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DSSSB TGT Maths Male Subject Concerned - 23 Sep 2018 Shift 2

Option 4 : C={{1, 1, 3],[0, 2, 1], [0, 0, 0]}

__Concept:__

A set of vectors **{v1, v2,…, vp}** in a vector space V is said to be

- linearly independent if the vector equation c1v1 + c2v2 +…+ cpvp = 0 has only
**one trivial solution c1 = 0, c2 = 0,…, cp = 0;** - linearly dependent if there exists weights c1, c2,…, cp not all 0, such that c1v1 + c2v2 +…+ cpvp = 0

__Calculation:__

Let us check the linear dependency for each set of vector individually,

**→ For A:**

A = {[1, 1, 1],[1, 1, 0],[1, 0, 0]};

The linear combination will be

a_{1}[1, 1, 1] + a_{2} [1,1, 0] + a_{3} [1, 0, 0] = 0

⇒ a_{1} + a_{2} + a_{3} = 0; a_{1} + a_{2} = 0, a_{1} = 0

**⇒ a _{1} = 0, a_{2} = 0, a_{3} = 0;**

**⇒ Linearly independent**

**→ For B:**

B = {[1, 0, 0], [0, 1, 1],[1, 0, 1]};

The linear combination will be

b_{1}[1, 0, 0] + b_{2} [0, 1, 1] + b_{3} [1, 0, 1] = 0

⇒ b_{1} + b_{3} = 0, b_{2} = 0, b_{2} + b_{3} = 0;

**⇒ b _{1} = 0, b_{2} = 0, b_{3} = 0;**

**⇒ Linearly independent**

**→ For C:**

C ={{1, 1, 3],[0, 2, 1], [0, 0, 0]}

The linear combination will be

c_{1}[1, 1, 3] + c_{2} [0, 2, 1] + c_{3} [0, 0, 0] = 0

⇒ c_{1} = 0, c_{1} + 2c_{2} = 0, 3c_{1} + c_{2} = 0;

**⇒ c _{1} = 0, c_{2} = 0, c_{3} = any value;**

**⇒ Linearly dependent **

**→ For D:**

D = {[1, 1, 1], [0, 1, 1], [2, 0, 1]}

The linear combination will be

d_{1}[1, 1, 1] + d_{2}[0, 1, 1] + d_{3} [2, 0, 1] = 0

⇒ d_{1} + 2d_{3} = 0, d_{1} + d_{2} = 0, d_{1} + d_{2} + d_{3} = 0;

**⇒ d _{1} = 0, d_{2} = 0, d_{3} = 0;**

**⇒ Linearly independent**

**∴ Among A, B, C, D, only C is the only linealry dependent set**