I would like to find all $x$ values such that the determinant of the following matrix is zero. $$ \begin{pmatrix} x & a & b & c\\ a & x & b & c\\ a & b & x & c\\ a & b & c & x \end{pmatrix} $$
I get that if $x= a, b, c$ you will get a pair of linearly dependent rows. The answer, however, also includes $x= -a, -b, -c$. I can't figure out why those values would make the determinant zero.
On
$$\begin{vmatrix} x&a&b&c\\ a&x&b&c\\ a&b&x&c\\ a&b&c&x\end{vmatrix}=\begin{vmatrix} x+a+b+c&a&b&c\\ x+a+b+c&x&b&c\\ x+a+b+c&b&x&c\\ x+a+b+c&b&c&x\end{vmatrix}=\begin{vmatrix} x+a+b+c&a&b&c\\ 0&x-a&0&0\\ 0&b-a&x-b&0\\ 0&b-a&c-b&x-c\end{vmatrix}=$$$${}$$
$$=(x+a+b+c)\begin{vmatrix} x-a&0&0\\ b-a&x-b&0\\ b-a&c-b&x-c\end{vmatrix}=(x+a+b+c)(x-a)\begin{vmatrix} x-b&0\\ c-b&x-c\end{vmatrix}$$$${}$$
$$=(x+a+b+c)(x-a)(x-b)(x-c)$$$${}$$
First step: add columns 2,3,4 to one...and etc.
I don't think that is true. You can use the cofactor method to find the determinant explicitly. It is given by $$ f(x) = (a-x)(b-x)(x-c)(a+b+c+x) $$ Clearly $x=a,b,c$ give $f(x) = 0$. But $$f(-a) = -2a(a+b)(a+c)(b+c)$$ for instance, which does not need to be zero.
EDIT: By the way, as pointed out in the comments, you can immediately see that $x=a, b, c, -a, -b, -c$ can't all be zero. This is because if $A$ is a an $n\times n$ matrix, $\det(A) = 0$ gives a degree $n$ polynomial (in this case, in $x$. A degree $n$ polynomial has exactly $n$ roots. So, the 4 values $x=a, b, c, -a-b-c$ are tho only $4$ that can be zeros of the determinant.