Range of a Matrix Transformation linear algebra

[SpiffyEh]

Homework Statement

Given,
A =
[ 1 -3 4;
-3 2 6;
5 -1 -8]

b =
[b_1;
b_2;
b_3]
Show that there does not exist a solution to Ax = b for every b in R3 and describe the set of all {b1,b2,b3} for which Ax = b does have a solution.

Homework Equations

row reduction

The Attempt at a Solution

I row reduced until I got the following augmented matrix:
[ 1 -3 4 | b_1;
0 1 (-18/7)| (b_2 + 3*b_1)/7;
0 0 8 | b_3-5*b_1 - 14*b_2]
I'm confused about this because I was lead to believe that since there is a pivot for every value then there should exist a solution for every b in R3. And There is in fact a pivot in every row. Can someone explain to me if there is a solution or not and why please. I'm just not seeing why there wouldn't be one. Thank you

 

[vela]

Are you sure you wrote down the matrix correctly? In particular, is the 4 in the first row supposed to be a -4?

 

[SpiffyEh]

Are you sure you wrote down the matrix correctly? In particular, is the 4 in the first row supposed to be a -4?

oops, wow I feel dumb. I even double checked it. So, I had the right idea at least right?

 

[SpiffyEh]

ok so with that -4 in place I got the matrix to be:

[ 1 -3 -4 | b_1;
0 1 (6/7)| (b_2 + 3*b_1)/7;
0 0 0 | b_3-5*b_1 - 14*b_2]

from this i can see that there isn't an answer for all b in R3. The 2nd part of the question asks "describe the set of all {b1,b2,b3} for which Ax = b does have a solution"

this would be where b_3-5*b_1 - 14*b_2 is equal to zero. Otherwise there would be no solution. Would I just write this or do i need to say something about b1 and b2? Would i need to say that b1 and b2 could be anything? Or how would I go about saying that?

 

[vela]

It depends on what "describe" means here. You might be expected to give a geometric interpretation of the solution set.

 

[HallsofIvy]

ok so with that -4 in place I got the matrix to be:

[ 1 -3 -4 | b_1;
0 1 (6/7)| (b_2 + 3*b_1)/7;
0 0 0 | b_3-5*b_1 - 14*b_2]

from this i can see that there isn't an answer for all b in R3. The 2nd part of the question asks "describe the set of all {b1,b2,b3} for which Ax = b does have a solution"

this would be where b_3-5*b_1 - 14*b_2 is equal to zero. Otherwise there would be no solution. Would I just write this or do i need to say something about b1 and b2? Would i need to say that b1 and b2 could be anything? Or how would I go about saying that?

Good: $b_3- 5b_1- 14b_2= 0$ and you can write that as $b_3= 5b_1+ 14b_2$ so that
$$\begin{pmatrix}b_1 \\ b_2 \\ b_3\end{pmatrix}= \begin{pmatrix} b_1 \\ b_2 \\ 5b_1+ 14b_2 \end{pmatrix}= b_1\begin{pmatrix}1 \\ 0 \\ 5\end{pmatrix}+ b_2\begin{pmatrix}0 \\ 1 \\ 14\end{pmatrix}$$.

Geometrically, what is that?