Eigenvalues of $\left[\begin{smallmatrix}1 & 1 & 2 \\1 & 2 & 1 \\2 & 1 & 1 \\ \end{smallmatrix}\right]$ I am trying to find the eigenvalues of $A$ where
$$A=
        \begin{bmatrix}
        1 & 1 & 2 \\
        1 & 2 & 1 \\
        2 & 1 & 1 \\
        \end{bmatrix}
$$
I'm stuck after writing out the equation $(1-λ)(2-λ)(1-λ)-6(1-λ) = 0$. I have tried solving this in two different ways (using characteristic equation and otherwise, both result in the above equation), but the solutions on the eigenvalue calculator for 3x3 matrices on WolframAlpha differ from mine. What am I doing wrong here?
Edit: My bad, I forgot to include the solution I have written. I took a picture of it: 

I also have another solution where I used the formula for the determinant of a 3x3 matrix and I got the same answer. WolframAlpha says that the eigenvalues for this are $λ = 1, -1$, and $4$.
 A: You have $(1-λ)(2-λ)(1-λ)-6(1-λ)$; before doing anything else, collect $1-\lambda$, getting
$$
(1-\lambda)\bigl((1-λ)(2-λ)-6\bigr)
=
(1-\lambda)(2-2\lambda-\lambda+\lambda^2-6)
=
(1-\lambda)(\lambda^2-3\lambda-4)
$$
and you can find separately the solutions of $1-\lambda=0$ (that is, $\lambda=1$) and of $\lambda^2-3\lambda-4=0$ (that is, $\lambda=4$ or $\lambda=-1$).
A: here is a way to see that $A$ has eigenvalues  $4, 1, -1.$  each row sums to $4.$ therefore $4$ is an eigenvalue and an eigenvector is $(1,1,1)^T.$ 
if you look at $A-I,$ the sum of the first and the last row is twice the second row, i.e., $(1,-2,1)$ is left eigenvector corresponding to the eigenvalue $1.$ now, use the fact that $4 = trace(A)$ is the sum of the eigenvalues, so the third eigenvalue must be $-1$
A: $$ 
\left|\begin{matrix}
1-\lambda & 1 & 2 \\
1 & 2-\lambda & 1 \\
2 & 1 & 1-\lambda 
\end{matrix} \right| \\
= (1-\lambda)[(2-\lambda)(1-\lambda)-1]-[1-\lambda-2]+2[1-2(2-\lambda)] \\
= (1-\lambda)(2-2\lambda-\lambda+\lambda^2-1)-(-\lambda-1)+2(1-4+2\lambda) \\
= (1-\lambda)(\lambda^2-3\lambda+1)-(-\lambda-1)+2(2\lambda-3) \\
= (\lambda^2-3\lambda+1-\lambda^3+3\lambda^2-\lambda)+\lambda+1+4\lambda-6 \\
= (-\lambda^3+4\lambda^2-4\lambda+1)+5\lambda-5 \\
= -\lambda^3+4\lambda^2+\lambda-4 \\
= \lambda^2(-\lambda+4)+\lambda-4 \\
= -\lambda^2(\lambda-4)+(\lambda-4) \\
= (\lambda-4)(1-\lambda^2) \\ 
= (\lambda-4)(1-\lambda)(1+\lambda)
$$
So  the root of the characteristic polynomial (eigenvalues of the original matrix) are $\lambda_1 = -1, \lambda_2=1, \lambda_3 = 4$.

My suggestion for finding eigenvalues by hand of arbitrary matrices is to express the characteristic polynomial in the form $\lambda^n+a_{n-1}\lambda^{n-1} + \cdots + a_2\lambda^2 +a_1\lambda + a_0$. First off, these polynomials are usually designed to be factored so go with your first instincts when finding their roots. If your teacher is cruel it will only be solvable numerically however. 
Secondly, you can figure out if you found the correct characteristic polynomial by looking at the values of certain coefficients for an $n\times n$ matrix. The value of $a_0$ should be the determinant of the original matrix. It is also the product of your eigenvalues: $\lambda_1 \lambda_2 \cdots \lambda_n$ with multiplicities. 
Also, the value of $a_{n-1}$ is the trace of the original matrix. The trace is also equal to the sum of your eigenvalues with multiplicities: $a_{n-1} = \lambda_1 + \lambda_2 + \cdots + \lambda_n$. Admittedly, the trace is the much easier value to compute so check that first if you're unsure whether you obtained the correct characteristic polynomial.
