in Linear Algebra
5,301 views
20 votes
20 votes

Consider the following set of equations

  • $x+2y=5$
  • $4x+8y=12$
  • $3x+6y+3z=15$

This set

  1. has unique solution
  2. has no solution
  3. has finite number of solutions
  4. has infinite number of solutions
in Linear Algebra
5.3k views

4 Comments

With this print mistake, if we solve the question we get Option A) has unique solution. :P

Please let me know if I'm wrong. :)
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@rohith1001 Yes I too got the same option.

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Constant matrix also linearly dependent with same factor as coefficient matrixYes (Then Inconsistent System of Equation- No Solution) If →No (Then consistent System of Equation- Infinite Solutions)

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2 Answers

38 votes
38 votes
Best answer

There are no solutions.

If we multiply $1^{\text{st}}$ equation by $4,$ we get

  • $4x + 8y = 20$

But $2^{\text{nd}}$ equation says

  • $4x + 8y = 12$

Clearly, there can not be any pair of $(x,y),$ which satisfies both equations.

Correct Answer: B.

edited by

2 Comments

I am getting rank of augmented matrix as 3 as well as rank of A =3 so according to this condition there must be a unique solution , although it shouldn't exist but then how can it contradict ?
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Rank of A is not 3. Recheck your method.
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12 votes
12 votes

First transformed the matrix into echelon form and rest can seen from the pic.

Answer:

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