$$\begin{vmatrix} a-E&-b&-b&-b&-b\\ -b&a-E&-b&-b&-b\\ -b&-b&a-E&-b&-b\\ -b&-b&-b&a-E&-b\\ -b&-b&-b&-b&a-E \end{vmatrix}=0$$
I'm asked to show by solving that the two solutions for $E$ are $E=a+b$ and $E=a-4b$. If I add all the rows using row operations I get every term as $a-E-4b$ and so for the matrix to have zero determinant $a-E-4b=0$ which gives we the second solution on rearrangement. How can I obtain the first solution (and indeed the second if my method is not sufficient)?
On
this matrix is an example of rank one perturbation of the identity matrix. if you take $u = (1,1,\cdots, 1)^\top,$ then the matrix $uu^T$ of all ones is a rank one matrix whose eigenvalues are $\{n, 0,0, \cdots 0\}$ and the corresponding eigenspaces are spanned by $u, u^\perp.$
it can be easily seen that the eiegnvalues of $(a+b-e)I -buu^\top,$ which is the matrix in question, are $\{(a+b-e)-nb, a+b-e, a+b-e, \cdots, a+b-e\}.$ therefore the determinant is $$det\left((a+b-e)I -buu^\top\right)= \left(a-(n-1)b-e\right)(a+b-e)^{n-1}.$$
If you look at the general form of this type of matrix, you have a element $\alpha$ on the diagonal, and some $\beta$ everywhere else. We can rewrite such a matrix as:
$$A=(\alpha - \beta)(I+uv^T)$$
Where $v^T=(\beta, \beta, \ldots)$, and $u^T = (1/(\alpha - \beta), 1/(\alpha - \beta), \ldots)$.
Now, by the Matrix determinant lemma, the determinant of such a matrix is simply: $$\det(A)=(\alpha - \beta)^n\left(1+n\frac{\beta}{\alpha - \beta}\right)$$ In your case, $\alpha=a-E, \beta=-b, n=5$. The above can be zero if one of the factors is zero, i.e. either: $$1-n\cdot\frac{b}{a-E+b}=0$$ $$\bf E=a-4b$$ Or: $$a-E+b=0$$ $$\bf E=a+b$$