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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.
in Linear Algebra by Boss (41.9k points)
edited by | 4.7k views
+3
$I.$ An underdetermined linear system has either no solution or infinitely many solutions.

 

The homogeneous (with all constant terms equal to zero) underdetermined linear system always has non-trivial solutions (in addition to the trivial solution where all the unknowns are zero).

$II.$ An overdetermined system is almost always inconsistent (it has no solution) when constructed with random coefficients. However, an overdetermined system will have solutions in some cases, for example if some equation occurs several times in the system, or if some equations are linear combination of the others.

 

The homogeneous case is always consistent (because there is a trivial, all-zero solution). There are two cases, depending on the number of linearly dependent equations: either there is just the trivial solution, or there is the trivial solution plus an infinite set of other solutions.

$III.$ If m=n, then there exists a system which has a solution.                                                              

 

$\therefore$ Ans: C

4 Answers

+45 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 $C$ !
by Boss (41.9k points)
edited by
+2
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.
+42

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 *

+2
will the answer change for homogeneous equations?
+2

thank you .... it helped .. ur 'there exists'

+4
Statement 3) is " there exists a system which has a solution "  not  " there exists a solution "
0
ans should be option d ,    there may be cases where m=n  but those m equations be linearly dependent and results in no solutions so case iii will also be wrong
+1
@sushmita in case of homogeneous equeations when m>n the max rank will be n and this boild down to r<=n hence solution exists in case of m<n max rank is m and r<n infinite solutions exist and when m=n r<=n hence solution exists in all the three cases
0
for linear equations

a+b+c=2

a+b+c=3

a+b+c=5

system is inconsistent

so how option C is correct
0

@Akash Kanase 

If m>n, will ALL such systems always have a solution?

If m<n will NO such systems have a solution?

0

@Shubham101    the reason is again - there exists a system.  ;)

@Sambhrant Maurya   both are false.

0

Correct me if I'm wrong.

1. not all such but few systems have a solution because suppose in a 3d space(3 unknown) we're given 2 line eqn.It's underdetremined so we can't solve this type of equation.

2.It's overdetermined sysytem.suppose 3 line eqn are given in a 2d space.Now this 3 line may or may not be intersect at a same point.so none of this system has a solution is false.but few system has a solution is true.

3.It's perfectly determined.2 eqn. in 2d space but same concept as point 3, can be parallel or intersect at a point.So there exists a system which has a solution is true but for all it will be false statement.

0
can anyone explain with respect to the ranks why first two options are false for homogeneous system of equation they are right
0

@Kaluti I think for a system of linear homogenous equation all the 3 options are correct.

 

+4 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..

by Active (1.5k points)
+2

i  think homogeneous systems always have at least one solution, namely, the case where all unknowns are equal to zero.  i.e.trivial solution to the homogeneous system, so whether m> n or m<n or m=n homogenous will have always a solution and can never be inconsistent.

0 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
ago by Loyal (6.9k points)
–2 votes

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.

by (165 points)
+5

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 *

Answer:

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