Reduced Row Echelon Form for an Arbitrary Matrix I understand the idea of rref and how to use it with real numbers. however, I wondered if it will be possible to apply to an arbitrary system of equations (no actual numbers).
for example, how should I go about solving (geometrically, finding the intersection of the two hyperplanes) this system of equations:
$$a_1x_1 + a_2x_2 +...+a_nx_n = d_1$$
$$b_1x_1 + b_2x_2 +...+b_nx_n = d_2$$
or even finding the null space if the above system does not make sense:
$$a_1x_1 + a_2x_2 +...+a_nx_n = 0$$
$$b_1x_1 + b_2x_2 +...+b_nx_n = 0$$
EDIT: I have tried to do the rref for the above matrices, but it becomes extremely messy extremely fast. is there a better way of going at it? or do I have to power through the math?
 A: One thing you can do is:
Consider the following matrix $A$
$$ A = \left[ \begin{array}{cccc} a_1& a_2&\cdots &a_n\\ b_1&b_2&\cdots&b_n\end{array}\right]$$
and compute the Gram matrix $AA^{t}$ which is of size $2x2$ and the condition: $$\det(AA^t)\neq 0$$ at least will give you the conditions to have linearly independents rows. More precisely, (assume $n>2$) since $A$ is of size $2\times n$ 
$$ \det(AA^{t})\neq 0 \ \Longleftrightarrow \ A\ \text{is of full rank}$$
Recall that any matrix $A$ of size $m\times n$ is said to be of full rank if $\ {\rm rank}(A)=\min\{m,n\}$ (that is, $\ A$ has it's maximum possible rank). 
In general:
Let $A$ be a matrix of size $m\times n$. You have two cases:


*

*If $m<n$, then $\det(A^{t}A)=0$ and we use the following result to see when $A$ is of full rank:
$$\det(AA^{t})\neq 0 \ \Longleftrightarrow \ A\ \text{is of full rank}$$


*If $m>n$, then $\det(AA^{t})=0$ and we use the following result to see when $A$ is of full rank:
$$\det(A^{t}A)\neq 0 \ \Longleftrightarrow \ A\ \text{is of full rank}$$



You can do a similar analysis with the augmented matrix:
$$ B = \left[ \begin{array}{cccc|c} a_1& a_2&\cdots &a_n & d_1\\ b_1&b_2&\cdots&b_n&d_2\end{array}\right] $$
A: So I did solve this eventually, and I thought it might be useful for someone one day, so I´m posting the solution:
without loss of generality, assume both hyperplanes path through the origin. i.e. $H_1:\{x:v^Tx = 0\}, H_2:\{x:u^Tx = 0\}$, where $v,u \in R^n$. define A to be: $$
  A =
  \left[ {\begin{array}{cc}
   v_1 & v_2 & \dots & v_n\\
   u_1 & u_2 & \dots & u_n\\
  \end{array} } \right]
$$
since u and v are assumed to be linearly independent, it results that A has a full row rank. now again, without loss of generality, we assume $v_i, u_j \neq 0$, such that: $$i \le j \le n$$ and $$v_k = 0, for \ 1\le k \lt i $$ $$u_k = 0, for \ 1\le k \lt j $$ in other words, $v_i$ and $u_j$ are the first non zero elements if v and u (respectively), and $v_i$ comes first. then it is needed to be splitted into two cases only, $i = j$ and $i \neq j$, and then the space of intersection can be found through row reduction of the augmented matrix of A.
