Combinatorics, Linear Algebra, n-Dimensional space. I am a volunteer agroecological grower or gardener who is currently studying Linear Algebra on M.I.T. courseware.
Gilbert Strang is currently talking about a 9-dimensional problem, which is impossible to imagine geometrically because we only have 3 axes to work with.
This got me thinking about Combinatorics because if you plug in 9 choose 3 into a Combinatorics calculator you get 84 Combinations of 3-dimensional space.  If you solved each Linear Algebra 3-dimensional combination you would presumably end up with 84 points in 3d space (if every combination lay on a discrete plane).
This could be efficiently solved with a computer program.
Is there anything to learn from this strategy?  Or is it just useless data?
Kind regards,
David.
 A: If I understand you correctly, your idea won't work because you won't get $84$ points.
Disclaimer: What I write below holds in general, but there are some exceptions (when equations are linearly dependent), which complicate the discussion.  I will skip the complications in order to present a clearer overall picture that (I think) is more relevant to your idea.


*

*A linear equation in $3$ unknowns defines a $2$-D subspace in the $3$-D space.  A $2$-D subspace happens to be a plane.

*If you have $3$ linear equations in $3$ unknowns, they define $3$ planes which (in general) intersect at exactly one point.  This is when the $3\times 3$ matrix is invertible, and you have exactly one solution.

*Similary, a linear equation in $9$ unknowns defines an $8$-D subspace in the $9$-D space.  An $8$-D subspace is obviously not a plane.

*If you have $9$ linear equations in $9$ unknowns, they define $9$ $8$-D subspaces which (in general) intersect at exactly one point.  This is when the $9 \times 9$ matrix is invertible, and you have exactly one solution.

*If you pick some $3$ of those $9$ linear equations (each in $9$ unknowns), they define $3$ $8$-D subspaces.  Their intersection will (in general) be a $6$-D subspace.  You will not get a single point.

*Your idea seems to be to investigate each of the ${9 \choose 3} = 84$ subsets of $3$ equations.  These will give you, not $84$ single points, but instead $84$ $6$-D subspaces.  Because they came from the same $9$ equations to begin with, they will (in general) intersect at exactly one point.  This is when the $9 \times 9$ matrix is invertible, and you have exactly one solution.
So in short, you won't get $84$ points.  
An easier (less formal?) way is to think of each equation as eliminating a "degree of freedom" (dof). You start with the entire $9$-D space with $9$ dof, and when you have $n$ equations you are left with $9-n$ dof.  When you use any $3$ equations, you are still left with $6$ dof ($6$-D subspace) in general.  But when you use all $9$ equations, you are left with $0$ dof, i.e. a single point.
Hope this helps instead of confuses even further?
