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Homework Help: Lineal Algebra: Inverse Matrix of Symmetric Matrix

  1. Aug 4, 2010 #1
    1. The problem statement, all variables and given/known data

    Hello,

    I need some help in the fist parts of two lineal algebra problems, specially with algebraic manipulation. I guess that if I rewrite the determinant nicely some terms get canceled and I can write the inverse nicely, but don't know how to do it...

    Problem 1:

    21k0cux.png

    Problem 2:

    nvz5v8.png

    3. The attempt at a solution

    Problem 1:

    (1) [tex] Det(A) = a(a^2-b^2) -b(ba-b^2) + b(b^2-ab) = b(b-a)(2b-\frac{a^2}{b})[/tex]

    (2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

    [tex]Adj(A) = \left[ \begin{matrix} a^2-b^2 & b(b-a) & b(b-a) \\ b(b-a) & a^2-b^2 & b(b-a) \\ b(b-a) & b(b-a) & a^2-b^2 \end{matrix} \right][/tex]

    [tex] A^-1 = ? [/tex]
    (I can write the terms outside the diagonal nicely because some parts get cancelled, but not the diagonal itself...)

    Problem 2:

    (1)

    Sum of eigenvalues:
    Trace(A) = a + b + c

    Product of eigenvalues:
    [tex] Det(A) = a(bc -b^2) -a(ac-ab) + 0 = abc - ab^2 - a^2c + a^2b [/tex]

    (2) [tex] A^-1 = \frac {adj(A)}{Det(A)} [/tex]

    [tex]Adj(A) = \left[ \begin{matrix} b(c-b) & a(b-c) & 0 \\ a(b-c) & a(c-a) & a(a-b) \\ 0 & a(a-b) & a(b-a) \end{matrix} \right][/tex]

    [tex] A^-1 = ? [/tex]


    Thanks in advance...
     
  2. jcsd
  3. Aug 4, 2010 #2

    vela

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    You made an algebra error when you simplified and factored. You should get (a-b)2(a+2b).
    That's correct. The diagonal elements already look pretty nice to me.
     
  4. Aug 4, 2010 #3

    vela

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    This is all correct. You can factor det(A) and get some cancellation when you calculate A-1. You can always check your answer by multiplying A by the inverse you calculate and verify you get the identity matrix.
     
  5. Aug 4, 2010 #4
    Thanks. However, my main problem is how to get the inverse, because I want to get some parts canceled but don't know how to do it...

    My main problem is on getting the inverse matrix, not the adjoin matrix ...
     
  6. Aug 4, 2010 #5
    I am sorry it may sound very basic but... how could I factor that ? I have been trying but reach nowhere...
     
  7. Aug 4, 2010 #6

    vela

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    I'm not sure where you're getting stuck. You noted that the inverse is just the adjoint matrix divided by the determinant. You have both. Just do the division.
     
  8. Aug 4, 2010 #7

    vela

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    [tex]abc-ab^2-a^2c+a^2b = a(bc-b^2-ac+ab) = a[b(c-b)-a(c-b)] = a(c-b)(b-a)[/tex]
     
  9. Aug 4, 2010 #8
    Thanks! By the way, which method did you follow to reach that factorization?

    Thanks!

    I used the factored determinants you showed to me to calculate the inverse matrix, but the calculations don't get as simplified as I thought at first. Are there any other properties of symmetric matrices that can be used for calculating the inverse faster ?

    Thanks for your help...
     
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