vectors and matrices (a tough one!)

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mattj35
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#1
Report Thread starter 16 years ago
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A rectangular plate has its edges maintained at temeratures 0 degrees, 5 degrees, 10 degrees and 20 degrees.

The temperature t1 is approximately the average of the four surrounding temperatures, and similarly for the teperature t2 so that,

t1 = 1/4(20+10+t2+0)

t2 = 1/4(t1+10+5+0)

Simplifying these equations gives,
4t1 - t2 = 30
t1 - 4t2 = -15

write these equations as a single matrix equation, and use an inverse matrix to find t1 and t2.

I would appreciate it if anyone knows the answer to this one, I have not got a clue!
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Gaz031
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#2
Report 16 years ago
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(Original post by mattj35)
A rectangular plate has its edges maintained at temeratures 0 degrees, 5 degrees, 10 degrees and 20 degrees.

The temperature t1 is approximately the average of the four surrounding temperatures, and similarly for the teperature t2 so that,

t1 = 1/4(20+10+t2+0)

t2 = 1/4(t1+10+5+0)

Simplifying these equations gives,
4t1 - t2 = 30
t1 - 4t2 = -15

write these equations as a single matrix equation, and use an inverse matrix to find t1 and t2.

I would appreciate it if anyone knows the answer to this one, I have not got a clue!
They can be written:
(4 -1) (t1) = (30)
(1 -4) (t2) (-15)

If the first matrix is called A then:
(t1) = A^-1 (30)
(t2) ......... (-15)
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dvs
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Generally, if you have two simultaneous equations of the type:
ax + b = c
dx + e = f

Then you could re-write them into the matrix equation:
Ax = b

Where A is called the matrix of coefficients, x is the coefficients vector and b is the solutions vector.

A =
a b
d e

x =
x
y

b =
c
f

To solve the matrix equation, multiply both sides by A^(-1):
A^(-1) . Ax = A^(-1) b
Ix = A^(-1) b, since A^(-1) . A = I
x = A^(-1) b

Make sure you multiply using the correct order for the RHS, since A^(-1).b doesn't equal b.A^(-1).

After you multiply, x = first line of RHS and y = second line of RHS.
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