(A) A
(B) B
(C) C
(D) D
Answer: (A)
Explanation: To be basis of subspace x, 2 conditions are to be fulfilled
1) They must span x
2) The vectors have to be linearly independent
1)the general solution of x1+x2+x3=0 is [-x2-x3 , x2 , x3]^T (Transpose)
Which gives two linearly independent solutions by assuming x2 = 1 and x3 = 0 and next x3 = 1 and x2 = 0 gives [-1,1,0]^T and [-1,0,1]^T respectively. Since both of these can be generated by linear combinations of [1,-1,0]^T & [-1,0,1]^T given in question, it span x.
2) Above set of column vector is linearly independent because one cannot be obtained from another by scalar multiplication
(second method rank is 2..that is why linearly independent)
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