Set of Linear equation has no solution or unique solution or infinite solution? For the system 
$$
\left\{
\begin{array}{rcrcrcr}
 x &+ &3y &-  &z &= &-4 \\
4x &-  &y &+ &2z &= &3 \\
2x &-  &y &- &3z &= &1
\end{array}
\right. 
$$
what is the condition to determine if there is no solution or unique solution or infinite solution? 
Thank you!
 A: The important concept here is linear dependence versus linear independence. As shown in the examples posted by others, linear dependence occurs when one equation in the system of equations can be shown to be a multiple of another. This is ultimately what Gaussian elimination or computing the determinant reveals. In this instance, there is no unique solution to the system of equations. 
Conversely, if the system of equations is linearly independent, then a unique solution does exist (though you still have to compute it, as is done in the examples in other answers). 
This can be visualized by graphing the equations, assuming a low order for the system. Linear dependence implies that two or more of the lines obtained when graphing the system are parallel or one on top of the other. Linear independence will result in a graph in which the various lines intersect at one point, that point being the solution to the system of equations.
A: There is an easier way to determine whether a system of equations has unique, infinite or no solution. It is as follows: calculate determinant $D$ of the coefficients of the three variables in three equations, then calculate $Dx$, where the x coefficients with the constant terms in the determinant $D$. Similarly calculate $Dy$ and $Dz$. Then, if $D$ is not equal to zero, the system has a unique solution. This is called consistent. If $D$ not equal to $0$, at least one of $Dx$, $Dy$, $Dz$ are not equal to zero, the system has no solution and this is called inconsistent. If $D=Dx=Dy=Dz=0$, the system has infinite solutions.
