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.