GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 7

Answer 1

Let’s see option by option …

• A. $\left\{\left[\begin{array}{I} 1\\2 \end{array}\right], \left[\begin{array}{I} 2\\1 \end{array}\right]\right\}$ 

These two vectors are not multiple of each other so, they are Linearly Independent . 

• A. $\left\{\left[\begin{array}{I} 1\\-1 \end{array}\right]\right\}$

These is a single vector which is not any multiple of other vector in the set and also there are not any zero vector which can lead the set to LD. Hence these single set of vector also linearly Independent.

100.  $\left\{\left[\begin{array}{I} 1\\2\\1 \end{array}\right], \left[\begin{array}{I} 2\\3\\4 \end{array}\right], \left[\begin{array}{I} 1\\-1\\2 \end{array}\right], \left[\begin{array}{I} 0\\1\\0 \end{array}\right]\right\}$ 

 These set is in $R^{3}$, So, here we can see that we have more than 3 vectors in the set which is leading to Linearly Dependent condition,

Also, we can see that  $0 \times \left[\begin{array}{I} 1\\2\\1 \end{array}\right] + 2 \times \left[\begin{array}{I} 1\\-1\\2 \end{array}\right] + 5 \times\left[\begin{array}{I} 0\\1\\0 \end{array}\right] = \left[\begin{array}{I} 2\\3\\4 \end{array}\right]$ 

So we can represent also 1 vector as a linear combination of other 3 vectors. So it is clear that these set is LD.

500. $\left\{\left[\begin{array}{I} 1\\1\\1 \end{array}\right], \left[\begin{array}{I} 1\\0\\1 \end{array}\right], \left[\begin{array}{I} 0\\0\\0 \end{array}\right]\right\}$ 

The set is already consisting a zero vector that means this set is Linearly Dependent. 

Question is asking for LI set, 

So, ANS is: A;B

Comments

for option c- even if there are 3 independent vectors the 4^th vector will always be combination of other 3 vectors because it is in R3

Answer 2

OPTION : A , B IS CORRECT ONE 

Answer 3

A.v1=[1  2],v2=[2  1] any of the vector cannot be written as linear combination of the others means if c1v1+c2v2=0 where c1 ,c2 are scalars then we are not able to find any non trivial solution. so {[ 1 2],[2 1]} this set is linearly independent .

B.{[1 -1]} this set consists of only one vector so here we also are not able to find any non trivial solution and for this the set also linearly independent.

C.{[1 2 1],[2 3 4],[1 -1 2],[0 1 0]}only seeing this we can say that this set is linearly dependent as in R3 there should be linearly independent set consisting of at most 3 vectors means among this three vectors any one of them cannot be written as the linear combination of others.and the given set consists of 4 vectors then any of this four vector always can be written as linear combination of other three vector.

D.{[1 1 1],[1 0 1],[0 0 0]} in this set zero vector is present here and we know that if zero vector is present in a set of vector then the set is always linearly dependent.[0 0 0]=0[1 1 1]+0[1 0 1].

So option A,B are correct.