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Can this problem be solved using Linear Algebra? Proof type problem.

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canis

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A, B and C are constants. It can be proven using trig that theta=-80 for all positive real values of A, B, C that make the three equations true. In other words, for the given conditions, theta must be equal to -80 for all three equations to be true, if theta is not equal to -80 at least one equation must be false. Degrees are not in radians. Can anyone show how to prove theta=-80 using linear algebra? Thank you.
 
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4 unknowns, 3 equations. I don't know any way to fully solve the equation without guessing and checking at some stage. You can substitute the third into the first to get 3 unknowns, sub the second into the resulting equation to get it into C and Theta, but at that point you'll have to do some sort of guess and check.
 
deadlyapp said:
4 unknowns, 3 equations. I don't know any way to fully solve the equation without guessing and checking at some stage. You can substitute the third into the first to get 3 unknowns, sub the second into the resulting equation to get it into C and Theta, but at that point you'll have to do some sort of guess and check.
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I am looking for a way to prove it using linear algebra. It can be proven with trig.
 
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canis said:
Theta can be proven to be -80 for all real values A,B,C. There should be a way to prove it using linear algebra.
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I hate proofs and I've forgotten a large amount of linear algebra.

A combination of elimination and trig identities maybe?
 
silverpig said:
I don't think so, because the last equation is not linear in A, B, or C.
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That was my thought process as well. Some sort of manipulation of the existing equations that will probably include trig identities (considering they have trig functions in them).
 
If you look at your equations they very closely resemble a rotation of the vector [B C] such that it is transformed into [A 0]. The last equation just cements this since the length of the vector will not change. Therefor cos(theta) = -sin(10) and sin(theta) = cos(10);
 
I would say no. You have three equations, and four unknowns. It cannot be solved through purely algebraic methods, that is, you must have some additional information (such as geometric knowledge) in order to solve it.
 
If (A,B,C,theta) is a solution, then so is (tA,tB,tC,theta). This means that it won't be possible to find A, B and C (unless A=B=C=0), but finding theta might be possible.

If C<>0 then you can set C=1 because it won't make a difference to theta, and then try to find A, B and theta because you now have 3 equations and 3 unknowns.
 
I spent a few minutes on this earlier & can see where the -80 degrees is popping out of the equation for theta. But, to get there, I had to use some trig identities; not linear algebra.

Since it results from a term with (cos(10-theta)), I'm not positive (yet) that it's a unique solution for theta, or if it's simply a trivial solution for theta.
 
You need trig identities, linear algebra won't solve for 4 unknowns with 3 equations. You could probably use matrices though.
 
Matrices are only reductions of linear equasions VG101.

If you can't solve it by linear, you can't put them into a matrix and go all NEO on their asses.
 
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