Determinant of Union Jack matrix without Scottish diagonal Let $n \ge 1$ be an integer , and let $A_n$ be the matrix in $M_{2n-1}\mathbb(F)$ with entries $(a_{ij})$ where $a_{ij}=1 $ if $i+j=2n$  or $i=n$ or $j=n,$ and $a_{ij}=0$ otherwise . Find det$(A_n)$
my idea:
for $n=4$ the matrix of the form from the given data
$$\begin{bmatrix}
 &  &  & 1 &  &  &1 \\ 
 &  &  &  1&  &  1& \\ 
 &  &  &  1& 1 &  & \\ 
 1&1 & 1 & 1 & 1 &1  &1 \\ 
 &  &  1&  1&  &  & \\ 
 &  1&  &  1&  &  & \\ 
 1&  &  &1  &  &  & 
\end{bmatrix}$$
How to find genearl formula for determent :

Take the mirror image of the matrix, so you get a factor $(-1)^{n-1}$. Now the mirror image is
$$
I
+\pmatrix{
&&&1\\ 
&&&\vdots\\ 
&&&1\\ 
1&\cdots&1&0&1&\cdots&1\\
&&&1\\ 
&&&\vdots\\ 
&&&1}=I+B\quad \text{(say)}.
$$
Clearly, the matrix $B$ has rank $2$ and $(1,\ldots,1,\pm\sqrt{2n-2},1,\ldots,1)^T$ is an eigenvector for the eigenvalue $\pm\sqrt{2n-2}$. Hence $\det A=(-1)^{n-1}(1+\sqrt{2n-2})(1-\sqrt{2n-2})=(-1)^n(2n-3)$.
Edit. Alternatively, subtract the middle row by the sum of all other rows. Then subtract the middle column by the sum of all other columns. What remains is a matrix with $(1,\ldots,1,-(2n-3),1,\ldots,1)$ on its anti-diagonal and zeroes elsewhere. Therefore $\det(A)=-(2n-3)\times(-1)^{n-1}=(-1)^n(2n-3)$.
