Prove the determinant $[b_{ij}]_{n \times n}=(-1)^{n-1}(n-1)[a_{ij}]_{n \times n}$

Question

If $b_{ij}=(a_{i1}+a_{i2}+\cdots+a_{in})-a_{ij}$, show that:

$\begin{vmatrix} b_{11} & \cdots & b_{1n} \\ \vdots &&\vdots\\b_{n1} & \cdots & b_{nn} \end{vmatrix} = (-1)^{n-1}(n-1)\begin{vmatrix} a_{11} & \cdots & a_{1n} \\ \vdots &&\vdots\\a_{n1} & \cdots & a_{nn} \end{vmatrix}$

Answer 1

Hint: Note that $B = AM$, where $$ M = \pmatrix{0&1&\cdots & 1\\1&0&\ddots & \vdots\\ \vdots & \ddots & \ddots &1\\1 & \cdots & 1 & 0} $$