Are X & Y Equal in Matrix Equations?

[Mathman23]

Hi

I have this following problem:

Two matrix equations are given

$$C^{T} X = K \ \ Y C^{T} = K$$

where K, X,Y and C are square matrices. If I want to calculate X in equation 1 and Y in equation 2 I multiply with $${C^{T}}^{(-1)}$$ one both sides of each equation.

The resulting matrix X in equation is still equal to Matrix Y in equation two ??

/Fred

 

[arildno]

Not necessarily!
Substitute into equation 1 the expression for K from equation 2:
$$C^{T}X=YC^{T}$$
which, assuming invertibility of $$C^{T}$$ can be rewritten as:
$$C^{T}X(C^{T})^{(-1)}=Y$$
Why should we have X=Y?

 

[Mathman23]

Hi but how do I calculate Y in equation 2 ?

Hope You can help to understand why X could equal Y ?

Sincerley and Best Regards,

Fred

p.s.

Here are the matrices used in the equations..

$$C = \left[ \begin{array}{ccc} 1 & 1 & 2 \\1 & 2 & 4 \\ 2 & -5 & 2 \end{array} \right]$$ and $$K = \left[ \begin{array}{ccc} 1 & 2 & 4 \\-3 & 2 & 0 \\ -1 & -1 & 2 \end{array} \right]$$

Not necessarily!
Substitute into equation 1 the expression for K from equation 2:
$$C^{T}X=YC^{T}$$
which, assuming invertibility of $$C^{T}$$ can be rewritten as:
$$C^{T}X(C^{T})^{(-1)}=Y$$
Why should we have X=Y?

 

[OlderDan]

Hi but how do I calculate Y in equation 2 ?

Hope You can help to understand why X could equal Y ?

Sincerley and Best Regards,

Fred

p.s.

Here are the matrices used in the equations..

$$C = \left[ \begin{array}{ccc} 1 & 1 & 2 \\1 & 2 & 4 \\ 2 & -5 & 2 \end{array} \right]$$ and $$K = \left[ \begin{array}{ccc} 1 & 2 & 4 \\-3 & 2 & 0 \\ -1 & -1 & 2 \end{array} \right]$$

Given these two, you can calculate X and Y explicitly and compare them. They are not equal

 

[Mathman23]

Hi and Thank You for Your answer,

Does $$Y = K {C^{T}}^{(-1)}$$ ?

Given these two, you can calculate X and Y explicitly and compare them. They are not equal

 

[arildno]

Hi and Thank You for Your answer,

Does $$Y = K ]{C^{T}}^{(-1)}$$ ?

Correct; however, if you haven't got the explicit expression for $$(C^{T})^{(-1)}$$
it is better to solve the linear system for the 9 components of Y instead

(In order for two matrices to be equal, their components must be equal; this gives you 9 equations.)

 

[OlderDan]

Correct; however, if you haven't got the explicit expression for $$(C^{T})^{(-1)}$$
it is better to solve the linear system for the 9 components of Y instead

(In order for two matrices to be equal, their components must be equal; this gives you 9 equations.)

Good point. After a long period of doing other things my introduction to these calculators the students all now have has been fairly recent. Punching in a 3 by 3 and hitting the T and -1 buttons is now such a trivial exercise I didn't even think of doing it by hand

 

[arildno]

Calculators??
Are those the things with frills and pink ribbons about them?
I don't like that..

 

[Mathman23]

Hi

I got the correct result now.

Thanks for Your answers,

/Fred

Calculators??
Are those the things with frills and pink ribbons about them?
I don't like that..