Doubt on Solution to DSE 2011 Question no 4

[Granpa Simpson]

In the solution to question number 4 (DSE 2011) its given that the set of vectors {a1, a2,.......b} will be linearly dependent, but if that is the case then the row reduced echelon form of the augmented matrix will consist of a zero row, which implies that there will be no solution (since in that case Rank A(1)<n, n= number of unknowns) for the set of equation, however in the question its given that the vector x solves the system of equations, i find it contradictory, I think the answer will be "a" rather than "c". Because only if they are linerly independent the determinant of the row reduced echelon matrix will be non zero, which implies that there will be a solution.

[Amit Goyal]

Hi Subhayu,

We are given that there exists x = (x1, x2, ..., xn) that solves x1(a1) + x2(a2) + ... + xn(an) = b.
And, by definition, a set of vectors S are said to be linearly dependent if there exists a vector in the span of S which can be written as linear combination of vectors in S in more than one way.

{a1, a2,......, an, b} will be linearly dependent because b can be written as the linear combination of vectors in {a1, a2,......, an, b} in more than one way: one is x1(a1) + x2(a2) + ... + xn(an) + 0(b) and another is 0(a1) + 0(a2) + ... + 0(an) + 1(b).