Let $\ A $ be a $\ 4 \times 4 $ matrix over $\ \mathbf R $ with a rank of 3. $\ (1,2,0,-1) , (0,2,1,1) $ are solutions to $\ Ax = b $ and I need to find the solution space of the $\ Ax = 0$ and general solution of $\ Ax = b $
If the rank of the $\ A $ is $\ 3$ then the dimension of solution space of $\ Ax= 0 $ is $\ 1$ and I guess any vector that is a solution for $\ Ax = 0 $ can't be spanned by $\ (1,2,0,-1),(0,2,1,1)$ but how do I proceed from here?
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $\mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $\mathbf v$ and $\mathbf w$ are distinct solutions to your inhomogeneous equation, then $A\mathbf v-A\mathbf w = A(\mathbf v-\mathbf w)=0$, so $\mathbf v-\mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $\mathbf v + \lambda(\mathbf v-\mathbf w)$, $\lambda\in\mathbb R$, or, more symmetrically, $(1-\lambda)\mathbf v+\lambda\mathbf w$. This is just the line through $\mathbf v$ and $\mathbf w$, as above.