Question Regarding Span and Linear Combination I believe I understand both topics individually:  When asked if a linear system spans a certain R^n, the question is, "can any point be reached in that dimensional plane?"  Linear combination is multiplying a vector with a scalar and adding that to another vector being multiplied by a scalar and creating a linear system to solve to see if theres a unique solution or set of solutions.
My question is, how is it that finding a set of solutions or a unique solution, through the use of linear combination, will tell you if the system spans that dimensional plane?  I'm having a hard time understanding how know that theres a set of numbers that solved all 3 equations means that it spans the plane.  If anyone can explain this part in layman's terms, that would be great.
 A: It seems like your question is about the geometry of the relation between invertibility of a matrix and the linear independence of its columns. Invertibility means that one can always solve $Ax=b$ for $x$, no matter the $b$. Indeed, $A^{-1}$ exists iff the columns (or rows) are linearly independent.
(Note that $Ax=b$ can be solvable (depending on $b$) even when $A$ does not have linearly independent columns; i.e. not spanning the space).
One can consider that $A$ is a map: it takes in a vector $x$ and produces a vector $Ax$.
The question of solving $Ax=b$ is the same as asking whether $b$ is in the column space (i.e. span of the matrix's column vectors). 
This is by definition of matrix multiplication, which produces $Ax$ by taking a linear combination of the columns of $A$. 
This is the connection between span and solving linear systems. We can only guarantee that the system is solvable if the span hits all the space of $\mathbb{R}^n$; if there are parts of the space that cannot be reached, then choosing a $b$ in those areas means we cannot solve the system.
In terms of linear independence, if two vectors out of $n$ are linearly dependent, then they give no new information about the space (and thus won't span it).
Another way to see this is by the Rank-Nullity Theorem (albeit in a less layman way), which says $\text{rank}(A)+\text{nullity}(A)=n$. 
Only if the dimension of the span of the columns is the whole space can we always solve the system. 
