# Homework Help: Help with Gaussian Elimination

1. Sep 28, 2005

### jtm

S =

Columns 1 through 3

1.0000 0 0
0 1.0000 0
0 0 1.0000

Columns 4 through 5

0.2750 -0.2786
-0.1750 0.5929
0.2250 0.1357

Which is the rref of

A =

2 9 9 1 6
2 7 3 0 4
9 6 7 3 2

(i) Which columns of S are the pivot columns?
(ii) Which variables xi are the free variables?
(iii) What is the rank of R?
(iv) What is the rank of A?
(v) What is the nullity of A?
(vi) Why does the equation Ax = b have a solution?

i Columns 1 2 3 are pivot columns in S.
ii x4 and x5 are free variables because we are in R^3.
iii Rank of R is 3.
iv Rank of A is 3 also.
v Nullity of A is 0.
vi Ax = b has a solution because in the rref of [A b] is consistant, and since it was consistant we had 3 variables (x1-x3) (because of R^3 space) which depend on the free variables and 2 free variables (x4-x5).

I am having trouble with mostly vi. Just because this matrix in rref is supposed to have x4 and x5 in R^3 space? I'm lost. Help! Thanks!

2. Sep 29, 2005

### CrankFan

Since no one seems to want to touch this you're going to have to settle for an explanation from a B teamer like me.

i) Okay.
ii) True, x4 and x5 are free but it's not because you're "in" R^3. A 3 x 5 matrix could have pivot columns of 1, 4 and 5, which would make x2 and x3 free.
iii) Okay
iv) Okay
v) This is wrong. Check your definition of Nullity.
vi) [A b] is either consistent or inconsistent. If and only if it's inconsistent the matrix RREF(A b) has a special property. What is that property? Based on what we know of RREF(A) ... can that property every be satisfied? Hint: No it can't.

Last edited: Sep 29, 2005