 Solution

The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every vector in the null space as $t(2,-9,-3,21{)}^{\top }$. Thus a basis for the kernel of the linear map isDirect multiplication gives $A(2,-9,-3,21{)}^{\top }=0$. The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.