- #1

- 15

- 0

w*a = b -2a + e

w*b = c -2b + a

w*c = d -2c + b

of the style w*u = T*u where u is a vector, u = {a, b, c} and T is a matrix. w is just a number.

- Thread starter andybham
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- #1

- 15

- 0

w*a = b -2a + e

w*b = c -2b + a

w*c = d -2c + b

of the style w*u = T*u where u is a vector, u = {a, b, c} and T is a matrix. w is just a number.

- #2

mathman

Science Advisor

- 7,904

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What are d and e? If they are really different and not typos, then you will need T*u+v=w*u, where v is the vector (e,0,d) and ------------------------------------------------

------------------------------------------------------------------

T=

-2 1 0

1 -2 1

0 1 -2

--------------------------------------------------------------------------------

------------------------------------------------------------------

T=

-2 1 0

1 -2 1

0 1 -2

--------------------------------------------------------------------------------

Last edited:

- #3

- 15

- 0

no they are not typos, could I not use

T = -2 1 e/c

1 -2 1

d/a 1 -2

or is that just stupid?

T = -2 1 e/c

1 -2 1

d/a 1 -2

or is that just stupid?

- #4

radou

Homework Helper

- 3,115

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What exactly are the unknowns in your equations? a, b, c? If so, then

w*a = b -2a + e

w*b = c -2b + a

w*c = d -2c + b

of the style w*u = T*u where u is a vector, u = {a, b, c} and T is a matrix. w is just a number.

(2 + w)a - b = e

-a + (2 + w)b - c = 0

-b + (2 + w)c = d ,

which can be written as

[tex]\left(\begin{array}{ccc}2+w & -1 & 0\\-1 & 2+w & -1\\0 & -1 & 2+w\end{array}\right) \left(\begin{array}{ccc}a \\b\\c\end{array}\right)=\left(\begin{array}{ccc}e \\0\\d\end{array}\right)[/tex].

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