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anonymous
 4 years ago
hey im struggling with a qustion, can someone help please?
if A = the matrices:
−1 4 −2
−3 4 0
−3 1 3
Find an invertable matrix P and diagonal matrix D so that AP = PD.
thank you for any feedback x
anonymous
 4 years ago
hey im struggling with a qustion, can someone help please? if A = the matrices: −1 4 −2 −3 4 0 −3 1 3 Find an invertable matrix P and diagonal matrix D so that AP = PD. thank you for any feedback x

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1greenday will have to help you since they feel noone is capable of helping out ...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1D is just the eugene values on the identity

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1I cant recall for the life of me what to do with the P tho

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1do you recall how to find the eugene values?

phi
 4 years ago
Best ResponseYou've already chosen the best response.0It is asking for the eigenvalues (diagonal entries of D) and eigenvectors (cols of P)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1−1L 4 −2 −3 4L 0 −3 1 3L determinate of that and solve for L

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i spose we could echlon it too to find out but i never get a good result that way

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1eugene vectors; thats the P .... ill remember that someday lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1−1 4 −2 −3 4 0 −3 1 3 −1 4 −2 0 8 6 0 11 9 −1 4 −2 0 8 6 0 0 6/8 maybe?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1−1L 4 −2 −3 4L 0 −3 1 3L (1L)( (4L)(3L)0) +3 (4(3L)1()2) 3(0(4L)(2) (1L)(L^27L+12) 3612L+6 24+6L L^3+6L^211L+6 with any luck thats the eugene equations

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1L=1 1+6+11+6 not= 0 ugh!!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1maybe the eugenes are the negative diags? or i simply mismathed the whole thing

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1Yes, the matrix D has the eigenvalues of A down it's diagonal. So first you need to find the eigenvalues. Then the matrix P is the matrix of the eigenvectors as column vectors corresponding to those eigenvalues.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1if we echoln the matrix, dont we get the Evalues as well? or is that just me wishful thinking?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1No, we don't. There's no escaping solving the characteristic equation. @sammy_maths: I'm sure you have a worked example of a problem like this in your class notes, albeit for a 2x2 or 3x3 matrix. I can't find a good one to link to right now. But go back and have a look at it. And then if you're still stuck, as us again.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1bummer... http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B1%2C4%2C2%7D%2C%7B3%2C4%2C0%7D%2C%7B3%2C1%2C3%7D%7D you can use this for a check :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=characteristic+polynomial%7B%7B1%2C4%2C2%7D%2C%7B3%2C4%2C0%7D%2C%7B3%2C1%2C3%7D%7D yay, my char eq was right lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i know L=1 works by sheer luck, so whats left over is a quadratic

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have been following it, and I'm understanding where you get the values from so far The only notes that may help now is the answer to P = 1 2 1 3 3 1 4 3 1 thank you x

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yeah, ill have to go over finding Evectors again to refresh me cobwebs :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1it may be from the augmented  A LI gauss jordon stuff

phi
 4 years ago
Best ResponseYou've already chosen the best response.0Once you find an eigenvalue, subtract it from the diagonal, and find the null space of the resulting matrix. Use elimination to do this

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or simply −1+L 4 −2 −3 4+L 0 −3 1 3+L row reduced for each L

phi
 4 years ago
Best ResponseYou've already chosen the best response.0to find the evals solve −1L 4 −2 det −3 4L 0 = 0 −3 1 3L as previously posted

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1http://www.sosmath.com/matrix/eigen2/eigen2.html might be the same, hard for me to tell

phi
 4 years ago
Best ResponseYou've already chosen the best response.0it is easy with matlab or wolfram, and tedious by hand (1L)(4L)(3L)  4(3(3L)) 2(3 +3(4L)) = 0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1i see, AX = LX AX  LX = 0 (ALI)X = 0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1they had a L = 4 which threw me for a loop :)
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