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Your first calculation for $3\times3$ matrices applies in full generality: If your matrix writes blockwise $[1 \quad e^T ; e \quad M]$, with $e^T=(1,\ldots,1)$, and if $M$ is non-singular (implied by your assumption), then the property that $\det A=0$ is equivalent to $e^T\hat M e=\det M$, that is $e^TM^{-1}e=1$, or to $\det(J-M)=0$, with $J=ee^T$ the matrix with $1$s everywhere. Just use the Sherman-Morrison formula $\det(B+xy^T)=(\det B)(1+y^tB^{-1}x)$.

Edit after 3 hours. Take the $3\times3$ matrix $N:=[1 \quad i \quad -1;i \quad -1 \quad1;-1 \quad1 \quad-1]$. We have $N^{-1}e=(i-1,-2-1)^T$. N^{-1}e=(-i-1,-2i,-i)^T$. Let$D$be a diagonal matrix with unit entries on the diagonal, so that$M:=ND^{-1}$is still an admissible matrix. Then $$e^TM^{-1}e=e^TDN^{-1}e=(i-1)z_1-2z_2-z_3.$$$e^TM^{-1}e=e^TDN^{-1}e=(-i-1)z_1-2iz_2-iz_3.$$Claim: There exist unit numbers z_j such that the right-hand side equals 1. There exists actually a lot of them Consequence: the matrix A is singular. Yet it does not have two equal rows or columns. Proof of the claim: we may search for unit numbers y_j such that \sqrt2 y_1+2y_2+y_3=1. Taking y_3=y_1, we just have (1+\sqrt2)y+2y', which covers a corona (\sqrt2-1)\le |z|\le 3+\sqrt2. In particular the equation (1+\sqrt2)y+2y'=1 has a solution. If instead we choose y_3=e^{i\epsilon}y_1 with a small enough \epsilon, the number \sqrt2 y_1+y_3 covers a circle of raidus \rho close to \sqrt2-1, and the corona obtained by adding 2y_2 still contain 1. 3 An example of singular A for n=4. Your first calculation for 3\times3 matrices applies in full generality: If your matrix writes blockwise [1 \quad e^T ; e \quad M], with e^T=(1,\ldots,1), and if M is non-singular (implied by your assumption), then the property that \det A=0 is equivalent to e^T\hat M e=\det M, that is e^TM^{-1}e=1, or to \det(J-M)=0, with J=ee^T the matrix with 1s everywhere. Just use the Sherman-Morrison formula \det(B+xy^T)=(\det B)(1+y^tB^{-1}x). I Edit after 3 hours. Take the 3\times3 matrix N:=[1 \quad i \quad -1;i \quad -1 \quad1;-1 \quad1 \quad-1]. We have no time to continue nowN^{-1}e=(i-1,-2-1)^T. Let D be a diagonal matrix with unit entries on the diagonal, but I keep so that M:=ND^{-1} is still an admissible matrix. Then$$e^TM^{-1}e=e^TDN^{-1}e=(i-1)z_1-2z_2-z_3. Claim: There exist unit numbers $z_j$ such that the right-hand side equals $1$. There exists actually a lot of them Consequence: the matrix $A$ is singular. Yet it in minddoes not have two equal rows or columns. Proof of the claim: we may search for unit numbers $y_j$ such that $\sqrt2 y_1+2y_2+y_3=1$. Taking $y_3=y_1$, we just have $(1+\sqrt2)y+2y'$, which covers a corona $(\sqrt2-1)\le |z|\le 3+\sqrt2$. In particular the equation $(1+\sqrt2)y+2y'=1$ has a solution. If instead we choose $y_3=e^{i\epsilon}y_1$ with a small enough $\epsilon$, the number $\sqrt2 y_1+y_3$ covers a circle of raidus $\rho$ close to $\sqrt2-1$, and the corona obtained by adding $2y_2$ still contain $1$.

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Your first calculation for $3\times3$ matrices applies in full generality: If your matrix writes blockwise $[1 \quad e^T ; e \quad M]$, with $e^T=(1,\ldots,1)$, and if $M$ is non-singular (implied by your assumption), then the property that $\det A=0$ is equivalent to $e^TM^{-1}e=\det e^T\hat M e=\det M$, that is $e^TM^{-1}e=1$, or to $\det(J-M)=0$, with $J=ee^T$ the matrix with $1$s everywhere. Just use the Sherman-Morrison formula $\det(B+xy^T)=(\det B)(1+y^tB^{-1}x)$.

I have no time to continue now, but I keep it in mind.

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