GRE linear algebra question

The following is a question from the sample GRE Mathematics Subject Test found on the ETS website:

Let $M$ be a $5\times 5$ real matrix. Exactly four of the following five conditions on $M$ are equivalent to each other. Which of the five conditions is equivalent to NONE of the other four?

(A) For any two distinct column vectors $u$ and $v$ of $M$, the set {$u,v$} is linearly independent.

(B) The homogeneous system $Mx=0$ has only the trivial solution.

(C) The system of equations $Mx=b$ has a unique solution for each real $5\times 1$ column vector $b$.

(D) The determinant of $M$ is nonzero.

(E) There exists a $5\times 5$ real matrix $N$ such that $NM$ is the $5\times 5$ identity matrix.

Apparently, the correct answer is (A), but I can't figure out why this is true. If $M$ is nonsingular, as is implied by statements (B)-(E), then isn't that equivalent to its column vectors being linearly independent? And if the 5 column vectors are independent, then I can easily show that each pair of vectors are independent. What am I missing?

Linear independence of $n$ vectors, for $n>2$, is not equivalent to "pairwise" independence. Take three different vectors in the plane, for an example.
• Thanks. I realized right after I posted the question that although linear independence of the $n$ vectors implies pairwise linear independence, the other way around is false. I appreciate your quick answer though. – jake Oct 14 '10 at 3:35
HINT $\rm\ \ u,v\$ linear independent $\rm\ \Rightarrow\:\ u + v,\:u+2v,\: u+3v,\:\ldots\:$ pairwise independent
Suppose you have a vector $v$ for which is nonzero. Then $v \not= 2v$ but $M(2v) = 2M(v)$ and $M(v)$ are not linearly independent. Conditions B-E are all equivalent.