Finding unknown matrix a, when aB=C

  • Thread starter Thread starter forrestm
  • Start date Start date
  • Tags Tags
    Matrix
Click For Summary
To find the unknown symmetrical 2x2 matrix 'a' in the equation aB=C, where B and C are known 2x1 matrices, one can set up a system of linear equations by multiplying 'a' with 'B' and equating it to 'C'. The resulting equations indicate that there are infinitely many solutions for 'a', as any values assigned to two of its elements can yield corresponding values for the others. The discussion highlights that the non-invertibility of the 2x1 matrix B contributes to this infinite solution set. Additionally, the symmetry of matrix 'a' does not change the nature of the solutions. The conclusion is that without additional constraints, multiple valid matrices 'a' can satisfy the equation.
forrestm
Messages
3
Reaction score
0
[SOLVED] Finding unknown matrix a, when aB=C

Homework Statement


I know that aB=C, and want to find the matrix a.
B and C are known 2x1 matrices.
a is an unknown symmetrical 2x2 matrix.

This seams like something fairly simple.

--> Can someone point me in the right direction?


Homework Equations



--

The Attempt at a Solution



--
 
Physics news on Phys.org
let's say:

<br /> B = \begin{bmatrix}<br /> 1 \\<br /> 3 \\<br /> \end{bmatrix}<br />

and
<br /> C = \begin{bmatrix}<br /> 4 \\<br /> 6 \\<br /> \end{bmatrix}<br />

then, take 'a' to be:

<br /> a = \begin{bmatrix}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{bmatrix}<br />

Multiply with 'B' to get:

<br /> aB = \begin{bmatrix}<br /> a + 3b \\<br /> c + 3d \\<br /> \end{bmatrix}<br />

Equate it with 'C' to get the following linear equations:

<br /> \begin{bmatrix}<br /> a + 3b \\<br /> c + 3d \\<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 4 \\<br /> 6 \\<br /> \end{bmatrix}<br />

as you can see.. there are infinitely many such matrices possible. Put in any values of 'a' and 'c' and the corresponding values of 'b' and 'd' to get one such matrix.
 
Last edited:
  • Like
Likes CynicusRex
Are you sure? Either you're right and my math problem is wrong, or I haven't represented the matrix dimensions properly.

a = 2x2
|xx|
|xx|

B = 2x1
|x|
|x|
C = 2x1
|x|
|x|Anyway.. would this method work?
a.B = C
a.B.B^-1 = c.B^-1
a = c.B^-1

edit: No it wouldn't. a 2x1 matrix is not invertible.
 
Last edited:
rohanprabhu said:
let's say:

<br /> B = \begin{bmatrix}<br /> 1 \\<br /> 3 \\<br /> \end{bmatrix}<br />

and
<br /> C = \begin{bmatrix}<br /> 4 \\<br /> 6 \\<br /> \end{bmatrix}<br />

then, take 'a' to be:

<br /> a = \begin{bmatrix}<br /> a &amp; b \\<br /> c &amp; d \\<br /> \end{bmatrix}<br />

Multiply with 'B' to get:

<br /> aB = \begin{bmatrix}<br /> a + 3b \\<br /> c + 3d \\<br /> \end{bmatrix}<br />

Equate it with 'C' to get the following linear equations:

<br /> \begin{bmatrix}<br /> a + 3b \\<br /> c + 3d \\<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 4 \\<br /> 6 \\<br /> \end{bmatrix}<br />

as you can see.. there are infinitely many such matrices possible. Put in any values of 'a' and 'c' and the corresponding values of 'b' and 'd' to get one such matrix.

Thanks
That certainly helped.

Considering the fact that a is symmetric, would your outcome be any different?
 
forrestm said:
Are you sure? Either you're right and my math problem is wrong, or I haven't represented the matrix dimensions properly.

well.. sorry for that.. i first got confused over it. Anyways, I've edited my post as you can see :D

forrestm said:
edit: No it wouldn't. a 2x1 matrix is not invertible.

yes.. this is also the reason why there are infinite solutions to this problem. Because a property of matrices is that if a given matrix is invertible, then it has a unique inverse.

forrestm said:
Considering the fact that a is symmetric, would your outcome be any different?

i didn't get what u are trying to say...
 

Similar threads

Replies
9
Views
2K
  • · Replies 69 ·
3
Replies
69
Views
9K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K