GATE CSE 1998 | Question: 1.2

Question

Consider the following set of equations

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

This set

• A. has unique solution
• B. has no solution
• C. has finite number of solutions
• D. has infinite number of solutions

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. :)

---

@rohith1001 Yes I too got the same option.

---

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

Answer 1

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.

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 ?

---

Rank of A is not 3. Recheck your method.

Answer 2

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