,I'm trying to solve this problem.

I have two additive groups $$G, H$$. The first is the group of matrices $$4x1$$ with coefficients in $$\Bbb Z_{11}$$. The second is the group of matrices $$3x1$$ with coefficients in $$\Bbb Z_{11}$$.

Let a = $$\left(\begin{matrix} 2 & 7 & 2 & 4 \\ 3 & 5 & 1 & 2 \\ 5 & 1 & 8 & 5 \end{matrix}\right)$$ (coefficients in $$\Bbb Z_{11}$$) and $$f: G \rightarrow H$$ defined as $$x \rightarrow a*x$$.

Prove that $$f$$ is a homomorphism of additive groups. Tell which is the kernel and its order.

Is $$\left(\begin{matrix} 4 \\ 6 \\ 10 \end{matrix}\right)$$ $$\in H$$ in the image of $$f$$?

Since it is a very long and mechanical approach, I don't write my attempt to try that the function is a homomorphism (it is, maybe someone could confirm). Now I'm trying to find its kernel. I reasoned in a similar way to the approach used to verify the first point.

First of all, I consider a general matrix $$\in G$$: $$\left(\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix}\right)$$ then I need to do the multiplication a$$*$$b. I obtain the matrix:

$$\left(\begin{matrix} 2x_1+7x_2+2x_3+4x_4 \\ 3x_1+5x_2+x_3+2x_4 \\ 5x_1+1x_2+8x_3+5x_4 \end{matrix}\right)$$

Now, by the kernel definition, I need to find when for certain matrices $$\in G$$:$$\left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right)$$

It is sufficient to solve a system of three linear equations? In particular: $$2x_1+7x_2+2x_3+4x_4=0 \\ 3x_1+5x_2+x_3+2x_4=0\\ 5x_1+1x_2+8x_3+5x_4=0$$

If this is a good approach, how do I need to continue after I solved the system? How can I find the order of the kernel? Can I apply the same approach to answer the last question? If yes, which tells me that the element belongs or not the image of the function?

• Your approach is good, Once you solve the system, you will express it as a combination of matrices in $G$. If done correctly, then you will have linearly independent vectors that will form a basis of the kernel. With the same system and right hand side vector changed appropriately you can check if the given vector is in the range, – Anurag A Mar 14 at 19:12
• @AnuragA Hi, I tried to solve the system. I obtained the following matrix (assuming I didn't make a mistake in calculations) $\left(\begin{matrix} 1 & 9 & 0 & 0 &0 \\ 0 & 0 & 1 & 0 &0\\ 0 & 0 & 0 & 1 &0 \end{matrix}\right)$ so let $x_2=t, t\in \Bbb Z_{11}$ the solutions are $x_1=-9t, x_2=t, x_3=0, x4=0$. Now how I need to continue? – PCNF Mar 15 at 17:04

Edited: I didn't realize that the last matrix you had in your comments is an augmented matrix, hence I had added $$x_5$$.
So you should have $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=t\begin{bmatrix}-9\\1\\0\\0\end{bmatrix}, \quad t \in \mathbb{Z}_{11}.$$
Thus $$\ker(f)=\left\{t\begin{bmatrix}-9\\1\\0\\0\end{bmatrix} \quad | \, \quad t \in \mathbb{Z}_{11}\right\}.$$ This is a cyclic subgroup generated by the element $$\begin{bmatrix}2\\1\\0\\0\end{bmatrix}$$ whose order is $$11$$ (since both $$2$$ and $$1$$ have orders $$11$$.)
• Thanks for details. Why $x_5$? The system has only 4 unknowns... – PCNF Mar 15 at 19:37