linear independence of column vectors guarantees unique solution? In a recent test that I came across, there was a true/false statement. It was, a linear system $Ax=B$ has unique solution if and only if the column vectors of $A$ form a linearly independent set. I marked it as false but the answer was given as true. I still think it as false. consider a $3$ x $2$ matrix and $B$ as a vector which is not in the range space of $A$, then system will have no solution. So am I right?
 A: You are correct. Take $A$ a column matrix and $B\ne A$ for an even simpler example. Linear algebra classes have a bad habit of assuming matrices are square even when they're not. (In the square case, linearly independent columns do imply every $Ax = b$ has a unique solution.)
A: You are referring to what is known as an Overdetermined System, i.e. one which has more equations than variables.  This wiki page provides an interesting explanation on how each variable can be thought of as an available degree of freedom, and each equation as a restriction of a degree of freedom:
https://en.wikipedia.org/wiki/Overdetermined_system.
These systems are usually inconsistent, unless there are enough linearly dependent equations for the system to have a solution, so it looks like you won this round.
A: You are right... there are however some cases where this is true, which might have been what they meant.


*

*$B=0$. 

*$A$ is a $n\times n$ matrix.


My guess is that the exercise said that it was a $n\times n$ matrix (and you missed it), or maybe your teacher forgot to write it.
