Geometric interpretation of Gauss elimination Solution of a set of linear equations is finding a point of intersection of all planes represented by those equations. But how can we relate it to gauss elimination method ? Suppose we have equations like-
x + 3y = 4 (A)
2x - 6y = 8 (B)
We are trying ti find common value of x and y that will satisfy both equations. So we do B-2A and find the value of y. But how can we explain B-2A geometrically in context of solving these equations ?
 A: In 2D, the two equations represent two intersecting lines.
We cancel the coefficient of $x$ in the second equation in a way that doesn't alter the solution. This means that we rotate the second line around the intersection so that it becomes horizontal, immediately giving $y$.

In 3D, the three equations represent three intersecting planes meeting at a point and intersecting pairwise in three lines.
We cancel the coefficient of $x$ in the second plane in a way that doesn't alter the solution, by means of the first equation. This means that we rotate the second plane around the intersection line with the first plane, so that the second becomes a profile plane (parallel to the axis $x$).
Similarly, we cancel the coefficient of $x$ in the third plane by rotating it around the intersection line with the first so that it becomes a profile plane.
Now we can ignore the $x$ coordinate and just look at the traces of these two planes in the $yz$ plane and we have turned our 3D problem to a 2D one involving two lines.

More generally, in N dimensions, we "hyper"rotate all hyperplanes but the first around their "hyper"intersection with the first so that they become parallel to $x$. Doing so, we obtain a simpler problem involving N-1 hyperplanes in N-1 dimensions. And we iterate.

Note about the "rotation around" operation:
Assume we have two planes $ax+by+cz=d$ and $a'x+b'y+c'z=d'$. Their intersection is described by all the points $(x,y,z)$ that verify both equations. Now if we form the linear combination
$$ax+by+cz+\lambda(a'x+b'y+c'z)=(a+\lambda a')x+(b+\lambda b')y+(c+\lambda c')z=d+\lambda d',$$ we get another plane which also contains these $(x,y,z)$. Hence we have performed a rotation around the common line, and we can choose $\lambda$ in a way that cancels $a+\lambda a'$.
