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Matrices - Linear Equations 3

https://isaacphysics.org/questions/3x3_equation_3?board=d5bdf9cd-5c49-43cc-83ff-e7703d392d45&stage=further_a

I dont really get what im doing wrong with this question.
I set up my equation as:

0k + mw^2 =0 -> w1^2 =0
-k + mw^2 = 0 -> w2^2 = k/m
-2k + mw^2 = 0 -> w3^2 = 2k/m, but my equation for w3^2 was wrong and I dont really know why.

Any help greatly appreciated.
(edited 9 months ago)
Reply 1
Matrix:
Reply 2
Did you solve det = 0 (find eigenvalues of A) as it suggests?
Reply 3
That is incredible from me, I just recognized it as absolute value and didnt give it a second thought.
Thank you - though I have a question - why does this solution require the use of a determinant - aren't they only used to find the inverse of the matrix?
Is this basically a way of saying if the determinant is 0, then the inverse cant exist and then there must be no solutions?
My logic could be completely wrong here.
Reply 4
Original post by mosaurlodon
That is incredible from me, I just recognized it as absolute value and didnt give it a second thought.
Thank you - though I have a question - why does this solution require the use of a determinant - aren't they only used to find the inverse of the matrix?
Is this basically a way of saying if the determinant is 0, then the inverse cant exist and then there must be no solutions?
My logic could be completely wrong here.

I guess youve not done eigenvalues, but if you had a relationship like
Ax = lambda x
for a scalar lambda, which is basically what you have here, then
(A - lambda I) x = 0
so for a nonzero x, there must be value of lambda which maps it to zero. If (A-lambda I) was invertible then take inverses and youd get x = 0. So A must be singular so the det(A - lambda I) = 0 and solve for lambda. If det() is zero, when you try and form the inverse matrix you have a "divide by zero", so its similar to trying to solve 0*x = 0

For this question, its relatively straighforward and you can "forget" about the k and m as you can simply scale the numerical answer by k/m at the end.
(edited 9 months ago)
Reply 5
Oh I dont do eigenvalues until further pure yr 2, but thats cool to know about.
Thinking about a nonzero x makes the determinant make sense actually, since it has a contradiction otherwise.

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