For what values of λ does Rank(A) = 1? Rank(A)= 2? Rank(A) = 3? $$ A = \begin{pmatrix} λ & 1 & 1 \\ 1 & λ & 1 \\ 1 & 1 & λ \end{pmatrix} $$
My attempt: 
For Rank(A) = 1, the rows of matrix A must be scalar multiples of each other. Thus, the only value that works in this case would be λ = 1. 
For Rank(A) = 2, the determinant of A must equal zero in order to find the values of λ. 
The determinant is λ^3 -3λ +2 = 0 which gives values of λ = -2 and λ = 1. Since we know that λ = 1 makes Rank(A) = 1, we can disregard λ = 1 in this case. Thus λ = -2 would make Rank(A) = 2. 
For Rank(A) = 3, the determinant of A must not equal zero. Since we found values of λ for which det(A) = 0, the values of λ for which Rank(A) = 3 should be numbers that do not equal 1 or -2. 
Am I missing any λ values for each case? 

If the rank is not three, then it is not invertible and hence, its determinant must be $0$ (iff). Finding the characteristi polynomial, we can see when this happens. $$\left\vert\begin{pmatrix} λ & 1 & 1 \\ 1 & λ & 1 \\ 1 & 1 & λ \end{pmatrix}\right\vert=\lambda\cdot\left\vert\begin{pmatrix} \lambda & 1 \\ 1 & \lambda\end{pmatrix}\right\vert-\left\vert\begin{pmatrix} 1 & 1\\1&\lambda\end{pmatrix}\right\vert+\left\vert\begin{pmatrix} 1 & \lambda\\1&1\end{pmatrix}\right\vert=\lambda(\lambda^2-1)-\lambda+1+1-\lambda=\lambda^3-\lambda+2-2\lambda=\lambda^3-3\lambda+2=(\lambda-1)^2(\lambda+2)$$
which implies $\lambda=1,-2$ as you stated. Now, check these two values if you need to to see which give rank to be $1$ and which give rank to be $2$. All other values clearly don't give determinant equal to $0$ and hence give rank=$3$. So, your reasoning was just about right. Good job.
