How do you know that Ax=b can be solved for all b?

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SUMMARY

The discussion centers on solving the equation Ax = b for a 3x4 matrix A, where the only special solution to Ax = 0 is S = (2,3,1,0). The rank of A is determined to be 3, indicating that the nullity is 1. The row reduced echelon form (RREF) of A is provided as R = ((1, 0, -2, 0), (0, 1, -3, 0), (0, 0, 0, 1)). It is established that Ax = b can be solved for all b in R3 because the transformation represented by A is onto R3, given that dim(ker(A)) = 1.

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Homework Statement



Suppose you know that the 3 by 4 matrix A has the vector S = (2,3,1,0) as the only special solution to Ax = 0.
(a) What is the rank of A and the complete solution to Ax=0
(b) What is the exact row reduced echelon form R of A?
(c) How do you know that Ax=b can be solved for all b?

Homework Equations



r(A) = n - r

The Attempt at a Solution



a) r(A) = n - r(nullspace) = 4 - 1 = 3
x = a(2,3,1,0), a = any scalar
b) R has 3 pivots and Rx=0 with x = (2,3,1,0)
R = (
(1, 0, -2, 0),
(0, 1, -3, 0),
(0, 0, 0, 1))
c) Clueless. :( How should I attempt this?
 
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OK, what I posted here first was complete nonsense, so let me try again...

How would you normally like to solve Ax = b?
 
I don't know how to find the determinant of R, it's not a square matrix, it's 3x4.

Edit: Find a particular solution for Axp=b by setting free variables to 0, then find special solutions for Ax0 = 0 and express x as xp + x0.
 
A has to be non-singular if the vector equation Ax=b, is to have a unique solution for every column vector b.
 
sutupidmath said:
A has to be non-singular if the vector equation Ax=b, is to have a unique solution for every column vector b.
But since A is 3X4, there's no chance of A being nonsingular.

However, since dim(ker(A)) = 1, it should be easy to show that the transformation represented by A is onto R3. Hence for any vector b in R3, there is some vector x in R4 such that Ax = b.

Also, since Cade has the RREF form of the matrix, he/she can set up an augmented matrix that represents Ax = b (where b is an arbitrary vector in R3), and show b has a preimage vector x.
 
Last edited:
Mark44 said:
However, since dim(ker(A)) = 1, it should be easy to show that the transformation represented by A is onto R3. Hence for any vector b in R3, there is some vector x in R4 such that Ax = b.

Also, since Cade has the RREF form of the matrix, he/she can set up an augmented matrix that represents Ax = b (where b is an arbitrary vector in R3), and show b has a preimage vector x.

I am sorry for the late reply, my homework was just handed back to me today. I tried the approach second and it worked, now I also understand the first. Thanks for your help.
 

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