Finding determinant (using row reduction) I'm trying to calculate the following
$\det \begin{pmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{pmatrix}$
and I'm sure there must be a way to get this as an upper/lower triangular matrix. However whichever way I try looking at it, I can't see how to make it into one, if you minus the bottom row from each of the other rows you're still left with the bottom row and the same goes for columns. Is there some clever trick to note here? I've also briefly entertained the idea of proving a general form (since I'm sure one exists for a matrix of this form) but it seems a bit overkill...
Many thanks!
 A: Hints: make zeros below the entry 2-1 on the first column and develop wrt it and etc.:
$$\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{vmatrix}=\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ 0 & -x & x & 0 & 0 \\ 0 & -x &0 & x & 0 \\ 0 & -x & 0 & 0 & x \end{vmatrix}=$$
$$=(x-1)\begin{vmatrix} x-1 & -1 & -1 & -1 \\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\  -x & 0 & 0 & x \end{vmatrix}+\begin{vmatrix} -1 & -1 & -1 & -1\\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\ -x & 0 & 0 & x \end{vmatrix}\stackrel{\text{develop by Col. 2}}=$$
$$=(x-1)\left(\,\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\  -x  & 0 & x \end{vmatrix}+x\begin{vmatrix} x-1 & -1 & -1 \\ -x & x & 0 \\  -x & 0 & x \end{vmatrix}\,\right)+\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\  -x  & 0 & x \end{vmatrix}+x\begin{vmatrix}  -1 & -1 & -1 \\ -x & x & 0 \\  -x & 0 & x \end{vmatrix}=$$
$$=(x-1)\left(-x^3+x(x^2(x-1)-2x^2)\right)+(-x^4-3x^3)=$$
$$=(x-1)\left(-x^3+x(x^3-3x^2)\right)-x^3(x+3)=(x-1)(x^4-4x^3)-x^3(x+3)=$$
$$=x^3(x-1)(x-4)-x^3(x+3)=\ldots$$
A: By Gaussian elimination, using the the first entry of the fifth row as the first pivot, we get that
$$
 \begin{vmatrix} -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \\ -1 & -1 & -1 & -1 & -1+x \end{vmatrix} 
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \end{vmatrix} 
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & x & 0 & 0 & -x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix} 
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix} = -\begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & x & -x \end{vmatrix} 
= \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & 0 & x^2-5x \end{vmatrix} 
= (-1)(-x)^3(x^2-5x) = x^4(x-5).
$$
