I have found the determinant of the following matrix of order $n\ge3$ using some elemetary operations 
$$\begin{bmatrix}0& 1 & 1& \dots & 1 \\ 0 & 0 & 1& \ddots & 1\\ 1 & 0 & 0 & \ddots & 1 \\ \vdots & \ddots & \ddots & \ddots & 1 \\ 1 & 1 & 1 & \dots & 0 \end{bmatrix}\_{n\times n},$$ that is, a matrix which have diagonal, subdiagonal entries zero and rest of entries are equal to 1. Its determinant comes out to be equal to $$(-1)^n\Big(\lfloor\frac{n-2}{2}\rfloor\Big),$$ where, $\lfloor.\rfloor$ is the floor function. Now, I want to find determinant of the following matrix 
$$ 
\begin{bmatrix}0& -1 & 1& \dots & 1 \\ 0 & 0 & -1& \ddots & 1\\ 1 & 0 & 0 & \ddots & 1 \\ \vdots & \ddots & \ddots & \ddots & -1\\ 1 & 1 & 1 & \dots & 0 \end{bmatrix}\_{n\times n},$$ that is, a matrix having diagonal and subdiagonal entries zero. Superdiagonal elements are -1 and rest of the entries are equal to 1. 



Determinant of a matrix having diagonal and subdiagonal entries zero
I will try to give a different explicit solution. From the problem formulation of answer 1, $$\det(A+uu^T)=(1+u^TA^{-1}u).\det(A).$$ From [here](https://en.wikipedia.org/wiki/Tridiagonal_matrix), we need to solve some recursive expressions, in order to calculate the determinant and inverse of $A$. I solve these recursive expressions using roots of their characteristic equations. For the determinant of $A$, recursive expression is $$f\_n=-f\_{n-1}-2f\_{n-2},\ \ \ \ \ f\_0=1,\ f\_{-1}=0.$$ Roots of the resulting characteristic equation $x^2+x+2=0,$ are $$ s\_1=\frac{-1}{2}+ \frac{\iota \sqrt{7}}{2},\ \ s\_2=\frac{-1}{2}- \frac{\iota \sqrt{7}}{2}.$$ Hence, $$\det(A)=f\_n=r\_1.s\_1^n+r\_2.s\_2^n,$$ where, using initial conditions $$r\_1=\frac{1}{2} +\frac{\iota }{2\sqrt{7}},\ \ r\_2=\frac{1}{2} -\frac{\iota }{2\sqrt{7}}.$$ 
Now, to calculate $A^{-1}$ we need to solve following recursive expressions $$g\_i=-g\_{i-1}-2g\_{i-1}, \ \text{for} \ \ i=2, 3\dots,n,\ \ g\_0=1,\ g\_1=-1$$ $$ h\_i=-h\_{i+1}-2h\_{i+2}, \ \text{for} \ \ i=n-1,\dots,1, \ \ h\_{n+1}=1,\ h\_{n}=-1.$$ Similar to $f\_n$, solving these recursive expressions we get $$g\_i=r\_{1}.s\_1^i+r\_{2}.s\_2^i.$$ And, $$h\_i=r\_{h1}.s\_1^{n-i}+r\_{h2}.s\_2^{n-i},$$ where, $$r\_{h1}=\frac{-1}{2}-\frac{3}{2\iota\sqrt(7)},\ \ \ r\_{h2}=\frac{-1}{2}+\frac{3}{2\iota\sqrt(7)}.$$ Entries of $A^{-1}$ are clearly given by $g\_i, h\_i$ ([here](https://en.wikipedia.org/wiki/Tridiagonal_matrix)). 


$$A^{-1}\_{ij}=\begin{cases}\frac{2^{j-i}g\_{i-1}h\_{j+1}}{g\_n}
 & \mbox{if $i\le j$}
\\ \frac{g\_{j-1}h\_{i+1}}{g\_n} &\mbox{if $j< i$}
\end{cases}.$$


