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Given matrix $A\left(m \times n\right)$ such that $A x=b.$ Consider below statements :

  • $\text{S1}:$ If $b$ is Linearly Dependent on cols of matrix, then it need not always have a unique solution.
  • $\text{S2}:$ If columns of matrix are Linearly Dependent, then above system of equation can have a unique solution.

  1. $\text{S1}$ and $\text{S2}$ are true
  2. $\mathrm{S} 1$ and $\mathrm{S} 2$ are false
  3. $\text{S1}$ is true and $\text{S2}$ is false
  4. $\mathrm{S} 2$ is true and $\mathrm{S} 1$ is false
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$\mathrm{S} 1$ : In this case, there is always a solution. It will always have a unique solution iff cols of matrix are LI. Hence, $\mathrm{S} 1$ is true.

$\text{S2} :$ In this case, if the the column vectors cover $b$ in their subspace, then their will be infinitely many solutions(as multiple linear combinations can satisfy $b$ since, all the col vectors are LD), else, no solution case will happen. Hence, $\text{S2}$ is false.

It is based on an imp statement in lec :

If a vector can be represented as a lin comb of few vectors and further if those few vectors are LI then there is just one way to do it, otherwise there are many ways.

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Here the answer is option :C. 

Case 1:  for S1 there it is said that b is Linear combination of column of A so there will be solution

now weather the solution is unique or whether it is infinite it depends on  Columns of A are linear independent or linearly dependent 

IF the columns are L.I then there is a unique solution and if they are L.D there is infinite solution

so S1 is true 

 

​​​​​​CASE 2 : for S2 it is said that b is linear dependent on Columns of A and there is a unique solution whic is false because if b is L.D then there is always infinite solution 

 

 

 

Answer:

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