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

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Homework Help Overview

The problem involves a 3 by 4 matrix A and explores the conditions under which the equation Ax = b can be solved for all vectors b in R3. The original poster presents a specific scenario regarding the null space and rank of A, and seeks clarification on the implications for the solvability of Ax = b.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the rank of the matrix and its relationship to the null space. There are attempts to understand the implications of A being non-singular and how that affects the existence of solutions for Ax = b. Questions arise regarding the determinant and the nature of the transformation represented by A.

Discussion Status

Some participants have provided insights into the relationship between the kernel of A and the ability to find solutions for any vector b. There is acknowledgment of the original poster's progress in understanding the problem, and guidance has been offered regarding setting up an augmented matrix to explore solutions.

Contextual Notes

Participants note that A is a 3x4 matrix, which influences the discussion about its singularity and the dimensionality of its kernel. The original poster's confusion about the determinant is also highlighted, given the non-square nature of the matrix.

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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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