We know $\mathbb{Z}_5=\{0,1,2,3,4\}$ is field that its multiplication and addition are in hang of $5$. For exampple $4\times 2=3$. Alternatively we can say $\mathbb{Z}_5=\{[0],[1],[2],[3],[4]\}$ which its elements are equivalent classes. For example $[2]$ means all the numbers that their remainder in division to $5$ is $2$. For example $17\in [2]$. And we say $M_{2\times 3}(\mathbb{Z}_5)$ is the set that includes all $2\times 3$ matrices that its entries are elements of $\mathbb{Z}_5$. Now I want to know how many row-reduced echelon matrices exists in $M_{2\times 3}(\mathbb{Z}_5)$?
My approach:
- The matrix that its all entries are zero $\rightarrow 1$ case
- The case that we have only one zero row. So it will be forced to be the second row. If the first entry of the first line is 1 $\rightarrow 5\times5$ cases
- Like the previous case but the second entry of the first line is 1 $\rightarrow 5$ cases
- Like the previous case but the third entry of the first line is 1 $\rightarrow 1$ case
- We have no zero rows. In different cases we have $25+25+1$ cases.
So all the cases are $83$. Is it correct?