Cramer's Rule

  • Thread starter mathsTKK
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  • #1
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Hi, everyone, I am mathstkk, I am new to the Physics Forum, but I think, at my first sense, this forum is going to be helpful to me^^

Recently, I met one problem about matrix.
The problem is as follow:
Show that the matrix below has NO non-trivial solution if a+b+c=0
The matrix is
1 cos(c) cos(b)
cos(c) 1 cos(a)
cos(b) cos(a) 1

With the highlighted word 'NO', is it means that the only solution to the matrix is [0]?
If it is so, I would have to find the determinant of the matrix, right?
But still I find that the working was rather complicated with the trigonometric function as the element of the matrix.

So, can someone help me to solve this problem?
I appreciate those who offers help to me^^
Thank you very much to all of you!!!
 

Answers and Replies

  • #2
chiro
Science Advisor
4,790
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Hey mathsTKK and welcome to the forums.

In terms of a non-trivial solution I think, but I may be wrong, that the trivial solution to
a + b + c = 0 is a = 0, b = 0, c = 0 and any other solutions are the non-trivial ones.

If you put in the non-trivial solution you'll get a matrix full of 1's and since they are linearly independent you get det(A) = 0 in this case.
 
  • #3
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Thanks Chiro for that reply^^

Btw, I have tried to find its det too, but it ends up with a mess, the result was as follow:

det=1-cos^2(a)-cos^2(b)-cos^2(c)+2cos(a)cos(b)cos(c)

from a+b+c=0, do I need to substitute c=-a-b to the det equation? But I think that will end up messier,haha!

I have to prove that det=0 but no matter how I try to plug in the unknowns, it jz end up messier.

I dunoe how to continue the equation.

Can u help me? or someone else has any suggestion?

I appreciate all of you^^
 
  • #4
chiro
Science Advisor
4,790
132
Are you are aware of the trig rules like cos(a+b) cos(a-b) and so on? If you use the substitution c =-a-b you can get everything in terms of sin and cos of a and b. For things like -a you can use properties that sin(-a) = -sin(a) and cos(-a) = cos(a).

You'll get everything in terms of sin and cos in a and b and you should get something that is reducible so that you can solve it.
 
  • #5
31
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Oh, Okay, I will try and see and will type the solution here if I solve it.

Btw, from the equation of the det, is there a possible that the terms after the first term, i.e 1 is equals to -cos(a+b+c) which =-cos(0)=-1, then 1-1 will be equals to 0=det?

I try to expand the term cos(a+b+c), but it ends up to
cos(a)cos(b)cos(c)-sin(a)sin(b)cos(c)-sin(a)cos(b)sin(c)-cos(a)sin(b)sin(c),
how to simplify this expansion?

So complicated! haha!
 

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