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 $\det(A)=-(2n-3)\times(-1)^{n-1}=(-1)^n(2n-3)$.