Linear Algebra Matrix number of solutions I just want to make sure I understand this correctly. 
If I have 10 linear equations with 20 unknowns I will have 0 solutions, where as if I have 10 linear equations with 5 unknowns I will have infinite solutions?
 A: That understanding is overly simplistic.  If you have 10 linear equations with 20 unknowns, you will have either zero or infinitely many solutions.  For example, the first two equations might be $x_1+x_2=1, x_1+x_2=2$.  It doesn't matter what the other 8 equations are, there will be no solutions.
If you have 10 linear equations with 5 unknowns, you may have zero, one, or infinitely many solutions.
In general, a linear system may have zero, one, or infinitely many solutions.  The only way to have exactly one is if there are at least as many equations as unknowns.
A: I would like to give some details, and a global view of the situation, which is best understood in terms of vector spaces and linear maps.
A system of $m$ linear equations in $n$ unknownw can be seen matricially, as an equation:
$$Ax=b,$$
where $A$ is an $m\times n$ matrix, and $b$ is a vector in $K^m$, $K$ being the base field (whether $\mathbf Q, \mathbf R$ or $\mathbf C$). $A$ represents a linear map:
$$u\colon K^n\longrightarrow K^m.$$
This equation has solution(s) if $b$ is in the range of $u$. As the range of $u$ is the subspace of $K^m$ generated by the column vectors of $A$, this means, denoting $Ab$ the matrix $A$ bordered by the column-vector $b$, that
$$\color{red}{\operatorname{rank}A=\operatorname{rank}Ab}.$$
If $r=\operatorname{rank}A$, we know $r\le\min(m,n)$, and by the rank-nullity theorem, the solutions, if any, are an affine subspace of $K^n$, of dimension $\color{red}{n-r}$.
In particular, if $m\ge n$, and $A$ has maximal rank ($r=n$) and $Ab$ also has rank $r$, the subspace of solutions has dimension $0$, i.e. there is only $1$ solution.
