Can You Find Matrices B and C Where AB = AC but B ≠ C?

  • Thread starter Thread starter LaraCroft
  • Start date Start date
  • Tags Tags
    Matrix Suggestions
Click For Summary
To find matrices B and C such that AB = AC but B ≠ C, it is essential to understand that matrix A has rank 1, meaning its null space allows for multiple solutions. Any components of B and C that lie in the null space of A will result in the same product when multiplied by A, leading to the desired equality. Using variables for the elements of matrix B and performing row operations on A can help identify how these variables can change without affecting the product. Row reducing A while applying the same transformations to the identity matrix can clarify the relationships between B and C. This approach provides a systematic way to explore potential matrices that meet the criteria.
LaraCroft
Messages
14
Reaction score
0
Let A =

[ -3 9 ]
[ -1 3 ]

Find two 2x2 matrices B and C, such that AB = AC BUT B does not equal C.

I am stumped, but I am trying inverses and such, with no luck :(

Any insight would be great!:wink:
 
Physics news on Phys.org
LaraCroft said:
Let A =

[ -3 9 ]
[ -1 3 ]

Find two 2x2 matrices B and C, such that AB = AC BUT B does not equal C.

I am stumped, but I am trying inverses and such, with no luck :(

Any insight would be great!:wink:

Matrix A has rank 1. Any component of the column vectors in B and C in the null space of A will be mapped to zero. This leaves quite a few possibilities.

However, if you haven't learned about null spaces yet, Just use variables for the ellements of the B matrix, multiply A by B and see how these variables can be changed without affecting the product.

This might be more clear if you row reduce A. Remember that each row operation can be represented by a linear transformation. Doing the row operation on the identity matrix will give you the desired transformation. Therefor you can safely row reduce A without affecting the solution to the problem.
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K