# GRE practice prob

## Homework Statement

This is a GRE practice problem I was reading through that confused me.

36. Let M be a 5x5 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 5x1 column vector b.
(D) The determinant of M is nonzero.
(E) There exists a 5x5 real matrix N such that NM is the 5x5 identity matrix.

IMT

## The Attempt at a Solution

I was under the impression that all of these were equivalent via IMT. Am I misreading something? Thanks for the help!

I think it's A. Why? Consider the following matrix:
Code:
1 0 0 0 1
0 1 0 0 1
0 0 1 0 1
0 0 0 1 1
0 0 0 0 0

Take any two columsn. They are linearly independent.

Now take b = (0,0,0,0,1). There is no solution to Mx=b.

Savvy?

Yea, I see it now. One of the columns has to be a linear combination of all the others if not invertible, but not a linear combination of just one. Thanks for the response, I feel a little silly in overlooking that!

if the determinant is nonzero, that means there is a unique solution to any homogeneous or non-homogeneous system. And also you can't take the inverse of a matrix with a zero determinant. But A says something different. The linear independence of the columns don't really say anything about the determinant. Only if any two of the columns (or rows) are linearly dependent, you can say that the determinant is zero. But independence doesn't make any conclusions about the determinant (as seen in the example)

"Only if any two of the columns (or rows) are linearly dependent, you can say that the determinant is zero. But independence doesn't make any conclusions about the determinant (as seen in the example)"

Strictly speaking, pairwise linear independence for the columns or rows still isn't enough. You need all vectors in the column space to be linearly independent (not pairwise... but all together). This means that the solution to:

(a1)v1 + (a2)v2 + ... + (an) vn = 0

Must have only the trivial solution a1 = a2 = ... = an = 0.

This, however, is not equivalent to the pairwise linear independence of column vectors, which states that

(a1)v1 + (a2)v2 = 0 imples a1 = a2 = 0 for all v1, v2. My example demonstrates this.