<s>[INST] Find the determinant, inverse of a matrix and under what condition it is positive The matrix is 
$B=[(1-\rho)I_n+\rho\textbf{1}\textbf{1}']$ where $\textbf{1}=[1\;\cdots\;1]'$, an $n\times 1$ vector with every entry $1$.
So what's the determinant, inverse of this matrix and under what condition it is positive?
(I just know the matrix is equal to $\begin{bmatrix}
1 &\rho  & \cdots & \rho\\ 
 \rho& 1 &\cdots  & \rho\\ 
 \cdots&\cdots  &\cdots  &\cdots \\ 
\rho & \rho & \cdots & 1
\end{bmatrix}$, but what's then?)
 [/INST] If we subtract the first row from all of the remaining rows, we obtain the matrix
$$\begin{bmatrix}
1 &\rho  & \rho & \cdots & \rho\\ 
\rho-1& 1-\rho & 0 & \cdots  & 0\\ 
\rho-1 & 0 & 1-\rho & \cdots  & 0\\ 
 \cdots&\cdots  &\cdots  &\cdots \\ 
\rho-1 & 0 & 0 & \cdots & 1-\rho
\end{bmatrix}$$
which has the same determinant (link).
If we compute the determinant, we obtain:  $$(1-\rho)^{n-1}-(n-1) \times \rho(\rho-1)(1-\rho)^{n-2}$$ accounting for the diagonals marked with $(*)$ below:
$$\begin{bmatrix}
1 (*) & \rho & \rho & \cdots & \rho\\ 
\rho-1 & 1-\rho (*) & 0 & \cdots  & 0\\ 
\rho-1 & 0 & 1-\rho (*) & \cdots  & 0\\ 
 \cdots&\cdots  &\cdots  &\cdots \\ 
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}$$
for the first term
$$\begin{bmatrix}
1 & \rho (*)& \rho & \cdots & \rho\\ 
\rho-1 (*) & 1-\rho & 0 & \cdots  & 0\\ 
\rho-1 & 0 & 1-\rho (*)  & \cdots  & 0\\ 
 \cdots&\cdots  &\cdots  &\cdots \\ 
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}, \qquad \begin{bmatrix}
1 & \rho & \rho (*)& \cdots & \rho\\ 
\rho-1 & 1-\rho (*) & 0 & \cdots  & 0\\ 
\rho-1 (*) & 0 & 1-\rho  & \cdots  & 0\\ 
 \cdots&\cdots  &\cdots  &\cdots \\ 
\rho-1 & 0 & 0 & \cdots & 1-\rho (*)
\end{bmatrix}, \qquad \text{etc.}$$
for the second term (noting that there are $n-1$ ways this type of diagonal can be found).
The above equation simplifies to $$\det=(1-(n-1)\rho)(1-\rho)^{n-1}.$$
</s>