# GRE linear algebra inquiry

The following is an inquiry from the example GRE Mathematics Subject Test located on the ETS internet site:

Let $M$ be a $5\times 5$ actual matrix. Specifically 4 of the adhering to 5 problems on $M$ are equal per various other. Which of the 5 problems amounts NONE of the various other 4?

(A) For any kind of 2 distinctive column vectors $u$ and also $v$ of $M$, the set $u,v$ is linearly independent.

(B) The uniform system $Mx=0$ has just the unimportant remedy.

(C) The system of formulas $Mx=b$ has an one-of-a-kind remedy for each and every actual $5\times 1$ column vector $b$.

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

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

Evidently, the proper solution is (A), yet I can not identify why this holds true. If $M$ is nonsingular, as is indicated by declarations (B) - (E), after that isn't that equal to its column vectors being linearly independent? And also if the 5 column vectors are independent, after that I can conveniently show that each set of vectors are independent. What am I missing out on?

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2019-05-19 00:11:26
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Linear freedom of $n$ vectors, for $n>2$, is not equal to "pairwise" freedom. Take 3 various vectors in the aircraft, for an instance.
HINT $\rm\ \ u,v\$ straight independent $\rm\ \Rightarrow\:\ u + v,\:u+2v,\: u+3v,\:\ldots\:$ pairwise independent