Let $A=(a_{ij}) \in M_n(\mathbb{C})$ and $A_{ij}$ the matrix obtained by $A$ changing $a_{ij}$ with $2-a_{ij}$. If $\det(A)=\det(A_{ij})$ for every $i,j$ and $B=(a_{ij}+(-1)^i)\in M_n(\mathbb{C})$ find $\det B(\det A-\det B)$.

I have no idea how to start. For $n=2$ it's easy because all the elements of $A$ are 1, so $\det A=\det B=0$

  • $\begingroup$ Do you mean $\det\left(B(\det(A)-\det(B)\right)$ or $\left(\det(A)-\det(B)\right)\det(B)$? $\endgroup$ – Michael Hoppe Jun 13 '20 at 11:06

Let $C$ be the cofactor matrix of $A$ and let $m_{ij}$ be the minor of $A$ obtained by deleting row $i$ and column $j$ (i.e. $c_{ij}=(-1)^{i+j}m_{ij}$). Since $a_{ij}$ and $2-a_{ij}$ cannot be both zero, the condition that $\det(A)=\det(A_{ij})$ implies that $m_{ij}=0$ whenever $a_{ij}\ne1$. Therefore $a_{ij}m_{ij}$ is always equal to $m_{ij}$. In turn, $a_{ij}c_{ij}$ is always equal to $c_{ij}$ and $\sum_jc_{ij}=\sum_ja_{ij}c_{ij}=\det(A)$. Hence $Ce=\det(A)e$ or equivalently, $$ e^T\operatorname{adj}(A)=\det(A)e^T.\tag{1} $$ Let $u=(-1,1,-1,\ldots,(-1)^n)^T$ and $e=(1,1,\ldots,1)^T$. Then $B=A+ue^T$. It follows from $(1)$ that \begin{aligned} \det(B)&=\det(A+ue^T) =\det(A)+e^T\operatorname{adj}(A)u =\det(A)(1+e^Tu)\\ &=\frac{1+(-1)^n}{2}\det(A) =\mathbb1_{n \text{ is even}}\det(A), \end{aligned} where $\mathbb1_{P}$ denotes the indicator function for a predicate $P$ (i.e. $\mathbb1_P$ is equal to $1$ when $P$ is true or $0$ otherwise). Hence \begin{aligned} \det(B)\left(\det(A)-\det(B)\right) &=\mathbb1_{n \text{ is even}}\det(A) \left[\det(A)-\mathbb1_{n \text{ is even}}\det(A)\right]\\ &=\mathbb1_{n \text{ is even}}\mathbb1_{n \text{ is odd}}\det(A)^2 =0. \end{aligned}

  • $\begingroup$ There is a problem. What happend if $a_{ij}=1$? $\endgroup$ – alexb Jun 16 '20 at 19:38
  • $\begingroup$ @alexb Please see my new edit. $\endgroup$ – user1551 Jun 16 '20 at 23:24

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