Discussion
\[ \begin{aligned} x+2y+z &= 4 \\ 2x+y+2z &= 5 \\ x-y+z &= 1 \end{aligned} \]
The system of algebraic equations given above has
Answer: C
The system has an infinite number of solutions.
Subtract the third equation from the first:
\[ (x+2y+z)-(x-y+z) = 4-1 \Rightarrow 3y = 3,\ y = 1 \]
With y = 1, the first equation gives x + z = 2 and the second gives 2x + 2z = 4, which is the same condition. So y = 1 and x + z = 2, and every pair x, z summing to 2 works: the coefficient matrix has rank 2, equal to the rank of the augmented matrix, with three unknowns.
(1, 1, 1) and (2, 1, 0) are both solutions, but they are only two of infinitely many, so 'unique' and 'only the two solutions' are wrong. A linear system cannot have exactly two solutions, and the system is consistent, so 'no feasible solution' is wrong too.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development