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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.

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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 $a,b \in \mathbb{R}$.

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