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Is there a geometric consequent to the number of free variables in a solved, consistent linear system represented as a row-echelon matrix (I use the row-echelon terminology just for an easy way of getting at the number of free variables)? That is, if we have a system of equations with 1 solution and (therefore no free variables), our solution is just a point and exists in 0 dimensions. If, as another example, we have two equivalent lines (and thus one free variable), we have infinitely many solutions, and the solution set is of course a line.

What I'm wondering is if this pattern generalizes: If we have 2 free variables in a consistent, row-echelon form linear system, is our solution set a (2d) plane? Clearly this is the case when we might have 3 equivalent planes, but what about the case where we might have a system of 5 variables, 2 of which being free? Still a plane? Similarly, if we have 3 free variables in a consistent row-echelon matrix, is our solution set a 3d plane? (and etc...?)

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Yes precisely. The number of variables you have (call this $m$) tells you the dimension of the vector space you are in. As you constrain more of the variables, the possible solution set reduces. If you have $n$ free variables at the end, then this is an $n$ dimensional surface in your $m$ dimensional space. ($n\le m$ always)

So when $n=2$, i.e. you have $2$ free variables, you will have a plane of solutions. If, say, $n=3$,$m=5$ then you have a $3D$ plane (referred to as a hyperplane) in a $5D$ space (this isn't really possible for us to visualise however).

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