Simple explanation for number of solutions of system of linear equations So a system of linear equations can be represented as:
$$Ax=d$$ 
where
$A$ is a $n\times n$ matrix and
$x$ and $d$ are $ n\times 1$ vectors.
Now in my notes it says the number of solutions are either:
$$0, 1 \text{ or } \infty$$
I understand if $\det(A) \neq 0$ then there is a unique one-to-one solution but could someone explain simply the other two cases in a easily understandable fashion?
Thank you.
 A: 0: 
For simplicity let's assume 2x2 matrix. Assume that after multiplication you get system of equations:
$x_1 + x_2 = 5$
$x_1 + x_2 = 10$
infinity:
Assume that after multiplication you get system of equations:
$x_1 + x_2 = 5$
$2x_1 + 2x_2 = 10$

To extend this on any $n$ x $n$ matrix: 


*

*if the determinant of the non-homogeneous is zero, then


*

*the system has infinite number of solution if it is not conflicted

*the system has $0$ solutions if it is conflicted


*if the determinant of the non-homogeneous is different from zero, then the system has unique solution
A: The statement remains true even when $A$ is a matrix with different numbers of rows and columns.  Say that $A$ has $m$ rows and $n$ columns.  This corresponds to a system of $m$ equations in $n$ variables.  
If $n=2$ then all of the equations represent lines.  If all $m$ lines cross in a single point, then there's a unique solution.   If all $m$ equations represent the same line, then every point on that line is a solution to the system, so there are infinitely many solutions.  If there is no point common to all $m$ lines, then there is no solution.
Could there be any other cases?  Could there, for example, be $2$ solutions?  If there were two equations, this would correspond to a pair of lines that intersect twice, but not infinitely many times, a clear impossibility.  In the case of $m$ lines, all $m$ of the lines would have to pass through the same $2$ points, but couldn't all be the same line.  Again, this is geometrically impossible.
If $n=3$, the equations represent planes.  It becomes harder to visualize all cases.  If all of the planes have a point in common, then there is a unique solution.  Two non-parallel planes will intersect in a line.  A third plane not parallel to that line will cross the line in a unique point—the unique solution.  One can think of many configurations where a set of planes have no point common to all planes—the no solution case.  There are essentially two ways to have infinitely many solutions: (1) have all of the equations represent the same plane, so that all points on that plane are solutions; (2) have all of the planes intersect in a common line, so that all points on that line are solutions.  In case (1), we have a $2$-parameter family of solutions; in case (2), we have a $1$-parameter family of solutions.  You can think about why, geometrically, it is impossible for a set of plane to have some other number of points in common.
Geometry, of course, only takes you so far.  The better way to understand why there are $0,$ $1,$ or infinitely many solutions is to understand row-reduction thoroughly.  If an inconsistency arises in the process of row-reduction, then there are no solutions; otherwise there are $1$ or infinitely many solutions.  Which one it is depends on whether there remain any undetermined parameters after row-reduction is complete.
