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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

edited
A|B matrix will be given as -

$\begin{bmatrix} 1 &2 &0 &|5 \\ 4& 8 &0 & |12 \\ 3& 6 & 3 & |15 \end{bmatrix}$

After the row-operations -   R3->R3-3R1  and

R2->4R1-R2

We obtain the matrix -

$\begin{bmatrix} 1 &2 &0 &|5 \\ 0 & 0& 0& |8\\ 0& 0 & 3 & |0 \end{bmatrix}$

Now since, $Rank(A)\neq Rank(A|B)$

Hence, this is an inconsistent solution, that is, no solution.
@shweta how did you get 0000 oin the last row?
After performing row operation R3 -> R3 - 3R1
how 3 -0=0?
My mistake it should be 3. I edited it.

Formatting Error in PDF

There is an error in the formatting of this question in the soft copy of the GO PDF (screenshot attached for reference).  ..

The error is still there in GO hardcopy- 2019 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.