Is Ax = b Consistent Given This RREF?

[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 put my answers as:
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!

 

[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.

 

[quicknote]

I would like to clarify and provide a more detailed explanation for the answers given above.

i) The pivot columns in S are columns 1, 2, and 3. These are the columns that contain the leading entries (or pivots) in the rref of S. These pivots determine the basic variables in the system.

ii) In this case, the free variables are x4 and x5. These are the variables that do not have a pivot in their respective columns in the rref of S. They are also known as the non-basic variables.

iii) The rank of R is 3 because it has 3 pivot columns. The rank of a matrix is equal to the number of pivot columns in its rref.

iv) The rank of A is also 3 because it is equal to the number of pivot columns in its rref. This means that the columns of A are linearly independent and span a 3-dimensional space.

v) The nullity of A is 0 because it is equal to the number of free variables in the system. Since there are no free variables, the null space (or set of solutions to Ax = 0) is only the trivial solution.

vi) The equation Ax = b has a solution because the rref of [A b] is consistent. This means that there exists at least one solution to the system of equations. In this case, there are 3 basic variables (x1-x3) and 2 free variables (x4-x5) which determine the values of the basic variables. The solution to the system is given by [x1, x2, x3, x4, x5] = [a, b, c, d, e], where a, b, and c are determined by the pivots in the rref of [A b] and d and e can be chosen arbitrarily as they correspond to the free variables. This is possible because the columns of A are linearly independent and span a 3-dimensional space, allowing for a unique solution to the system.