# 2x2 matrix with factor group elements

by jostpuur
Tags: elements, factor, matrix
 P: 2,068 We fix some $N=1,2,3,\ldots,$ and define the factor group $\mathbb{Z}_N$ as $\mathbb{Z}/N\mathbb{Z}$, and denote the elements $x+N\mathbb{Z}$ as $[x]$, where $x\in\mathbb{Z}$. My question is that how do you solve $[x_1]$ and $[x_2]$ out of $$\left(\begin{array}{c} \lbrack y_1\rbrack \\ \lbrack y_2\rbrack \\ \end{array}\right) = \left(\begin{array}{cc} \lbrack a_{11}\rbrack & \lbrack a_{12}\rbrack \\ \lbrack a_{21}\rbrack & \lbrack a_{22}\rbrack \\ \end{array}\right) \left(\begin{array}{c} \lbrack x_1\rbrack \\ \lbrack x_2\rbrack \\ \end{array}\right)$$ when $[y_1],[y_2],[a_{11}],[a_{12}],[a_{21}],[a_{22}]\in\mathbb{Z}_N$ are known constants. First edit: To clarify what I mean by "a solution", I'll show a one for the 1x1 case, where we seek to solve $[y]=[a][x]$. If $[a]=0$ and $[y]\neq 0$, then there are no solutions. If $[a]=0$ and $[y]=0$, then all $[x]\in\mathbb{Z}_N$ are solutions, and the number of different solutions is $N$. If $[a]\neq 0$, we choose representatives $a,y\in\mathbb{Z}$ such that [itex]0
 Sci Advisor P: 3,313 Giving the problem a technical name, I'd called it "solving simultaneous linear equations in a finite field". A useful digression would be to ask: What's the rellation between the solutions to to simultaneous linear Diophantine equations and the soluitons to linear equations in a finite field?
P: 59
 Quote by Stephen Tashi Giving the problem a technical name, I'd called it "solving simultaneous linear equations in a finite field".
What? Original poster didn’t say N must be prime. One can solve linear systems over rings with zero divisors (Ī mean ℤN for composite N), but it is a theory different from linear algebra over a field. Even solving one linear equation is a non-trivial problem since we, in general, can’t divide.

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