GATE CSE 2004 | Question: 71

Question

How many solutions does the following system of linear equations have?

• $-x + 5y = -1$
• $x - y = 2$
• $x + 3y = 3$

• A. infinitely many
• B. two distinct solutions
• C. unique
• D. none

Comments

What will be the ans in case of rank is Greater than the number of variables in the matrix?

-x +5y = -1

3x -7y = 2

X + 8 y = 3  

In this case rank (3) > number of variables (2)  .

Now what will be the ans?

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that will create [ 0 0 | nonzero ] situation and hence no solution.

Answer 1

answer = C

rank = r(A) = r(A|B) = 2

rank = total number of variables
Hence, unique solution

Comments

Thanks for reply bro.

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So b is out of question right? Coz here we have linear functions. If meeting exists only at one point.

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NICE

Answer 2

C unique solution..
3 equation , 2 variable.
solve any two equation and check 3rd equation by putting values in 3rd equation.
x = 9/4 , y = 1/4

Comments

@Digvijay Pandey sir, for given this type of equation i.e non-homogeneous eqn, how can I distinguish b/w infinitely many & no solution.

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When rank(A) != rank(AB) ==> No Solution since (AX=B ) is inconsistent.

When [ rank(A) = rank(AB) ] < n (no of unknown variables) ==> Infinitely many Solution

Answer 3

DIFFERENT APPROACH

all the equations are linearly independent. i.e non of the equations can be obtained by multiplying one of equation with a number (this means that no two vectors overlap each other leading to a rank of 3). therefore there will be a unique solution. 

Correct me if wrong :)

Comments

$2X2$  Minor with Determinant non - zero.

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how all equations are linearly independent.

multiplying -2 with eqn no. 2 & then add with eqn. no. 3 we'll get eqn no. 1

-2*(eqn2)+(eqn3) = (eqn1)

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Yes rank is coming 2 $\implies$ we have only 2 independent equations not 3.

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All three line are intersect at only point X=9/4 and y=1/4 so unique solution exist

Answer 4

rank[A] = 2 and rank[A|B] = 2

it is unique solution .

Comments

In this question many people get trapped.

Method-1:

For given equations Augmented matrix[A|b] is  :

-1   1   5

5   -1   3

-1   2   3

R2→ R2+R1

R3→R3+R1 

Then

R3→R3-2R2

After converting into echelon form we obtain:

-1   0   0

5   4   0

-1   1    0

After seeing [00…0] we think that infinite solution is the correct answer 

But the catch here is for infinite solution there must me atleast one free column. But in this case there is no free column so the answer here is unique solution

Method-2:

The matrix is 3X2 matrix

Find rank[A] and rank[A|b] 

rank[A] = 2 

rank[A|b]=2 

So number of free columns = 0

So there will be unique solution

therefore the option-C is correct