Determinant with unknown parameter. I'm given 4 vectors: $u_1, u_2, u_3$ and $u_4$. I'm going to type them in as points, because it will be easier to read, but think as them as column vectors. 
$$u_1 =( 5, λ, λ, λ), \hspace{10pt} u_2 =( λ, 5, λ, λ), \hspace{10pt} u_3 =( λ, λ, 5, λ), \hspace{10pt}u_4 =( λ, λ, λ, 5)$$
The task is to calculate the value of λ if the vectors where linearly dependent, as well as linearly independent. 
I managed to figure out that I could put them in a matrix, let's call it $A$, and set $det(A) = 0$ if the vectors should be linearly dependent, and $det(A) \neq 0$ if the vectors should be linearly independent. 
Some help to put me in the right direction would be great!

If $\lambda=0$, the vectors are clearly linearly independent.
If $\lambda\ne0$, we can divide through by $\lambda$ without affecting whether the determinant vanishes; this yields
$$
\pmatrix{\frac5\lambda&1&1&1\\1&\frac5\lambda&1&1\\1&1&\frac5\lambda&1\\1&1&1&\frac5\lambda}\;.
$$
Thus the values of $\lambda$ for which the determinant vanishes are those for which
$$
\frac5\lambda=1-\mu_i\;,
$$
where $\mu_i$ is an eigenvalue of
$$
\pmatrix{1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1}\;.
$$
This matrix annihilates all vectors whose components sum to $0$, so it has an eigenspace of dimension $4-1=3$ corresponding to the eigenvalue $0$, and thus a triple eigenvalue $0$. Since it is symmetric, its four eigenvectors can be chosen to form an orthonormal system, so the fourth eigenvector is a vector orthogonal to that eigenspace, e.g. $(1,1,1,1)$, which corresponds to eigenvalue $4$.
Thus we have the two possibilities $5/\lambda=1-0$, corresponding to $\lambda=5$, and $5/\lambda=1-4$, corresponding to $\lambda=-5/3$. All other values of $\lambda$ lead to linearly independent vectors.
