# Showing the determinant for a specific type of matrix is $\det(A) = (-1)^n 2^{n-1} \sum_{i=1}^{n} a_1 a_2 \ldots a_{i-1} a_{i+1} \ldots a_n$

I am trying to figure out the best way to prove a formula quickly because there is a chance it could be on exam with a lot of questions and on average we should spend two minutes per question. My tools for this problem have been the cofactor expansion theorem mainly but I have been starting to think there might be a way to apply induction although I have not been able to isolate the argument after going over computing the determinant for $n=3, n=4$. My guess is induction is the fastest way to prove the following formula.

$$A = \begin{pmatrix} 0 & 1 & 1 & 1 & \ldots & 1 \\ 1 & 0 & a_1+a_2 & a_1+a_3 & \ldots & a_1+a_n\\ 1 & a_2+a_1 & 0 & a_2+a_3 & \ldots & a_2 +a_n \\ 1 & a_3+a_1 & a_3 + a_2 & 0 & \ldots & a_3 +a_n \\ . & . & . & .& . &.\\ 1 & a_n+a_1 & a_n+a_2 & a_n+a_3 & \ldots & 0 \end{pmatrix}$$

How do you show $\det(A) = (-1)^n 2^{n-1} \sum\limits_{i=1}^{n} a_1 a_2 \ldots a_{i-1} a_{i+1} \ldots a_n$

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The title is the first thing people see; it might be worth spending a few seconds proof reading it... – Arturo Magidin Nov 7 '11 at 0:18
You can start by "simplifying" your matrix with the following operation on lines : $L_i \to L_i - a_i L_1$ for $i \ge 2$ (which does not alter the determinant). Then divide the column $i$ by $a_i$ (which changes the determinant by a factor $a_i$). Then use the cofactor expansion theorem. – Joel Cohen Nov 7 '11 at 0:47