GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 4

Answer 1

If we take a=0 then $v_1$ will become zero vector which in result will make {$v_1$, $v_2$, $v_3$} linearly dependent as we can then represent, v1 = 0.v2 + 0.v3 . Hence a$\neq$0.

If we take b=1 then $v_2$ and $v_3$ will become same vector. Hence b$\neq$1 for {$v_1$, $v_2$, $v_3$} to be linearly independent.

$\therefore$ b is correct.

Answer 2

For LI, we should NOT have linear combination/multiple/zero vectors. Checking this with options, b is the potential answer

Answer 3

v1=[a 0 0], v2=[0 b 1], v3=[0 1 1]

• A. a=0  and b=1 , v1=[0 0 0] , v2=[0 1 1], v3 =[0 1 1] and [0 0 0]=0[0 1 1]+0[0 1 1] also [0 1 1] =1[0 1 1]+k[0 0 0] where kεz.So {v1,v2,v3) linearly dependent.so A is not the correct answer.
• B. a≠0 and b≠1 let a=k ,b=m k,mεz v1=[k 0 0],v2=[0 m 1],v3=[0 1 1] any of them cannot be represented as linear combination of others even we take m=0 also.So linearly independent . So B is the correct answer .
• C. a=0,b≠1 let b=m mεz ,v1=[0 0 0] v2=[0 m 1] v3=[0 1 1] and [0 0 0]=0[0 m 1]+0[0 1 1] so { v1,v2,v3} linearly dependent,so option c is not the correct answer .
• D. a≠0 b=1 let a=k kεz v1=[k 0 0] v2=[0 1 1] v3=[0 1 1] and [0 1 1]=1[0 1 1]+0[k 0 0] so {v1,v2,v3} linearly dependent .So option d is not the correct answer.