Determinant of matrix using the n-linear function property. I am stuck computing the determiannt of a matrix using the n-linear function property of determinant. the matrix is : $\left|\begin{array}{cccc}a_{1}+b_{1} & a_{1}+b_{2} & \dots & a_{1}+b_{n} \\ a_{2}+b_{1} & a_{2}+b_{2} & \dots & a_{2}+b_{n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n}+b_{1} & a_{n}+b_{2} & \dots & a_{n}+b_{n}\end{array}\right|$
I dont really know how to gou about writing it down, could somebody please explain ?
Thanks.
 A: Let $V$ denote an $n$-dimensional vector space. A function $f: V^n \to \mathbb R $ is said to be $n$-linear iff for any ${\bf x_1}\ldots {\bf x_n} \in V$ the function
$$ {\bf v} \mapsto f({\bf x_1}, \ldots, {\bf x_{k-1}},{\bf v},{\bf x_{k+1}}, \ldots, {\bf x_n}) $$
is linear for any $k\in \{1,\ldots,n\}$. It can be proved that the determinant as $\det : \mathbb R^n \to \mathbb R$ function is an (antisymmetric or alternating) $n$-linear function.
Now let ${\bf a}$ denote the column vector $(a_1,\ldots,a_n)^T $ and let ${\bf b_k}$ denote the column vectors $(b_k,\ldots,b_k)^T$. Then the determinant you are looking for is
$$ \det({\bf a}+{\bf b_1}, {\bf a} +{\bf b_2},\ldots,{\bf a} + {\bf b_n}) = \det ({\bf a},{\bf a}+{\bf b_2},\ldots,{\bf a} +{\bf b_n}) + \det({\bf b_1},{\bf a}+{\bf b_2},\ldots,{\bf a} +{\bf b_n}) =... = \sum_{\epsilon \in\{0,1\}^n} \det(\epsilon_1{\bf a}+(1-\epsilon_1){\bf b_1},\ldots,\epsilon_n {\bf a}+(1-\epsilon_n){\bf b_n} )  $$
whenever there are two ${\bf a}$'s in the determinant its value is zero. Hence
$$ \det(\ldots) = \sum_{\epsilon\in\{0,1\}^n \\ \sum_{j}\epsilon_i = 1}\det(\epsilon_1{\bf a}+(1-\epsilon_1){\bf b_1},\ldots,\epsilon_n {\bf a}+(1-\epsilon_n){\bf b_n} ) = \sum_{i=1}^n\det({\bf b}_1,\ldots,{\bf b_{i-1}}, {\bf a}, {\bf b_{i+1}}, \ldots, {\bf b_n}) $$
