Calculating determinant of 100x100 matrix I was trying to calculate the determinant of 100x100 matrix:
$$A=\begin{bmatrix}5&5&5&...&5&5&-1\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&5\\ \\5&5&-1&...&5&5&5\\5&-1&5&...&5&5&5\\-1&5&5&...&5&5&5\end{bmatrix}$$
I thought that using LU decomposition would quickly show a pattern, but the numbers I'm getting after only few transformations quickly get out of hand. What approach might be good for calculating the determinant of such matrix?
 A: Hint: After rearranging rows, you get a matrix of the form $5 U-6I$ where $U$ is the all-ones matrix.  Note that $U$ has rank $1$.  Find the eigenvalues...
A: Let $D_n$ be the determinant of such a matrix that is $n\times n$. 
In any one column except the last column, the bottom $n-1$ entries sum to $5n-11$. So add $\frac{-5}{5n-11}$ times each of row 2, row 3, etc. to row 1, and you have 
$$\begin{bmatrix}0&0&0&...&0&0&-6\frac{5n-6}{5n-11}\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&5\\ \\5&5&-1&...&5&5&5\\5&-1&5&...&5&5&5\\-1&5&5&...&5&5&5\end{bmatrix}$$ which doesn't affect the determinant. (Note the upper right entry is $-1-\frac{5}{5n-11}5(n-1)$, simplified.) Now expanding across the top row:
$$\begin{align}D_n&=(-1)^{n+1}(-6)\frac{5n-6}{5n-11}D_{n-1}\\&=(-1)^{n}(6)\frac{5n-6}{5n-11}D_{n-1}\end{align}$$
which expands recursively as a telescoping product down to $$\begin{align}D_n&=(-1)^{n(n+1)/2-1}6^{n-1}\frac{5n-6}{-1}D_{1}\\&=-(-1)^{n(n+1)/2}6^{n-1}(5n-6)\end{align}$$ which perhaps looks nicer as $$\begin{align}D_n&=(-1)^{\binom{n+1}{2}+1}6^{n-1}(5n-6)\\\{D_n\}&=\{-1,24,-324,-3024,24624,\ldots\}\\D_{100}&=-494\cdot6^{99}\end{align}$$
A: 50  row transpositions produce the matrix $A:=5E-6 I$ of size $n\times n$, where $n:=100$, $E$ denotes the matrix of all ones, and $I$ is the identity matrix (cf. Robert Israel's answer). Choose an orthogonal basis $(f_i)_{1\leq i\leq n}$ with $f_1:=(1,1,\ldots,1)$. Then $Ef_i=0$ $(2\leq i\leq n)$ and therefore
$$Af_1=(5n-6)f_1,\qquad Af_i=-6f_i\quad(2\leq i\leq n)\ .$$
This shows that the matrix of $A$ (considered as a map ${\mathbb R}^n\to{\mathbb R}^n$) in terms of the basis $(f_i)_{1\leq i\leq n}$ is given by
$$[A]={\rm diag}(5n-6, -6,-6,\ldots,-6)\ ,$$ 
and one obtains
$$D:={\rm det}(A)=(5n-6)(-6)^{n-1}\ .$$
As $n=100$ in our case this amounts to $D=-494\cdot 6^{99}$.
