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Finding unknown matrix a, when aB=C

  • Thread starter forrestm
  • Start date
3
0
[SOLVED] Finding unknown matrix a, when aB=C

1. 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?


2. Homework Equations

--

3. The Attempt at a Solution

--
 

Answers and Replies

412
2
let's say:

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

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

then, take 'a' to be:

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

Multiply with 'B' to get:

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

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

[tex]
\begin{bmatrix}
a + 3b \\
c + 3d \\
\end{bmatrix} =
\begin{bmatrix}
4 \\
6 \\
\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:
3
0
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:
3
0
let's say:

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

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

then, take 'a' to be:

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

Multiply with 'B' to get:

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

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

[tex]
\begin{bmatrix}
a + 3b \\
c + 3d \\
\end{bmatrix} =
\begin{bmatrix}
4 \\
6 \\
\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?
 
412
2
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

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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