Finding unknown matrix a, when aB=C

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Homework Help Overview

The problem involves finding an unknown symmetrical 2x2 matrix 'a' given the equation aB=C, where B and C are known 2x1 matrices. The original poster expresses that this seems like a straightforward task and seeks guidance on how to approach it.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • One participant suggests a specific form for matrices B and C and derives linear equations from the multiplication of 'a' with 'B'. They note the existence of infinitely many solutions based on the values chosen for 'a' and 'c'. Another participant questions the dimensionality of the matrices and explores the implications of matrix invertibility, ultimately recognizing that a 2x1 matrix cannot be inverted.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the implications of matrix properties. Some guidance has been offered regarding the nature of solutions, but no consensus has been reached on the implications of symmetry in matrix 'a'.

Contextual Notes

Participants are considering the constraints of the problem, including the symmetry of matrix 'a' and the implications of having infinitely many solutions due to the non-invertibility of the 2x1 matrix B.

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[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



--
 
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let's say:

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

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

then, take 'a' to be:

[tex] a = \begin{bmatrix}<br /> a & b \\<br /> c & d \\<br /> \end{bmatrix}[/tex]

Multiply with 'B' to get:

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

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

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

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:
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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:

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

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

then, take 'a' to be:

[tex] a = \begin{bmatrix}<br /> a & b \\<br /> c & d \\<br /> \end{bmatrix}[/tex]

Multiply with 'B' to get:

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

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

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

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...
 

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