设A为5*4矩阵,若Ax=β有解,η1,η2是其两个特解,Ax=0的基础解系是X1,X2,则(A)Ax=β的通解是k1X1+K2X2+1/2(η1-η2);(B)Ax=β的通解是k1(X1+X2)+k2(X1-X2)+(2η1-η2);(C)Ax=0的通解是k1X1+k2(η1+η2);(D)Ax=0的通解是k1(X1-X

Question

设A为5*4矩阵,若Ax=β有解,η1,η2是其两个特解,Ax=0的基础解系是X1,X2,则
(A) Ax=β的通解是k1X1+K2X2+1/2(η1-η2);
(B) Ax=β的通解是k1(X1+X2)+k2(X1-X2)+(2η1-η2);
(C) Ax=0的通解是k1X1+k2(η1+η2);
(D) Ax=0的通解是 k1(X1-X2)+k2(η1-X1)

Answer 1

(B) 正确
(A) 的特解不对
(C),(D) 显然不对