Combinatorics and Matrices Find the number of $4\times4$ matrices such that $|a_{ij}| = 1 \forall i,j\in[1,4]$ , and sum of every row and column is zero.
I tried 'counting' the number of matrices that satisfy the above conditions, that is, elements are $1$ or $-1$ and sum of every row and column is zero.
In the attempt to generate a recursion I started off with a $2\times2$ matrix, for which case the answer is $2$. (First element is 1 or -1, other elements are decided accordingly) 
However, this method becomes cumbersome and mathematically disappointing for $3x3$ and larger matrices.
Could someone please explain the method, or post a solution to the problem? 
Is it possible to generalise the result to an nxn matrix? 

The first row has two $1$s and two $-1$s. There are $\binom{4}{2} = 6$ ways they can be arranged in the first row. We'll count the number of matrices with first row $(1,1,-1,-1)$, then multiply by $6$ to account for all the other arrangements of the first row.
The second row can be one of the three following things: $(1,1,-1,-1)$, or $(-1,-1,1,1)$, or $(1,-1,1,-1)$. In the third case this counts for $4$ possibilities: the first two columns might be switched, or the last two columns might be switched. We'll count the number of matrices in each of these three cases (multiplying the third cases's count by $4$, to account for the switches).
Case 1: The matrix looks like
$$
  \begin{pmatrix}
    1 & 1 & -1 & -1 \\
    1 & 1 & -1 & -1 \\
    * & * & * & * \\
    * & * & * & *
  \end{pmatrix}
$$
There is only one possible matrix:
$$
  \begin{pmatrix}
    1 & 1 & -1 & -1 \\
    1 & 1 & -1 & -1 \\
    -1 & -1 & 1 & 1 \\
    -1 & -1 & 1 & 1 \\
  \end{pmatrix}
$$
Case 2: The matrix looks like
$$
  \begin{pmatrix}
    1 & 1 & -1 & -1 \\
    -1 & -1 & 1 & 1 \\
    * & * & * & * \\
    * & * & * & *
  \end{pmatrix}
$$
In the last two rows, there is precisely one $-1$ in each column and two $-1$s in each row. There are $\binom{4}{2}=6$ ways to choose the locations of two $-1$s in the third row; then the fourth row is determined (it has $-1$s in the complementary positions).
Case 3: The matrix looks like
$$
  \begin{pmatrix}
    1 & 1 & -1 & -1 \\
    1 & -1 & 1 & -1 \\
    * & * & * & * \\
    * & * & * & *
  \end{pmatrix}
$$
The first and last columns are determined:
$$
  \begin{pmatrix}
    1 & 1 & -1 & -1 \\
    1 & -1 & 1 & -1 \\
    -1 & * & * & 1 \\
    -1 & * & * & 1
  \end{pmatrix}
$$
There are two solutions: the remaining block can be $\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}$ or the opposite. That gives two solutions when the second row is as displayed, but because of the possibility of switching the first two or last two columns, we count $8$ solutions.
Subtotal: With this first row, we have $1+6+8=15$ matrices.
Total: There are $6$ equivalent arrangements of the first row, so we have $6 \cdot 15 = 90$ matrices in total.
It seems unlikely that this kind of approach could generalize to larger matrices.
