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GO Classes CS/DA 2025 | Weekly Quiz 3 | Fundamental Course and Linear Algebra | Question: 8

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$$a\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4 \\ 5\end{array}\right]+b\left[\begin{array}{c}-1 \\ 2 \\ -3 \\ 4 \\ -5\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0 \\ 0 \\ 0\end{array}\right]$$
How many number of pairs $(a, b)$ are there, that satisfy the above equation?

  1. $0$
  2. $1$
  3. Infinite
  4. $2$
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Only one possible condition is to (0,0) as coefficient to get the Zero Vector on the other hand of the equation, there is no other way to get the zero vector except (0,0) as coefficient, so the answer is 1 possible choice

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v1=[1 2 3 4 5] , v2=[-1 2 -3 4 -5] and {v1,v2} is linearly independent as any of the two vector cannot be written as the linear combination of the other vector.so if av1 +bv2=0 then we are not able to find any non trivial solution  hence only trivial solution (0,0) exists for (a,b).So option B is correct.
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A good approach to this problem would be to check if the given vectors, say  v1= [1 2 3 4 5]T and  v2= [-1 2 -3 4 -5]T are linearly Independent or Linearly dependent.

In case they are LI, then av1+bv2=0 will have only trivial solution, i.e., only (0,0) would satisfy.

If they are LD, then they can have infinitely many solutions. 

As there are only two vectors it is very easy to check there linear dependence. They are visibly LI as one is not a multiple of the other in any way. 

Hence, Answer to this question would be OPTION B

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