How to calculate the determinant of all-ones matrix minus the identity? How do I calculate the determinant of the following $n\times n$ matrices
$$\begin {bmatrix}
0 & 1 & \ldots & 1 \\
1 & 0 & \ldots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & ... & 0
\end {bmatrix}$$
and the same matrix but one of columns replaced only with $1$s?
In the  above matrix all off-diagonal elements are $1$ and diagonal elements are $0$.
 A: I will compute the determinant of the matrix
$$
A = \left(
\begin {matrix}
b & a & \ldots & a \\
a & b & \ddots & \vdots \\
\vdots & \ddots & \ddots & a \\
a & \cdots & a & b
\end {matrix}
\right),
$$
where $a, b \in \mathbb{K}$. To obtain your case, put $a=1$ and $b=0$.
First proof. This works if $\mathrm{char}(\mathbb{K}) = 0$ or $n$ is prime to $\mathrm{char}(\mathbb{K}) > 0$. If $a =0$, then $\det A = b^n$. Suppose $a \neq 0$ and consider the vector $v  = (1, \dots, 1) \in \mathbb{K}^n$; it is clear that $v$ is an eigenvector of $A$ with eigenvalue $\alpha = (n-1) a + b$. Now consider $\beta = b-a$. $\beta$ is an eigenvalue of $A$ because
$$
B = A - \beta I_n = \left(
\begin {matrix}
a & a & \ldots & a \\
a & a & \ldots & a \\
\vdots & \vdots & \ddots & \vdots \\
a & a & ... & a
\end {matrix}
\right)
$$
has rank $1$.
Let $E_\alpha, E_\beta \subseteq \mathbb{K}^n$ the eigenspaces of $A$ of the eigenvalues $\alpha, \beta$. We have $\alpha \neq \beta$, $E_\alpha \cap E_\beta = \{ 0 \}$ and $\dim E_\beta = n-1$, thus $\mathbb{K}^n = E_\alpha \oplus E_\beta$ and $E_\alpha = \langle v \rangle$. This proves that $A$ is similar to the matrix $\mathrm{diag}(\alpha, \beta, \dots, \beta)$, therefore $\det A = \alpha \beta^{n-1} = [(n-1)a +b] (b-a)^{n-1}$. (Notice that this formula holds also when $a=0$.)
Second proof. The characteristic polynomial of the matrix $B = A - (b-a) I$ is
$$
\chi_B(t) = (-t)^n + c_{n-1} (-t)^{n-1} + \cdots + c_1(-t) + c_0,
$$
where $c_i$ is the sum of the principal $(n-i)$-minors of $B$. It is clear that all principal minors of $B$ are zero, except on $1$-minors. Thus
$$
\chi_B(t) = (-t)^n + na (-t)^{n-1}.
$$
From $A = B + (b-a) I$, we have $\chi_A(t) = \chi_B(t-b+a)$. Thus $\det A = \chi_A(0) = \chi_B(a-b) = (b-a)^n + na (b-a)^{n-1}$.
Now consider the matrix
$$
C = \left(
\begin {matrix}
a & a & \ldots & a \\
a & b & \ddots & \vdots \\
\vdots & \ddots & \ddots & a \\
a & \cdots & a & b
\end {matrix}
\right).
$$
For every $i=2,\dots,n$, replace the $i$th row $C_i$ of $C$ with $C_i - C_1$, where $C_1$ is the first row of $C$. Obtain
$$
\det C = \det \left(
\begin{matrix}
a & a & a & \cdots & a \\
0 & b-a & 0 & \cdots & 0 \\
0 & 0 & b-a & \ddots & 0 \\
\vdots & \vdots & \ddots & \ddots & 0 \\
0 & 0 & \cdots & 0 & b-a
\end{matrix}
\right) = a (b-a)^{n-1}.
$$
A: Here's an approach using Sylvester's determinant theorem, which says that for any rectangular matrices of mutually transposed shapes $A\in\mathrm M_{n,m}(K)$ and $B\in \mathrm M_{m,n}(K)$ one has $$\det(I_n+AB)=\det(I_m+BA).$$
If $N$ is your matrix then $-N=I_n-AB$ where $A\in\mathrm M_{n,1}(K)$ is a one column all-one matrix and $B$ is its transpose. Then 
$$
\det(N)=(-1)^n\det(-N)=(-1)^n\det(I_1-BA)=(-1)^n(1-n).
$$
A: Let us denote $A_n$ to be that matrix of order $ n \times n $.
Subtract each of the columns multiplied by $1/(n-2)$ from the first column.
You determinant then becomes:
$
|A_n|=
\left|
\begin {matrix}
\frac {n-1}{n-2} & 1 & \ldots & 1 \\
0 & 0 & \ldots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 1 & ... & 0
\end {matrix}
\right| = \frac{n-1}{n-2}|A_{n-1}|
$
Inductively you get that $|A_n| = \frac{n-1}{n-2} \frac{n-2}{n-3}...\frac{2}{1}|A_2|=(n-1)|A_2| $.
Yet $|A_2|=-1 $ so you get that $|A_n|=1-n$.
A: Replace the first 0 with $x$ and the other zeros with $y$.
Now, your determinant is a polynomial in $x$ and $y$ with dominant term $xy^{n-1}$.
If $y=1$, then $n-1$ rows are equal which immediately gives $n-2$ independent kernel vectors, so $(y-1)^{n-2}$ divides the determinant. If $x=(n-1)/(y+n-2)$, then the sum of the $n-1$ bottom lines is a multiple of the first line, so $(x(y+n-2)-(n-1)$ divides the determinant.
Therefore, the determinant is $(x(y+n-2)-(n-1))(y-1)^{n-2}$.
Let $x=y=0$ to get $(-1)^{n-1}(n-1)$.
Let $x=1$ and $y=0$ to get $(-1)^{n-1}$.
A: Let $E$ be the $(n\times n)$-matrix with all ones and put $f_1:=(1,\ldots,1)$. Then $Ex=(f_1\cdot x)f_1$ for all $x\in{\mathbb R}^n$. Since $A=E-I$ we therefore have
$$Ax\ =\ (f_1\cdot x)f_1 -x\qquad(x\in{\mathbb R}^n)\ .$$
Let $(f_2,\ldots, f_n)$ be a basis of the orthogonal complement of $\langle f_1\rangle$. One has
$$Af_1=(f_1\cdot f_1)f_1 - f_1=(n-1)f_1$$
and
$$Af_i=(f_1\cdot f_i)f_1- f_i=-f_i\qquad(2\leq i\leq n)\ .$$
Therefore the matrix of $A$ with respect to the basis $(f_1,f_2,\ldots, f_n)$ is the diagonal matrix ${\rm diag}(n-1,-1,-1,\ldots,-1)\phantom{\Bigl|}$ and has determinant $(-1)^{n-1}(n-1)$.
A: This is very close to Christian Blatter's solution: 
Let $E$ be the $n$ by $n$ matrix with all coefficients equal to $1$. 
Let $H\subset K^n$ be the hyperplane formed by the vectors whose coordinates add up to $0$, and set $v:=(1,\dots,1)$. 
Then $H=\ker E$ and $Ev=nv$. This implies
\begin{align}
\det(E-X)=(-1)^n\ X^{n-1}\ (X-n), 
\end{align}
and thus 
\begin{align}
\det(E-1)=(-1)^n\ (1-n). 
\end{align}
A: $$D_n(a,b)=
\begin{vmatrix}
a & b & b & b \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
($n\times n$-matrix).
$$D_n(a,b)=
\begin{vmatrix}
a & b & b & b \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
$$=[a+(n-1)b]
\begin{vmatrix}
1 & 1 & 1 & 1 \\
b & a & b & b \\
b & b & a & b \\
b & b & b & a
\end{vmatrix}$$
$$=[a+(n-1)b]
\begin{vmatrix}
1 & 1 & 1 & 1 \\
0 & a-b & 0 & 0 \\
0 & 0 & a-b & 0 \\
0 & 0 & 0 & a-b
\end{vmatrix}$$
$$=[a+(n-1)b](a-b)^{n-1}
$$
(In the first step we added the remaining rows to the first row and then "pulled out" constant out of the determinant. Then we subtracted $b$-multiple of the first row from each of the remaining rows.)
You're asking about $D_n(0,1)=(-1)^{n-1}(n-1)$.

If you replace one column by 1's, you can use this result to get the following. (I've computed it for $n=4$, but I guess you can generalize this for arbitrary $n$.)
$$
\begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
=
\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
+
\begin{vmatrix}
1 & 0 & 0 & 0 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}=
\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{vmatrix}
+
\begin{vmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 
\end{vmatrix}
$$
Note that both these determinants are of the type you already handled in the first part.
A: Let $A_n$ denote the $n\times n$ matrix of the form you give: $0$ on the diagonal and $1$ everywhere else. I'll find $\det A_n$ by computing the eigenvalues of $A_n$ and multiplying them together. 
First, let $B_n = A_n + I_n$, so that $B_n$ consists of all $1$'s. Since $B_n$ has rank $1$ it has an eigenvalue $0$ of multiplicity $n-1$; since $\operatorname{tr} B_n = n$ and the trace is the sum of the eigenvalues, the other eigenvalue of $B_n$ must be $n$. Now $v$ is an eigenvector for $B_n$ with eigenvalue $\lambda$ if and only if $v$ is an eigenvector for $A_n$ with eigenvalue $\lambda - 1$ (why?). Hence the eigenvalues of $A_n$ are
$$
\underbrace{-1,-1,\dots,-1}_{n-1\text{ times}},n-1
$$
and $\det A_n = (-1)^{n-1}(n-1)$.
This is similar to a few of the other answers, but I thought it elegant enough to warrant inclusion.
