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I need to solve the following problem,

In this problem, the eigenvalues of the coefficient matrix can be found by inspection and factoring. Apply the eigenvalue method to find a general solution of the system.

$x'_1=2x_1+x_2-x_3$

$x'_2=-4x_1-3x_2-x_3$

$x'_3=4x_1+4x_2+2x_3$

Now I know how to find the eigenvalues by using the fact that $|A-\lambda I|=0$, but how would I do it by inspection? Inspection is easy for matrices that have the sum of their rows adding up to the same value, but this coefficient matrix doesn't have that property.

EDIT: Originally I didn't understand what inspection meant either. After googling it this is what I found. Imagine you have the matrix, $A = \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix}$

By noticing (or inspecting) that each row sums up to the same value, which is 0, we can easily see that [1, 1, 1] is an eigenvector with the associated eigenvalue of 0.

  • 1
    The "inspection and factoring" refers to finding the zeros of the polynomial. You still have to calculate it.2011-11-07
  • 0
    "how would I do it by inspection" is an oxymoron2011-11-07
  • 0
    Ask your teacher what they mean by "inspection and factoring" if you aren't sure.2011-11-07

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