# Find $\det B(\det A-\det B)$.

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 $$\det A=\det B=0$$

• Do you mean $\det\left(B(\det(A)-\det(B)\right)$ or $\left(\det(A)-\det(B)\right)\det(B)$? – 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}
• There is a problem. What happend if $a_{ij}=1$? – alexb Jun 16 '20 at 19:38