=\frac{ det \left( \begin{array}{cc} y & D\\ z & F \end{array} \right)}{det \left( \begin{array}{cc} C & D\\ E & F \end{array} \right)} = \frac{yF-zD}{CF-ED}; (3)$$

$$v=\frac{ det \left( \begin{array}{cc} C & y\\ E & z \end{array} \right)}{det \left( \begin{array}{cc} C & D\\ E & F \end{array} \right)} = \frac{zC-yE}{CF-ED}; (4)$$

From (1) and (3) we obtain, after simplifying, the equation of the plane:

$$(CF-ED)Dx-(FA-BE)Dy-(AD-CB)Dz=0$$

and it's easy to see that from (2) and (4) we get the same equation of the plane.

But the same would happen if the equations were non-linear? And is there a way to prove it?