<s>[INST] 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
 [/INST] 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
$$ N_{tot}=36+24+24-4=80$$
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