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

  • Thread starter Cade
  • Start date
In summary, A has a rank of 3, a complete solution to Ax = 0, and can be solved for all b using the row reduced echelon form R.
  • #1
Cade
92
0

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?
 
Physics news on Phys.org
  • #2
OK, what I posted here first was complete nonsense, so let me try again...

How would you normally like to solve Ax = b?
 
  • #3
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.
 
  • #4
A has to be non-singular if the vector equation Ax=b, is to have a unique solution for every column vector b.
 
  • #5
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:
  • #6
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.
 

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

1. How do you determine if Ax=b can be solved for all b?

To determine if Ax=b can be solved for all b, we can use the rank-nullity theorem. This theorem states that the dimension of the null space of a matrix A is equal to the number of linearly independent columns of A. Therefore, if the rank of A is equal to the number of columns in A, then the null space is equal to 0 and Ax=b can be solved for all b.

2. Can you solve Ax=b for all b if A is a square matrix?

Yes, if A is a square matrix and has full rank (i.e. its columns are linearly independent), then Ax=b can be solved for all b. This is because the dimension of the null space of A is equal to 0, and therefore the only solution to Ax=0 is the trivial solution.

3. What is the significance of the rank of a matrix in solving Ax=b?

The rank of a matrix A represents the number of linearly independent columns in A. In order for Ax=b to have a unique solution, the number of linearly independent columns in A must be equal to the number of columns in A. This is because it ensures that the null space of A is equal to 0, and therefore there is only one solution to Ax=b.

4. Are there any cases where Ax=b cannot be solved for all b?

Yes, if the rank of A is less than the number of columns in A, then the null space of A is greater than 0 and Ax=b cannot be solved for all b. This means that there are infinitely many solutions to Ax=0, making it impossible to find a unique solution for Ax=b.

5. Can you explain how Gaussian elimination is used to solve Ax=b?

Gaussian elimination is a method used to reduce a matrix A to its row echelon form. This process involves using elementary row operations such as swapping rows, multiplying rows by a scalar, and adding multiples of one row to another. Once A is in its row echelon form, we can easily solve Ax=b by back substitution. This method works for all b as long as the rank of A is equal to the number of columns in A.

Similar threads

Replies
4
Views
592
  • Calculus and Beyond Homework Help
Replies
24
Views
980
  • Calculus and Beyond Homework Help
Replies
3
Views
446
  • Calculus and Beyond Homework Help
Replies
1
Views
420
  • Calculus and Beyond Homework Help
Replies
2
Views
510
  • Calculus and Beyond Homework Help
Replies
3
Views
435
  • Calculus and Beyond Homework Help
Replies
1
Views
875
  • Calculus and Beyond Homework Help
Replies
1
Views
531
  • Calculus and Beyond Homework Help
Replies
1
Views
310
  • Calculus and Beyond Homework Help
Replies
1
Views
593
Back
Top