A question on proving an equation to be an $n$-linear system in linear algebra

For the first one, look that $D_{ij}(A)$ does not depend on the i-th row of its argument (without loss of generality since D is [n-1] - linear) then $D_{ij}(\lambda A + B)) = \lambda^{n-1} D_{ij} A + D_{ij} B$ look that $D_{ij} (\cdot)$ just ignores the i-th rows of $A$ and $B$ and operates linearly on the rest. Since $A_{ij}$ doesn't get affected by $D_{ij}$, because it's in the i-th row, we can just multiply it with $D_{ij}(A)$ and we preserve the linear relationship with $A$, just write down the adecuate expresión for $\lambda A_{ij} D_{ij}(\lambda A)$ and you'll get that $A_{ij} D_{ij} (A)$ is in fact n-linear in A.

For the second one we assumed that the rows $\alpha_k = \alpha_{k+1}$ and that $D$ is alternating, which means that any (n-1)-matrix containing the rows $\alpha_k$, $\alpha_{k+1}$ will yield 0 in being operated by $D$. Those matrices are sub-matrices $A(i|j)$ of $A$ where $i \neq k$ and $i \neq k+1$. Observe that $j$ can be any column. This tell us that in the definition of $E$ (the sum over the rows), all terms not deleting the row $k$ nor $k+1$ must be zero (again since D is alternating). Therefore, we have only 2 terms left in our sumation, namely: $$ (-1)^{k+i} A_{k j} D_{k j}(A)+(-1)^{k+1+i} A_{(k+1) j} D_{(k+1) j}(A) $$

I hope it helps!