Let $A$ be a $k \times 3$ matrix with $rank(A) = 2$ and let $B = \begin{pmatrix} 1 & 2 & 3 \\ 2 & a & b \\ 3 & c & d \end{pmatrix}$. Suppose $AB = 0$. Find $a, b, c, d$.
I tried to do the following: First, I know that $AB=0$, thus meaning that $$Ab_1 = 0, Ab_2 = 0, Ab_3 = 0 \implies$$ $$A\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = 0, A \begin{pmatrix} 2 \\ a \\ c \end{pmatrix} = 0, A \begin{pmatrix} 3 \\ b \\ d \end{pmatrix} = 0$$ Which means that $\forall v \in Col(B), Av = 0 \implies v \in Nul(A)$. Also, because $rank(A) = 2$, that implies that $dimNul(A) + rank(A) = n \implies dimNul(A) +2 = 3 \implies dimNul(A) = 1$, so it appears that the basis for $Col(B)$ is also the part of the basis for $Nul(A)$, which would imply that $dimCol(B) \geq 1$. This is the furthest I got in trying to solve to problem. My intuition points me to believing that $dimCol(B) = 1$, but I can't prove this. Also, I can't seem to understand what kind of linear dependency would allow me to find the values of $a, b, c, d$
Was I on the right track? Also, what information am I missing in order to prove this?
Hint: Since the three columns of $B$ belong to $\ker A$ and since this is a $1$-dimensional space, the rank of $B$ must be $1$. Therefore, the second and the third columns must be multiples of the first one.