Exercise 3.5.19

(a) Elimination on $Ax=0$ leads to $0=b_3-b_2-b_1$ so $(-1,-1,1)$ is in the left nullspace. (b) 4 by 3: Elimination leads to $b_3-2b_1=0$ and $b_4+b_2-4b_1=0$, so $(-2,0,1,0)$ and $(-4,1,0,1)$ are in the left nullspace. Why? Those vectors multiply the matrix to give zero rows in $vA$. Section 4.1 will show another approach: $Ax=b$ is solvable ( $b$ is in $C(A)$ ) exactly when $b$ is orthogonal to the left nullspace.