How to find the eigenvalues of the following matrix. So I have to find the eigenvalues of this matrix: $\begin{bmatrix}-7/9&-4/9&8/9\\-4/9&-1/9&2/9\\2/9&-4/9&8/9\end{bmatrix}$. 
What I did is start by writing it like this: $\begin{bmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{bmatrix}$ and then row reducing it to this: $\begin{bmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\0&-1+\lambda&1-\lambda\end{bmatrix}$. Then I found the determinate of this by expanding the bottom row. So I did this: $(-7/9 - \lambda)(2/9) + 32/81$ = $(-1)(2/9 - 2/9\lambda)(-1+\lambda)$ = $(2/9\lambda - 2/9)(-1+\lambda)$
Then for the 3rd column: $(-7/9 - \lambda)(-1/9 - \lambda)$ = $\lambda^2 + 8/9\lambda + 7/81 - 16/81$ = $(\lambda^2 + 8/9\lambda -1/9)(1-\lambda)$
So then I combine the two equations: $-(1+\lambda)((2/9\lambda - 2/9) + (\lambda^2 + 8/9\lambda-1/9))$ = $-(1+\lambda)(\lambda^2+10/9\lambda-1/3)$ and then I factor the polynomial to find out what lambda is. 
The problem is I don't think my answer is correct since I can't factor my degree 2 polynomial where lambda equals two whole numbers. So where did I mess up in my answer? 
Sidenote: I did try this as well by factoring out 1/9 first and I STILL GOT THE WRONG ANSWER. So can someone please help me and tell why I keep getting the wrong answer because this is very confusing and I really need some help. 
 A: Note: I recommend you factor out a $\dfrac{1}{9}$ to reduce the possibility of errors, so repeat this approach.
We want to find the determinant of 
$$\begin{vmatrix}-7/9 - \lambda&-4/9&8/9\\-4/9&-1/9-\lambda&2/9\\2/9&-4/9&8/9-\lambda\end{vmatrix}=0$$
Expanding along the top row since there are no zeros to take advantage of, we have:
$$\left(-\dfrac{7}{9}-\lambda \right)\begin{vmatrix}-\dfrac{1}{9}-\lambda &\dfrac{2}{9}\\-\dfrac{4}{9}&\dfrac{8}{9}-\lambda\end{vmatrix} + \dfrac{4}{9} \begin{vmatrix}-\dfrac{4}{9}&\dfrac{2}{9}\\\dfrac{2}{9}&\dfrac{8}{9}-\lambda\end{vmatrix} + \dfrac{8}{9} \begin{vmatrix}-\dfrac{4}{9} &-\dfrac{1}{9}-\lambda\\\dfrac{2}{9}&-\dfrac{4}{9}\end{vmatrix}$$
Hopefully, you can take it from here and arrive at:
$$\lambda - \lambda^3 = 0 \implies \lambda_{1,2,3} = -1, 0, 1$$
Update Expanding the three determinants, we have
$$\left(\lambda-\frac{7}{9}\right) \left(\lambda^2-\frac{7 \lambda}{9}\right)+\frac{4}{9} \left(\frac{4 \lambda}{9}-\frac{4}{9}\right)+\frac{8}{9} \left(\frac{2 \lambda}{9}+\frac{2}{9}\right)$$
This reduces to
$$\dfrac{49 \lambda}{81}-\lambda^3 + \dfrac{16 (\lambda-1)}{81} + \dfrac{16 (\lambda+1)}{81} = \dfrac{49 \lambda}{81}-\lambda^3 + \dfrac{32 \lambda}{81} = \lambda - \lambda^3$$
A: Modifying the matrix with row operations can indeed make finding its determinant easier, but you have to do it correctly. It looks like you tried to add $1/2$ times the second row to the third, but you got the wrong answer. The second element should be $-(1+\lambda)/2$.
You do have to be careful when doing this because some row operations change the determinant by a constant factor. For the purpose of finding eigenvalues, however, you can ignore this since all of those constant multipliers will factor out at the end.
