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:

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

Ben Grossmann · Accepted Answer · 2018-10-03T16:42:39.067

3

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

edited Oct 03 '18 at 16:42

answered Oct 03 '18 at 16:31

Ben Grossmann

225,327

It implies that $det(M)=(-1)^{n-1}(n-1)$ and gotta! Thanks! I think I have an idea to prove it! – weilam06 Oct 03 '18 at 16:41
@WeiLam I had the multiplication backwards, but the same idea applies. You're welcome. – Ben Grossmann Oct 03 '18 at 16:43

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

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