Consider the additive groups $\mathit{G}:= \mathfrak{M}_{\mathrm{4x1}}(\mathbb{Z_{11}})$ and $\mathit{H}:= \mathfrak{M}_{\mathrm{3x1}}(\mathbb{Z_{11}})$ (column matrices with height 4 and 3 respectively). Suppose that $\mathbf{a}$ is a fixed matrix such that $\mathbf{a} :=$$\left(\begin{matrix}2&7&2&4 \\3&5&1&2 \\ 5&1&8&5\end{matrix}\right) \in \mathfrak{M}_{\mathrm{4x3}}(\mathbb{Z_{11}})$, consider the function:
$\mathit{f}_a:\mathit{G} \to \mathit{H}$
$\mathbf{x} \to \mathbf{a \cdot x }$
($\mathbf{a \cdot x}$ is, of course, product of matrix $\mathbf{a}$ by column matrix $\mathbf{x}$).
1) Prove that $\mathit{f}_a$ is a homomorphism of additive groups and determine kernel and its cardinality;
2) demonstrate if the element $\mathbf{y} := $$\left(\begin{matrix}4\\6 \\10 \end{matrix}\right) \in \mathit{H}$ is in the image of $\mathit{f}_a $.
So, I had this problem during my Discrete Mathematics test months ago (which I passed), but i couldn't solve this particular problem, I solved many exercises with Group Theory, Matrices etc, but this one is a merge of two different argument and I was mentally imprepared for that. I never ever tried to solve it, I just wanted to know the answer and the process behind it. Thanks!
It should be straight forward to check that a linear mapping is a homomorphism. Row-reduce in $\Bbb Z_{11}$ to find the kernel. The cardinality should be $11^{\operatorname {nullity}}$.
For part b, row-reduce the augmented matrix in $\Bbb Z_{11}$.
So, for starters, $2^{-1}\cong6\pmod {11}$. So multiply the first row by $6$. Get a $1$ on the pivot, and use it to clear the first column. Now do the same for the second pivot and column. Continue...
Here's what I got: $\left(\begin{array}{rrrr|r}2&7&2&2&4\\3&5&1&2&6\\5&1&8&5&10\end{array}\right)\to\left(\begin{array}{rrrr|r}1&9&1&2&2\\0&0&9&7&0\\0&1&8&5&10\end{array}\right) \to\left(\begin{array}{rrrr|r}1&0&6&1&0\\0&1&8&5&10\\0&0&9&7&0\end{array}\right)\to\left (\begin{array}{rrrr|r}1&0&0&0&0\\0&1&0&4&10\\0&0&1&2&0\end{array}\right) $.
Now we back-substitute: get $\{\begin{pmatrix}0\\-4t+10\\-2t\\t\end{pmatrix}, t\in\Bbb Z_{11}\}$.
So the answer to the last question is yes. For instance, $f_a\begin{pmatrix}0\\10\\0\\0\end{pmatrix}=\begin {pmatrix}4\\6\\10\end{pmatrix}$.
And, the rank is $3$. So the nullity is $1$ and the cardinality of the kernel is $11$. The kernel is $\{\begin{pmatrix}0\\-4t\\-2t\\t\end{pmatrix}\,,t\in\Bbb Z_{11}\}$