Solving Linear Equations: Understanding Matrices and Equality"

[Ry122]

in this system of matrices, ignoring the N M matrix, is matrix [A B B D] equal to [C] given that the matrices represent a system of linear equations in the form Ax = b?

I'm just wanting to know whether it's like algebra, where you can divide both sides by the same thing and keep equality.

[PLAIN]http://img294.imageshack.us/img294/7428/matrw.jpg

 

[lkh1986]

Sounds to me it's something to do with eigenvalue thingy.

[C] might be a real value or scalar and [A B B D] is a matrix with 4 elements.

For example, [C], or simply C = 4, and [A B B D] = [4 0 0 4]. Also, [e k] = [1 2].

So, [C][e k] = 4 [1 2] = [4 8]
Also, [A B B D] [e k] = [4 8]

So, [C][e k] = [A B B D] [e k] = [4 8]

 

[Ry122]

I have the values for the [A B B D] matrix, but not for anything else, and I'm expected to find [C].

How would I do that?

 

[lkh1986]

I have the values for the [A B B D] matrix, but not for anything else, and I'm expected to find [C].

How would I do that?

If your C is a real number, then you can find the value, or we call it the eigenvalue by solving
\left|A-\lambda I\right|=0, where A is the matrix [A B B D], \lambda is the eigenvalue, or C in this case, and I is identity matrix. And |.| is the determinant.

 

[Ry122]

Is the eigenvalue in that equation you've given, the eigenvalue of matrix A?

 

[lkh1986]

Yes. that's correct. In particular, since your A is a 2x2 matrix, you will find 1 (repeated), or at most 2 eigenvalues.

 

[Ry122]

Why is it that the eigenvalue needs to be calculated in this situation anyway?
Is it true for any system where the matrices are in the format
Ax=Bx
that the equation you gave above needs to be applied to find the unknown?

 

[lkh1986]

Why is it that the eigenvalue needs to be calculated in this situation anyway?
Is it true for any system where the matrices are in the format
Ax=Bx
that the equation you gave above needs to be applied to find the unknown?

The equation \left|A-\lambda I\right|=0 can actually be derived from Ax=Bx, where A is a matrix, x is column matrix and B is a scalar value.

You see, from Ax=Bx, you will get Ax-Bx=0. Factor out the x ill give you x\left(A-BI\right)=0.

For the rest of the explanation, you can refer this article.
http://en.wikipedia.org/wiki/Characteristic_polynomial