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This is a question given in an assignment I'm working on:

If the coefficient matrix $A$ in a homogeneous system of 33 equations with 28 unknowns is known to have rank 12, how many parameters are there in the general solution?

I've deduced that, since this is a homogeneous system with fewer variables than equations, the only solution is the trivial solution; I'm unsure, however, how to find the number of parameters in this solution.

How would I go about finding the number of parameters in this kind of abstract situation?

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To the gentleman who posed this question, please refrain from using such sites and see me during my office hours or at the math help center if you require help. –  user104422 Oct 31 '13 at 0:33
    
@BarryJessup Was this for a current test/takehome assignment? I can delete my answer for a week or so if the assignment is still open... (Also, if this user was using the site to cheat, please let one of the moderators know--I'm 99% sure the user can be banned for a certain number of days.) –  anorton Oct 31 '13 at 12:31

1 Answer 1

up vote 2 down vote accepted

You add a parameter for every column without a pivot in the REF form of the coefficient matrix. That is, there are the same number of parameters as the dimension of the null space of $A$. What is the relationship between $\dim NS(A)$ and $\text{rk}(A)$?

Mouse over for relationship...

$$\dim NS(A) = \text{columns} - \text{rk}(A)$$

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Yeah, but he has more equations than variables. –  Kaster Oct 29 '13 at 22:51
    
@Kaster wait a second... my answer still stands, I believe; the first paragraph was wrong, but the main idea was right... –  anorton Oct 29 '13 at 23:05
    
Thanks for the reply, @anorton. What, exactly, is the $NS$ of a matrix (I'm assuming it's the number of parameters)? –  at least three characters Oct 29 '13 at 23:15
    
@user2615799 $NS(A)$ is the null space of $A$. It is the set of all vectors $\vec{x}$ such that $A\vec{x} = \vec{0}$. The dimension of the null space is the same as the number of parameters in a system. If you haven't covered null space yet in your linear algebra course, I can edit my answer to phrase it another way... –  anorton Oct 29 '13 at 23:18

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