Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 3x3 Matrix Differential Equations

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to the system of differential equations.

    2. Relevant equations

    3. The attempt at a solution
    I uploaded the original equation and my work so see the attachment. I want to know how they got the vectors the got typically when I have done 2x2 systems I row reduce them and then use the top row that is left over and get the transverse and that is my vector however this did not work this time, what do I do?

    Attached Files:

  2. jcsd
  3. May 15, 2012 #2


    User Avatar
    Gold Member

    From my understanding, you need to find the eigenvector/s corresponding to each eigenvalue and the required answer is the sum of the basis vectors (eigenvectors) within the matrix containing all the eigenvectors. The constants, a, b and c are real constant multiples (can take any value) of the eigvenvectors.
  4. May 15, 2012 #3
    Ok but how did they get the answers they got because I am getting totally opposite
  5. May 15, 2012 #4


    User Avatar
    Gold Member

    OK, i can confirm your eigenvalues are correct: 0, 3 and 1.

    For eigenvalue = 0, i get the same eigenvector as in the expected answer: (1,1,1)^T

    What i did was reduce the matrix with that eigenvalue = 0, and found the special solution (using linear algebra, it's the nullspace).

    Now, for the other two eigenvalues, i would expect to get the rest of the answers.

    You have clearly miscalculated the eigenvectors.

    For eigenvalue = 3, i get the eigenvector: (1, -2, 1)^T.

    For eigenvalue = 1, i get the eigenvector: (-1, 0, 1)^T, which is equivalent (multiply by scalar value of -1) to (1, 0, -1)^T.

    Since there is a multiple of e^t involved with the eigenvectors, i assume that you have not posted the entire problem.

    Note: If you are familiar with finding the nullspace in linear algebra, you should be able to work it out. Also, in your work you take the trace of the matrix as λ-{value} but i guess it's just a matter of different method, since i personally use {value}-λ as the trace elements of the matrix. I guess either way you'll get the same results, since you were able to get the same eigenvalues.
    Last edited: May 15, 2012
  6. May 15, 2012 #5
    ok cool I think I can get it now ill let you know soon.
  7. May 15, 2012 #6


    User Avatar
    Science Advisor

    If a 3 by 3 matrix has three independent eigenvectors, then it can be "diagonalized". Here, the eigenvalues are, as you say, 0, 1, and 3. An eigenvector corresponding to eigenvalue 0 is <1, 1, 1>, an eigenvector corresponding to eigenvalue 1 is <-1, 0, 1>, and an eigenvector corresponding to eigenvalue 3 is <1, -2, 1>. The matrix having those eigenvectors as columns is
    [tex]P= \begin{bmatrix}-1 & 1 & 1 \\ 0 & 1 & -2 \\ 1 & 1 & 1\end{bmatrix}[/tex]
    and its inverse is
    [tex]P^{-1}= \begin{bmatrix}-\frac{1}{2} & 0 & \frac{1}{2}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{6} & -\frac{1}{3} & \frac{1}{6}\end{bmatrix}[/tex]
    and now it is easy to check that
    [tex]D= P^{-1}AP= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3\end{bmatrix}[/tex]
    the diagonal matrix with the eigenvalues on the diagonal.

    Now go back to Y'= AY, the differerential equation. We can multiply both sides, on the left, by [itex]P^{-1}[/itex] to get [itex]P^{-1}Y'= P^{-1}AY[/itex]. And, since P has only constant entries, and [itex]PP^{-1}= I[/itex], we can write that as [itex](P^{-1}Y)'= P^{-1}APP^{-1}Y= D(P^{-1}Y)[/itex] or, letting [itex]z= P^{-1}Y[/itex], [itex]z'= Dz[/itex] which is just
    [tex]\begin{bmatrix}z_1(t) \\ z_2(t) \\ z_3(t)\end{bmatrix}= \begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3\end{bmatrix}\begin{bmatrix}z_1(t) \\ z_2(t) \\ z_3(t)\end{bmatrix}[/tex]
    which is equivalent to the three separate equations [itex]z_1'= z_1[/itex], [itex]z_2'= 0[/itex], and [itex] z_3'= 3z_3[/itex] which, of course, have solutions [itex]z_1(t)= C_1e^t[/itex], [itex]z_2= C_2[/itex], [itex]z_3= C_3e^{3t}[/itex].

    Now, because [itex]Z= P^{-1}Y[/itex], [itex]Y= PZ[/itex] just multiply
    [tex]Y= \begin{bmatrix}-1 & 1 & 1 \\ 0 & 1 & -2 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix}C_1e^t \\ C_2 \\ C_3e^{3t}\end{bmatrix}[/tex] to get
    [itex]y_1(t)= -C_1e^t+ C_2+ C_3e^{3t}[/itex]
    [itex]y_2(t)= C_2- 2C_3e^{3t}[/itex] and
    [itex]y_3(t)= C_1e^t+ C_2+ C_3e^{3t}[/itex]
    Last edited by a moderator: May 15, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook