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?
Best Answer
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$