GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 5

Answer 1

(A) True. $\left\{v_1, v_2, v_3\right\}$ linearly dependent means there exists weights $a_1, a_2$, and $a_3$ not all zero such that
$$
a_1 v_1+a_2 v_2+a_3 v_3=0
$$
This means
$$
a_1 v_1+a_2 v_2+a_3 v_3+a_4 v_4=0
$$
for $a_4=0$ and the same $a_1, a_2, a_3$ as above. Since not all of $a_1, a_2$, and $a_3$ are zero, that means not all of of $a_1$, $a_2, a_3, a_4$ are zero. And that means $\left\{v_1, v_2, v_3, v_4\right\}$ are linearly dependent.

(B) False. Here's a simple counterexample: choose $v_1=e_1$ and $v_2=v_3=v_4=e_2$. Then $v_1$ is not a linear combination of $\left\{v_1, v_2, v_3\right\}$, yet $\left\{v_1, v_2, v_3, v_4\right\}$ are linearly dependent (because, for example, $0 v_1+0 v_2+v_3-v_4=0).$ 

(C) True. To show that $\left\{v_1, v_2, v_3\right\}$ is linearly independent, we must show that if $a_1 v_1+a_2 v_2+a_3 v_3=0$, then $a_1=$ $a_2=a_3=0 \;\text{(“only the trivial solution...")}.$

So, assume $a_1 v_1+a_2 v_2+a_3 v_3=0$. Then $a_1 v_1+a_2 v_2+a_3 v_3+0 v_4=0$ must also be true (we simply added 0 to the left-hand side). But because $\left\{v_1, v_2, v_3, v_4\right\}$ are linearly independent, that means all the weights in this expression must be zero.

In particular, $a_1=a_2=a_3=0$. Which is what we wanted.

Comments

Haha...Nice Paradox.
We have changed option D.

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Thank you, sir.. 😀🙏

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Sir @Sachin Mittal 1, the changed option is also incorrect .

Earlier Option-D was : All Statements are true ( And it was wrong )

So the Complement of it, 

i.e. (All Statements are true)$^{c}$ =  (At least one Statement is False) Should be the Correct option…  

Answer 2

option b is the right one . 

Answer 3

A.Let’s take  v3=c1v1+c2v2 then v3=c1v1+c2v2+0v4 so {v1,v2,v3,v4} is linearly dependent .So Option A is True.

B.May be v2 can be written as linear combination of {v1,v3,v4} So we cannot assure {v1,v2,v3,v4} is linearly independent .So option B is False.

C.{v1,v2,v3,v4}  linearly independent means any of the vector cannot be written as linear combination of others then obviously {v1,v2,v3}  linearly independent.So option C is True .

D. lets take the situation that {v1,v2,v3,v4,v5} linearly independent and we know that in R5 this set is minimum and also enough  to fill the space of R5.so If we add a vector v6 in the set then v6 can be represented as linear combination of v1,v2,v3,v4 and v5.So {v1,v2,v3,v4,v5,v6} linearly dependent.So Option D is true .