determinant and submatrix 
I'm trying to see how can we get that Laplace expansion in terms of the first $p$ rows will give us $\det A=0$. Is it a proof by induction over the size of the matrix or just a direct observation? 
I know that the Laplace expansion is given by
$\det A=\sum_{i=1}^n(-1)^{i+1}a_{i1}\det A_{i1}$. But form this, I cannot see why $\det A=0$.
 A: I think the confusion here stems from ambiguity in what is meant by ''Laplace expansion''.
Initially, I thought the argument was implicitly using induction: if $p \geq q$ and we take the expansion of the determinant along the first column, using the formula given in the question, then the first $p$ terms are 0 immediately and the following terms should be 0 by some appropriate induction argument.
However,  ''Laplace expansion'' can in fact describe a process of computing the determinant that is more general than expanding along a single row or column: we can instead expand along 2 rows (or 2 columns, or 3 rows, etc.) in which we take a sum over all $2\times2$ minors of the 2 chosen rows, where each term up to sign is the $2\times 2$ determinant times the complementary $(n-2)\times(n-2)$ determinant (or ''cofactor'').
An example calculation is given on the wikipedia page for Laplace expansion.
Using this interpretation of ''Laplace expansion'', each term must equal 0 because there is a row (or column) of all zeros in each minor, so the corresponding determinant is 0, and no induction is needed.
