I have a matrix $A=\begin{bmatrix} 4 &-1 &-1 &-\alpha &-\alpha\\ -1 &4 &-1 &-\alpha &-\alpha\\ -1 &-1 &4 &-\alpha &-\alpha\\ -\alpha &-\alpha&-\alpha& 4\alpha^2 &-\alpha^2\\ -\alpha &-\alpha&-\alpha& -\alpha^2 &4\alpha^2 \end{bmatrix}$ and Eigenvector is $\begin{bmatrix}r_1\\r_2\\r_3\\r_4\\r_5\end{bmatrix}$.
One of Eivenvalues is $5$. $$\begin{bmatrix} 4-5 &-1 &-1 &-\alpha &-\alpha\\ -1 &4-5 &-1 &-\alpha &-\alpha\\ -1 &-1 &4-5 &-\alpha &-\alpha\\ -\alpha &-\alpha&-\alpha& 4\alpha^2-5 &-\alpha^2\\ -\alpha &-\alpha&-\alpha& -\alpha^2 &4\alpha^2-5 \end{bmatrix}\begin{bmatrix}r_1\\r_2\\r_3\\r_4\\r_5\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix}$$
I would like to call Row 1 is $Eq. 1$, Row 2 is $Eq. 2$ and so on.
$\textbf{Eq. 6} : Eq. 4-Eq. 5$ give $$r_4-r_5=0.$$
$\textbf{Eq. 7} : Eq. 1+Eq. 2+Eq. 3 $ give $$(r_1+r_2+r_3)+\alpha(r_4+r_5)=0.$$
From $Eq. 6$, I know that if I choose $r_4=r_5=0$, I will get Eivenvector from $Eq. 7 $ $$\begin{bmatrix}r_1\\r_2\\-r_1-r_2\\0\\0\end{bmatrix}\rightarrow \begin{bmatrix}1\\1\\-2\\0\\0\end{bmatrix}$$
My question is How I know I have to choose what the values of $r_4$ and $r_5$? because the $Eq. 6 $ told only that $r_4 = r_5$, It is not necessary to be $0$.