# Find the matrix with given kernel

Find one matrix $$A\in M_{3,4}(\mathbb{R})$$ such that $$Ker(A)=L(\begin{bmatrix} -3\\ 1\\ 0\\ 0 \end{bmatrix},\begin{bmatrix} -2\\ 0\\ -6\\ 0 \end{bmatrix})$$ and then describe all matrices with this properity.

What I tried to do so far is the following: Let $$Ker(A)=V$$. Then $$V^{\perp }$$ can be determined with the system:$$\begin{matrix} -3x_{1} & + x_{2} & =0\\ -2x_{1}& - 6x_{3} & =0 \end{matrix}$$ Now, since $$V=(V^{\perp })^{\perp }$$, I find that the system that has $$V$$ as its set of solution is the system: $$\begin{matrix} 3x_{1}-9x_{2}-x_{3}=0 & & \\ x_{4}=0 & & \end{matrix}$$, so one matrix is:$$A=\begin{bmatrix} 3 & -9 & -1 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 &0 \end{bmatrix}$$.

My question is how can I describe all matrices that have $$V$$ as its kernel.

The rows of your matrix should span the complement $$V^\perp$$. So if you take a basis for $$V^\perp$$, like $$\{(3,-9,-1,0),(0,0,0,1)\}$$, than every row vector should be a linear combination of these two. Hence every row is of the form $$(3a,-9a,-a,b)$$, with $$a,b \in \mathbb{R}$$.