I am trying to prove the following property of the determinants:
$$\begin{vmatrix} a_1\newline a_2 \newline \vdots \newline a_k=a_k'+a_k'' \newline \vdots \newline a_n \end{vmatrix} = \begin{vmatrix} a_1\newline a_2 \newline \vdots \newline a_k' \newline \vdots \newline a_n \end{vmatrix}+\begin{vmatrix} a_1\newline a_2 \newline \vdots \newline a_k'' \newline \vdots \newline a_n \end{vmatrix}$$
What I've figured out so far.
We have $$a_{kj}=a_{kj}'+a_{kj}'', j=1,2,\dots,n,$$ so the LHS of the equality we're trying to show is equal to $$\sum (-1)^{[i_1,i_2,\dots,i_n]}a_{1i_1}\dots a_{ki_k}\dots a_{ni_n} \\ =\sum (-1)^{[i_1,i_2,\dots,i_n]}a_{1i_1}\dots (a_{ki_k}'+a_{ki_k}'')\dots a_{ni_n}$$ May someone show me in details how is this sum equal to the sum of the determinants we have in the RHS as it's not obvious for me?
I don't see how we can write the given sum as two sums. What does happen with the $(-1)^{[i_1,i_2,\dots,i_n]}$ terms?
May I also ask why do we call this property 'linearity of determinants in terms of their rows and colums'? How should I understand this?