Excercise: matrices whose rows form numbers divisible by a prime

Let $p$ be a prime, and $(a_{ij})$ be a matrix such that $p$ divides $\sum_{j=1}^na_{ij}10^{n-j}$, for $j=1,\dots,n$. Prove that $p$ divides $\det(a_{i,j})$. If each $a_{ij}$ is only formed by a digit, this means that the numbers formed by the digits in each row are dividible by $p$.

Solution (show):
The divisibility condition means that the image of the vector $w=(10^{n-j})_{1\leq{}j\leq{}n}$ is $0$, when reduced modulo $p$. In the other hand, the vector $w$ is never zero when reduce modulo $p$, for each possible prime $p$. Consequently modulo $p$ the matrix has a non-trivial kernel, and its determinant must be zero.

Excercise: sum of determinants

Let $A$ be a linear operator on $\R^n$, and $p_1,\dots,p_n$ be (column) vectors in $\R^n$. Calculate $$ det(Ap_1,\dots,p_n) + det(p_1,Ap_2,\dots,p_n) + \dots + det(p_1,p_2\dots,Ap_n). $$
Solution (show):
Assume $p_1,\dots,p_n$ linearly independent. Write each $Ap_i$ as a combination of $p_1,\dots,p_n$, then decomposing $det(p_1,\dots,Ap_i,\dots,p_n)$ the only component of that gives a non-zero contribute is the component in $p_i$, and its contribute is $A_{ii}\cdot det(p_1,\dots,p_n)$, where $A_{ii}$ is the $(i,i)$ entry of $A$ written in the basis of the $p_i$. Summing over all $i$ we obtain that the answer is $$ Tr(A)\cdot det(p_1,\dots,p_n).$$ The formula is a polynomial in terms of $A$ and of the $p_i$, so an expression that is valid in an open set (where the $p_i$ are linearly independent) is always verified.