Finding the number of $2\times2$ matrices whose sum of element is zero If the number of non skew symmetric matrices $A = [a_{ji}],$ where $a_{ji}\in\{-2,-1,0,1,2\} $ and the sum of elements is zero then find the number of such matrices.
I tried setting up a recursion using (-n....0....n) and adding -n-1 and n+1 but not able to build further
The answer given is 80 can anyone please help in how to approach such problems

split into cases based on the number of zeros in your 4 elements, there can only be 0, 1, or 2.
The biggest case will be for 0 zeros, for which the elements would be from one of 3 sets,
{1, 1, -1, -1}, {2, 2, -2, -2} or {1, -1, 2, -2}
To find the number of matrices you just need the permutations of each of the sets
$$ n(0)  = \frac{4!}{2!2!} + \frac{4!}{2!2!} + 4! \\ = 6 + 6 + 24 = 36$$
For 1 zero the sets have to be made up from
{1, 1, -2, 0} and {-1, -1, 2, 0}
$$n(1) =  \frac{4!}{2!}  + \frac{4!}{2!}  = 12 + 12 = 24$$
for 2 zeros the sets have to be
{0, 0, 1, -1} or {0, 0, 2, -2}
$$n(2) = 12 + 12 = 24$$
out of these only 4 are skew-symmetric
$$
\begin{bmatrix} 
 0 & 1 \\
 -1 & 0 \\
 \end{bmatrix}  
 \begin{bmatrix} 
 0 & -1 \\
 1 & 0 \\
 \end{bmatrix} 
 \begin{bmatrix} 
 0 & 2 \\
 -2 & 0 \\
 \end{bmatrix}  
 \begin{bmatrix} 
 0 & -2 \\
 2 & 0 \\
 \end{bmatrix}  
 $$
So total number of non-skew symmetric matrices is
$$ N_{tot}=36+24+24-4=80$$
