Does infinite solutions imply the row vectors are linearly dependent?

[mitch_1211]

if i have a 4x3 matrix, this means there are more equations than unknowns and so there are no solutions to the system.

does this mean that the row vectors are linearly dependent?

 

[henry_m]

The rows must be linearly dependent because they are four vectors living in a three-dimensional vector space. It doesn't necessarily mean there are no solutions to Ax=b (A the matrix, b a known 4x1 vector and x an unknown 3x1 vector) though: for example, the last two rows of A may be equal and the last two entries of b also equal, in which case it is reduced to 3 equations with 3 unknowns, which may or may not have a solution.

 

[mitch_1211]

The rows must be linearly dependent because they are four vectors living in a three-dimensional vector space.

Of course! I don't know why I didn't realize this straight away.

Thank you very much!

 

[td21]

Your question is about overdetermined system(4x3 in your example). It is mostly inconsistent but it can be consistent(as stated in post 2). If consistent, it may even have infinite solution.

For underdetermined system, it is mostly consistent but it can be inconsistent. If consistent, it must have infinite solution.So, "Does infinite solutions imply the row vectors are linearly dependent?"
Ans: Yes for overdetermined system(4x3 in your example). No for underdetermined system,
e.g
x+y+z=2
x+y+2z=3

have infinite solutions but the row vectors are linearly independent.