Let S = \(\displaystyle{\left\lbrace{x}∈{R}^{{4}}{\mid}{A}{x}={0}\right\rbrace}\) where A = [-1,1,2,1,1,0,2,3]. Now

PSKS={(x1,x2,x3,x4) ∈R^4: [-1,1,2,1,1,0,2,3](x1,x2,x3,x4)^T=0} ={(x1,x2,x3,x4) ∈R^4: -x+2x2+x3+2x4=0,x1+x2+x4=0} ={(x1,x2,x3,x4) ∈R^4: x3=-3x2-5x4} ={(x1,x2,-3x2-5x4,x4):xi ∈R} ={x1(1,0,0,0)+x2(0,1,-3,0)+x4(0,0,-5,1):xi ∈R} =span{(1,0,0,0),(0,1,-3,0),(0,0,-5,1)}ZSK

Again {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a linearly independent set. Hence {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a basis of S.

PSKS={(x1,x2,x3,x4) ∈R^4: [-1,1,2,1,1,0,2,3](x1,x2,x3,x4)^T=0} ={(x1,x2,x3,x4) ∈R^4: -x+2x2+x3+2x4=0,x1+x2+x4=0} ={(x1,x2,x3,x4) ∈R^4: x3=-3x2-5x4} ={(x1,x2,-3x2-5x4,x4):xi ∈R} ={x1(1,0,0,0)+x2(0,1,-3,0)+x4(0,0,-5,1):xi ∈R} =span{(1,0,0,0),(0,1,-3,0),(0,0,-5,1)}ZSK

Again {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a linearly independent set. Hence {(1,0,0,0),(0,1,-3,0),(0,0,-5,1)} is a basis of S.