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Consider the systems, 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. $\text{I, II}$ and $\text{III}$ are true.
  2. Only $\text{II}$ and $\text{III}$ are true.
  3. Only $\text{III}$ is true.
  4. None of them is true.
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Best answer
64 votes
64 votes
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 $C$ !
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7 votes
7 votes

I think here we required rank(A | B) and rank(A) to conclude clearly . Becoz;

FOR HOMOGENEOUS SYSTEM :-

1) INCONSISTENT ;- Not possible 

2) Unique solution :- Rank == No of variables

3) Many Solution :- Rank < No of variables.

FOR NON - HOMOGENEOUS SYSTEM :-

1) INCONSISTENT ;- when Rank(A|B) !=Rank (A) 

2) Unique solution :- when Rank(A|B) =Rank (A) and Rank == No of variables

3) Many Solution :- when Rank(A|B) =Rank (A) and Rank < No of variables.

So It has m linear eqn so maxm rank can be = "m"

and no of variables are = "n"

and to conclude anything Further we need to assume Rank (A|B)=rank(A).

After assuming this only Case (III) can be true.

Becoz m can not be greater than n (Becoz if so , then we can not conclude anything) , hence (II) is false.

When m<n , then all such system has Many solution, Hence (I) is false..

4 votes
4 votes

Note: This isn't a direct answer to the question. I've covered all the possible cases I could. Don't read this if you want a straight answer to the given question.


The number of equations is $m$.

Hence, there can be a $m*m$ matrix.

Hence the maximum rank can be $m$.

Hence, $m$ is the upper bound of $r$.
We conclude that $r \leq m$ ALWAYS.

Statement I

For homogeneous system of equations; if $m < n$, then $r < n$. Because $m$ is the upper bound of $r$. We'll have infinite solutions.

For non-homogeneous system of equations; maybe $r(A) \neq r(A|B)$. We'll have no solutions in this case.

So, all such systems have a solution? No. This statement is false.


Statement II

$m > n$. The only thing we know is $r$ lies below $m$. It can be between $m$ and $n$. Or it can be lower than $n$.

ie

  1. $m > r > n$ is possible.
  2. $m > n > r$ is possible.

For case 2, $r<n$. It's good.

For case 1, $r > n$. Which is inconsistent and we'll have no solutions in case of non-homogeneous equations.

We always want $r \leq n$ be it any system of equations.

 

So, this statement is false, because we have a favourable case.


Statement III

When $m = n$

$r$ could be $= m$, and hence, $r = n$.

$r$ could be $< m$, and hence $r < n$.

So, yeah, there exists favourable cases. This statement is correct.

 

Suppose $r$ is such that $r(A) \neq r(A|B)$. Then, we'll have no solution. But we need just one favourable case to make this statement True, and we did it.


Option C
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0 votes
We can convert the system of linear equations in the augmented matrix form to analyse.

For option 1 : If there is only one linearly independent vector and rest are multiples of this and the vector b is also linearly independent with the  vector in the A matrix then there cant be any solution in other words rank of the augmented matrix is 2 and rank of A matrix is one, there would be no solution thus option 1 is false.

For option 2 We can always construct a system  where the b (Ax=b) vector is a multiple of one of the vectors of the mXn A matrix and then obtain a  solution in other words we can always take n+1 (m*1) linearly dependent vectors then put those n vectors in the augmented matrix and we can always get a solution thus option 2 is false.

For option 3 For m=n which gives a square matrix A we can always have it filled with m linearly independent vectors and any vector in R3 will always be a linear combination of these m linearly independent vectors thus there exists a system which will have solution and that system is the one with m linearly independent vectors.

Hence the answer would be C.
Answer:

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