Let C be a 3-by-5 matrix. The linear system $Cx = b$ has solutions for $b_1= \begin{bmatrix} 2 \\ 1 \\ 1 \\ \end{bmatrix}, b_2=\begin{bmatrix} 1 \\ 0 \\ 3 \\ \end{bmatrix}$, and does NOT have solutions for $b_3=\begin{bmatrix} 3 \\ 0 \\ 2 \\ \end{bmatrix}$ Determine the nullity of $C$.
Given that $b_1$ is a solution, I determined that the rank of of $C$ is $3$, since if you were to augment the reduced row echelon form of $C$ with $b_1$, you would need a leading $1$ in each of the $3$ rows otherwise you would get a contradiction of $0$ equaling a constant. On the other hand, $3$ is the highest possible number for the rank of $C$.
I believe that my error is somewhere is that reasoning above, as the rank of $C$ would be all that I need to then solve for the nullity, but the answer is incorrect. Any help would be greatly appreciated.
The rank is the dimension of the image. Since $b_1$ and $b_2$ are independent, the image has dimension at least two. On the other hand the image doesn't have dimension three, since $b_3$ isn't in it.
Thus the rank of $C$ is $2$. So the nullity is...