Parameter Matrix Determinant $A=\begin{bmatrix} 7x+42&x-21&x-21&x-21&x-21\\x-21&7x+42&x-21&x-21&x-21\\x-21&x-21&7x+42&x-21&x-21\\x-21&x-21&x-21&7x+42&x-21\\x-21&x-21&x-21&x-21&7x+42\end{bmatrix}$
I am trying to find the determinant of $A$, so far without success. I tried reducing it to a triangular matrix but wasn't successful. I would greatly appreciate any help.
 A: HINT: Calculate the eigenvalues of $A=(x-21)O + (6x+63)I$, where $O$ is a matrix full of $1$s and $I$ is the unit matrix...
A: Let me write $a=7x+42$ and $b=x-21$. Then by subtracting the first columns from all other columns we obtain
$$
\det(A) = \det\pmatrix{ a & b & b & b & b \\ b & a & b & b & b \\ b & b & a & b & b \\ b & b & b & a & b \\ b & b & b & b & a }
= \det\pmatrix{ a & b-a & b-a & b-a & b-a \\ b & a-b & 0 & 0 & 0 \\ b & 0 & a-b & 0 & 0 \\ b & 0 & 0 & a-b & 0 \\ b & 0 & 0 & 0 & a-b }
\\
= (a-b)^4 \det\pmatrix{ a & -1 & -1 & -1 & -1 \\ b & 1 & 0 & 0 & 0 \\ b & 0 & 1 & 0 & 0 \\ b & 0 & 0 & 1 & 0 \\ b & 0 & 0 & 0 & 1 }
$$
Now developing with respect to the first column gives
$$
\det(A) = (a-b)^4(a+b+b+b+b) = (a-b)^4 (a+4b) = (6x+63)^4(11x-42).
$$
If you instead develop with respect to the last row (or column), you wil find a recursive way to compute the determinant, which can be generalized to matrices of any size.
A: I am linear algebra beginner and I don't know how to calculate determinant using eigenvalues, which was suggested by @draks, but...
It seems that this is typical example for Laplace formula, because adding one row to other rows makes a lot of zeros. (Adding multiple of another row does not change the determinant.)
A: This is an answer but it probably not nearly as instructive as what draks's answer has to offer. This approach is entirely "empirical" in the sense that I conjecture what the determinant should be.
Using WolframAlpha, I have found that the eigenvalues of matrices in the form as the one in OP's question are almost all equal to $3(2x+21)$, but there is always one exception to this. The "exceptional" eigenvalue is of a different form. 
If we multiply all the eigenvalues of matrices of the form in the question, we get their determinants. When I refer to the "form" of the matrix, specifically I define $$A_n = (x-21)(O_n-I_n) + (7x+42)I_n$$ where $O_n$ is an $n\times n$ matrix full of 1's and $I_n$ is the $n\times n$ identity matrix.
Here we go:
When $n=1$, $\det(A_1) = 7x+42$.
When $n=2$, $\det(A_2) = [3(2x+21)](8x+21)$.
When $n=3$, $\det(A_3) = [3(2x+21)]^2(9x)$.
When $n=4$, $\det(A_4) = [3(2x+21)]^3(10x-21)$.
If we're in the mood to conjecture, I would bet that the general form is given by $$\det(A_n) = [3(2x+21)]^{n-1}[(n+6)x+(3-n)21].$$
So in the case of the OP, we have $n=5$ and we get $$\det(A_5) = [3(2x+21)]^4(11x-42).$$

However, I'd be very interested to see draks's approach because it sounds like it would provide more insight on the eigenvalues (and by extension the determinant) of a sum of matrices in general.
