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Consider the system, each consisting of $m$ linear equations in $n$ variables.

  1. If $m < n$, then all such systems have a solution.
  2. If $m > n$, then none of these systems has a solution.
  3. If $m = n$, then there exists a system which has a solution.

Which one of the following is CORRECT?

  1. $I, II$ and $III$ are true.
  2. Only $II$ and $III$ are true.
  3. Only $III$ is true.
  4. None of them is true.

 

 

asked in Linear Algebra by Veteran (41.3k points)  
edited by | 1.2k views

2 Answers

+12 votes
Best answer
Correct answer => C)

why ?

I) This is false. Consider a system with m < n, which are incosistent like

a+b+c = 2

a+b+c = 3

Here m < n but no solution because of inconsistency !

II) m > n but no solution for none of system => What if this system of equations have 2 equations which are dependent ?

ex => a+b = 2

2a + 2b = 4

a-b = 0

Then a = 1, b= 1 is solutions . II) Is false.

III) this is true, M = 2, N = 2

a+b = 2

a-b = 0

Then m= 1, n= 1 Now there exists system which has solution . III) is correct. Answer is C !
answered by Veteran (41.3k points)  
selected by
but for the case of two parallel lines example y=x+5 and y=x+6 for these equations no solution so c should also be false.

You need to read the statement III) Clearly. What you are deducing is incorrect ! If m = n , the there exists a system which has a solution

* there exists *

Counter example is used for disproving, for all, not * there exists *

will the answer change for homogeneous equations?
–1 vote

Answer D:

I & II are false already but similarly for III also 

 for the case of two parallel lines example y=x+5 and y=x+6 for these equations no solution so c should also be false.

Hence D is correct ans.

answered by (163 points)  

You need to read the statement III) Clearly. What you are deducing is incorrect ! If m = n , the there exists a system which has a solution

* there exists *

Counter example is used for disproving, for all, not * there exists *



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