GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 10

Answer 1

As per the definition of $\text{LD},$ there should be a nontrivial solution to show Linear Dependence among vectors of a set. $\text{C}$ gives the trivial solution, i.e., all $0$ coefficients.

Comments

Guys how option A and B both can be correct? In the equation c1.v1+c2.v2+…..cn.vn=0 , RHS is a scalar or vector? In option A , RHS is a vector. In option B if we re-write it as 0u+0v-1w=0 then RHS is scalar. According to my knowledge if we add vectors after multiplying with a scalar then the result should be a vector. I marked option B because option A is a bit ambiguous to me as we introduced the zero vector ourself in the equation in RHS. Can anyone explain it? @Sachin Mittal 1

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@Valiant_ 
In option B, to bring w to LHS, we have to subtract w in both LHS, RHS. Therefore w-w = zero vector, not scaler.
But these small details can be ignored safely. Since 0 can be treated as vector and scaler both, wherever necessary.

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 @Cultured_Noob

If we can find at least one coefficient which is not equal to 0 then it is LD.

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Answer 2

Linear Dependence can be proved if the Linear Combination of Vectors equate to zero without having all the Coefficients of the Vectors as zero.

$ c1V1 + c2V2 + …. + cnVn = 0 $, where at least one ci ! = 0

i.e. Let c1 != 0, then :  $V1 = – (\frac{c2}{c1}V2 + \frac{c3}{c1}V3 + ... + \frac{cn}{c1}Vn )$ => L.D

Since u, v, w can already be written as Option A or Option B to prove L.D, Option C violates the above rule.

Option D: LHS ! = RHS as 7u – 1v + 1w = 2v and not 0.

Therefore, Option A,B are correct

Answer 3

Linearly dependent : A set of vector is linear dependent if atleast one of the vector can be written as linear combination of other.

lets suppose we have {v1,v2,v3,v4,v5} as set of vector and in this one vector say v1 is the linear combination of other ,v1 = c2v2+c3v3+c4v4+c5v5 then we can say set is linearly independent .

or 

c1v1+c2v2+c3v3+……..+cnvn=0 and atleast one ci is non zero and it satisfy the equation ?? what does this mean ? It means we can write a vector of non zero coefficient as a linear combination of other .

these above two things are equal to each other .

Now come to options .

• A. ou+ov+1w=[0 0 0]  →  can you give me one ci which is non zero and satisfy the equation , yes i have which is the coeffiecient of vector w .
  0*[1 -2 7]+0*[-7 14 -49]+1*[0 0 0]=[0 0 0]   now this option is correct .
   
• B. 0u+0v=w →  it is also correct w is a linear combination of u and v . we can also write this equation like this 
  0u+0v+(-1)w=0 now here we have one non zero ci which is -1 and still satisfy the equation. this option is also correct 
   
• C. Trivial solution 
• D. 7u+(-1)v+1(w)=[0 0 0] → even this equation does not satisfy after placing value of u,v,w.

So , finally option A ,B are correct .

Comments

or we can say that in option D is vector [0 0 0] is not linear combination of matrix A ,so  it is not a lineraly dependent (linearly independent )

Answer 4

c1u+c2u+c3w=0 where c1,c2,c3 are scalars from this finding at least one ci≠0 such that the equation satisfies is the necessary and the sufficient condition for saying {v,u,w) is linearly dependent. Option A and B are sufficient To say that {v,u,w} is linearly dependent.option c is neither necessary nor sufficient as the solution set is trivial .

Option D: LHS ! = RHS as 7u – 1v + 1w = 2v and not 0.

Therefore, Option A,B are correct

Comments

you mean it is equal to "-2v", hence option D is not even a correct option to begin with.