Find matrix $A$ over $\Bbb Z_3$ such that $\ker A = \left\langle\begin{pmatrix} 1\\ 2\\ 0\\ 1\\ 2\\ \end{pmatrix},\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix},\begin{pmatrix} 1\\ 0\\ 2\\ 1\\ 0\\ \end{pmatrix}\right\rangle$.
I am confused by this problem. I guess I am not sure what the task is. I know how to calculate $\ker A$ for a matrix but what is this telling me? $\ker A$ means $Ax = 0$ but how do you proceed when you have the set of solutions like this instead of the matrix?
Thank you.
Let $B$ be the matrix whose columns are the given vectors. Then you want to find a matrix $A$ satisfying $AB=0$, or equivalently $B^{\intercal}A^{\intercal}=0$. So the columns of $A^{\intercal}$ must be in the kernel of $B^{\intercal}$. Can you take it from here?