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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

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. :)

@rohith1001 Yes I too got the same option.

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)

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.

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.

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