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+3 votes
391 views

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

 

asked in Linear Algebra by Veteran (59.8k points)   | 391 views

2 Answers

+11 votes
Best answer
There are no solutions.

If we multiply 1st equation by 4, we get

4x + 8y = 20

But 2nd equation says

4x + 8y = 12

Clearly, there can not be any pair of (x,y), which satisfies both equations.
answered by Veteran (10.7k points)  
selected by
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 ?
Rank of A is not 3. Recheck your method.
+2 votes

The rank of augemented matrix(|AB|=3) and coefficient matrix|A|=2 is not same! hence there is no solution.


 

answered by Veteran (14.7k points)  


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