# n equation with n+1 unknown!

Hello

I know: An equation with 2 unknowns has infinite solutions which lies on a line. Two equation with three unknowns again have infinite solution which lies on a line (the intersection of 2 planes), and so on....

what about n equation with n+1 unknowns? I know it has infinite solutions which lies on a line again but how can I write a general equation for that line? I want to find a point on the line but I need to know the equation of that line first:)

• Hmmm...how many solutions do you think the following system (two equations, three variables, as you mention) has?: $$\begin{cases}x+y+z=0\\x+y+z=1\end{cases}$$ – DonAntonio Apr 16 '17 at 13:11
• But this is different... these two planes do not have any intersection and solution. One of them has no intercept but the other has! I think I should wrote homogeneous equations... – Mahdieh Apr 16 '17 at 13:30
• Ah, homogeneous is a huge difference from what you wrote in your question... – DonAntonio Apr 16 '17 at 13:34
• I am really sorry. Actually I am chemistry student and not completely familiar with this... – Mahdieh Apr 16 '17 at 13:36
• The solution to the following system of equations is not a line: $$\begin{cases}x+y+z=0\\2x+2y+2z=0\end{cases}$$ – Hurkyl Apr 16 '17 at 14:59

Generically, you're right that $n$ equations in $n+1$ unknowns determine a line. Consider the homogeneous equations $$a_{i,0}x_0+a_{i,1}x_1+\cdots+a_{i,n}x_n=0,\qquad i=1,2,\ldots,n.$$ Take any non-zero solution $(X_0,X_1,\ldots,X_n)$ to this system. Then any multiple of this solution is again a solution, so the line parametrized by $(x_0,x_1,\ldots,x_n)=(tX_0,tX_1,\ldots,tX_n)$ consists of solutions to the system of equations. Generically (there's that word again), this line contains all solutions to the system.
The exceptions to this are when the equations are linearly dependent, as in this example: \begin{aligned} x-y+z&=0\\y-z+w&=0\\x+w&=0 \end{aligned} Here, the last equation is the sum of the first two. Linear dependence is a bit more general than this, but the point is that, if the equations are linearly dependent, you get more solutions than just the line.
In a sense, linear dependence happens rarely. That is why I wrote generically above, twice. There are different ways to make this notion precise, but for applications to science, we may note that if the coefficients $a_{i,j}$ are drawn at random from a continuous probability distribution, there is zero probability that the resulting equations are linearly dependent. That does not mean that linear dependence doesn't happen in scientific applications! But when it does, it is a safe bet that this is because of some symmetry of the problem at hand, or a dependence of variables dictated by an underlying physical law. (Almost linear dependence is a different matter; not uncommonly, it creates trouble for numerical methods, with possibly catastrophic consequences for the accuracy of computed answers.)
• You may want to add the word "homogeneous" after "Generically, you're right that $\;n\;$ ..." , in the third line. Otherwise the comments already contain a counterexample to this claim. – DonAntonio Apr 16 '17 at 20:41
• @DonAntonio With the reservation of genericity, I think my claim is just fine. After all, a generic linear map $\mathbb{R}^{n+1}\to\mathbb{R}^n$ is surely surjective? (Also, I don't want to bump this to the front page unless absolutely necessary.) – Harald Hanche-Olsen Apr 17 '17 at 8:07
• Perhaps you're using the term "generic" in some way I just don't know. Anyway, lots and lots of linear maps $\;\Bbb R^{n+1}\to\Bbb R^n\;$ aren't surjective, of course, which in matrix language means that there are lots (=infinitely many) matrices of order $\;n\times (n+1)\;$ whose rank is less than $\;n\;$ ... – DonAntonio Apr 17 '17 at 8:14