Calculate the following determinant $$\det\begin{bmatrix} a_1 & 0 & ... &0&b_1 \\ 0&a_2 & ... &b_2&0 \\ &&... \\ 0 & b_{2n-1} & ... &a_{2n-1}&0\\ b_{2n} & 0 & ... &0&a_{2n}\end{bmatrix}$$
We thought about using the idea of a determinant of blocks. Using that way we got $a_1a_2..a_2n - b_1b_2...b_{2n}$ . Another thought was using the definition using permutations, but we are having trouble understanding why the rest of the permutations (except the identity permutation and the permutation that returns the last to the first etc) eventually become 0. We would like to verify correctness/get an explanation on how to solve this one using permutations and blocks.
Laplace expansion should give you the simplest way to solve the problem, but if you insist on using permutations and block matrices, you may first move the last column forward to the second column, and then move the last row upward to the second row. So, the required determinant is equal to the determinant of $$ \pmatrix{a_1&b_1\\ b_{2n}&a_{2n}\\ &&a_2&\cdots&b_2\\ &&&\cdots\\ &&b_{2n-1}&\cdots&a_{2n-1}} $$ and you may proceed recursively.