I'm having a hard time solving an exercise that seems fairly easy:
Given a linear map $$ f: (Z/5)^2 \mapsto (Z/5)^2 $$ and $$ f(\bar{2}, \bar{3}) = (\bar{1},\bar{1})$$ $$f(\bar{1}, \bar{4}) = (\bar{3}, \bar{3})$$
calculate $f(x,y)$.
I know I have to describe $\begin{pmatrix} x\\y \end{pmatrix}$ as a linear combination of the two function arguments.
$$\begin{pmatrix} x\\y \end{pmatrix} = a\begin{pmatrix} \bar{2} \\ \bar{3} \end{pmatrix} + b \begin{pmatrix} \bar{1} \\ \bar{4} \end{pmatrix}$$
Then I end up with the following system of linear equations: $$ x = \bar{2}a + b \rightarrow a = \bar{3}(x - b)$$ $$ y = \bar{3}a + \bar{4}b$$
Replace a in the second equation: $$ y = \bar{3}(\bar{3}(x -b)) + \bar{4}b =$$ $$ y = \bar{4}(x -b) + \bar{4}b =$$ $$ y = \bar{4}x - \bar{4}b + \bar{4}b = (\text{or } y = \bar{4}x + b + \bar{4}b) ? =$$ $$ y= \bar{4}x$$
Which doesn't get me any further. I'm not sure if $\bar{4}(x - b) = \bar{4}x + b$ since $\bar{4} * -1 = -4 = \bar{1}$ or if it's simply $ \bar{4}(x + b) = \bar{4}x +\bar{4}b$. Either way I end up with $y = \bar{4}x$.
Or maybe I'm interpreting this the wrong way?
The problem is that you have not enough information. I will ommit the bars, since we are making nothing with 'authentic' integers. The set $\{(2,3),(1,4)\}$ is not linearly independent, that is, it's not a basis: $$\left| \begin{array}{r} 2&3\\ 1&4 \end{array} \right|=8-3=5=0$$
so there are infinitely many linear functions that satisfy the requirements.
May this one be a trapped exercise ;)?