Correct Answer - Option 2 : S

_{1} can be enlarged but S

_{2} cannot be enlarged to a basis for R

3
__Concept:__

Let S = {v_{1}, v_{2}, . . . , v_{n}} be a set of n vectors in a vector space V .

If **S is linearly independent and the dimension of V is n, then S is a basis of V.**

If the **dimension of S is less than Dimension of V, then S cannot be basis**, but **can be extended to a basis if S is linearly independent.**

A set of vectors {v1, v2,…, v_{n}} in a vector space V is said to be linearly independent if the vector equation c1v1 + c2v2 +…+ c_{n}v_{n} = 0 has only one trivial solution c1 = 0, c2 = 0,…, c_{n} = 0;

The set is said to be linearly dependent if there exists weights c1, c2,…, c_{n} not all 0, such that c1v1 + c2v2 +…+ c_{n}v_{n} = 0

__Calculation:__

Given S1 = {[1, -1, 2], [3, 2, -1]} and S2 = {[2, 7, -3],[-6, -21, 9]}

Let S1 = (v1, v2) and the vector equation be av1 + bv2 = 0

S1 = {[1, -1, 2], [3, 2, -1]} ⇒ v1 = [1, -1, 2], v2 = [3, 2, -1];

⇒ a + 3b = 0, -a + 2b =0, 2a - b = 0 ⇒ a = 0, b = 0

Since, both the constants are zero, S1 will be linearly independent.

Hence S1 can be extended to basis.

Let S2 = (v1, v2) and the vector equation be av1 + bv2 = 0

S2 = {[2, 7, -3],[-6, -21, 9]} ⇒ v1 = [2, 7, -3], v2 = [-6, -21, 9];

⇒ 2a - 6b = 0, 7a - 21b = 0, -3a + 9b = 0 ⇒ a = 3b, a and b can be any real number

Since many trivial solutions, S2 will be linearly dependent.

Hence S2 cannot be extended to basis.