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This seems pretty simple but I am having a hard time figuring it out.
I need to find four variables, $\,a,\, b,\, c,\, d,\,$ and I have four equations:

1) $\;u_1a -v_1b + c = u_1' $

2) $\;u_1b -v_1a + d = v_1' $

3) $\;u_2a -v_2b + c = u_2' $

4) $\;u_2b -v_2a + d = v_2' $

Note that $\,u_1,\, v_1,\, u_1', \,v_1',\, u_2, \,v_2, \,u_2',\, v_2'\,$ are all known.

I tried to do substitution (setting one equation to one variable and plugging in) but it got messy and then I got lost.

Any suggestions to how to solve for the four variables, $a,b,c,d$?
This is for similarity transform

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3  
Welcome! Thanks for stating what you've tried; yes substitution can get messy; the more unknowns, the messier. That's where row operation come in to play, for solving a system of equations, in this case, of four equations in 4 unknowns. Have you dealt with row operations, or matrices for representing such a system of equations? –  amWhy Nov 3 '12 at 22:53
    
Like linear algebra? It's has been awhile since I have taken that class. Could you give me some details as to how to approach the problem the way you described? –  ealeon Nov 3 '12 at 22:59

1 Answer 1

up vote 6 down vote accepted

Four equations, four unknowns: $a, b, c, d.$

1) $u_1a -v_1b + c = u_1' $

2) $u_1b -v_1a + d = v_1' $

3) $u_2a -v_2b + c = u_2' $

4) $u_2b -v_2a + d = v_2' $

Try subtracting row 3 from row 1 (the "c's" cancel out):

5) $(u_1 -u_2)a - (v_1 - v_2)b = (u_1^\prime - u_2^\prime)$

Subtracting row 4 from row 2 (the "d's" cancel out):

6) $(u_1 -u_2)b - (v_1-v_2)a = (v_1^\prime - v_2^\prime)$

Rearranging 6):

6)$^\prime (v_2 -v_1)a + (u_1-u_2)b = (v_1^\prime - v_2^\prime)$

Now you're down to two equations (5, 6') in two unknowns!

Can you take it from here?

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yes, thank you!!! –  ealeon Nov 3 '12 at 23:05
    
@ealeon You're welcome! –  amWhy Nov 3 '12 at 23:09

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