# 4 equations with 4 unknowns analysis

I was curious to know the analysis of a $4$ equation with $4$ unknowns. I know bunch of solutions like substitution, Cramer, etc. for solving this problem but couldn't find any analysis on this form.

For example I know three assumptions for a $2$ equations with $2$ unknowns:

1. two lines may cross ($1$ answer - which is point),
2. two lines may overlap (unlimited answers),
3. two lines incoherent (no answers).

Is there any graphical analysis for a $4\times 4$ equation? Or an analysis which could explain the behavior of it?

In $2D$, the solution is intersection of $2$ lines (in every case!). In $n$-dimensional case, the solution is intersection of $n$ hyperplanes.

For $n=3$, hyperplanes are just ordinary planes, for $n=2$, hyperplanes are lines, for general $n$, hyperplane is a $(n-1)$-dimensional affine subspace of $\mathbb{R}^n$.

Every hyperplane devides $\mathbb{R}^n$ on two parts, since $(n-1)$-dimensional hyperplane is a solution of equation $$a_1x_1+\cdots +a_nx_n=b\text{,}$$ where at least one $a_i \neq 0$. Let be $a = (a_1,\dots, a_n)$. Then the upper equation simply says $\langle a,x \rangle = b$ and these two parts can be given explicitly as

$A^+ = \{(x_1,\dots,x_n)\in \mathbb{R}^n \; |\; \langle a,x \rangle > b\}$ and

$A^- =\{(x_1,\dots,x_n)\in \mathbb{R}^n \; |\; \langle a,x \rangle < b\}$.

If you can visualize hyperplane in $\mathbb{R}^4$, good for you, I cannot do that:)

• Thanks, That explains things very well. But is a hyperplane divide the space into 2 sections either? – Novin Shahroudi Feb 1 '14 at 11:55
• @NOVIN I have edited the answer. – Antoine Feb 1 '14 at 12:03
• @Antoine. Could you with 12 ? Joke, be sure. Cheers. – Claude Leibovici Feb 1 '14 at 12:07
• @ClaudeLeibovici Of course I can, $12$ is not a perfect square. – Antoine Feb 1 '14 at 12:08
• @Antoine. +1 for good answer and good sense of humor !! – Claude Leibovici Feb 1 '14 at 12:11