Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| System of equations | Solution | ||
| (A) | \(2x + 3y = 5,\ 4x + 6y = 10\) | (I) | Unique solution \(x = 1,\ y = 1\) |
| (B) | \(2x + 3y = 5,\ 4x + 6y = 9\) | (II) | No solution |
| (C) | \(2x + 3y = 5,\ 3x - y = 2\) | (III) | Infinitely many solutions |
| (D) | \(3x - 2y = 4,\ x + y = 3\) | (IV) | Unique solution \(x = 2,\ y = 1\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
For AX = B: if \(|A| \ne 0\) there is a unique solution. If \(|A| = 0\), the system has either no solution or infinitely many, depending on whether the equations agree.
(A) \(2x + 3y = 5,\ 4x + 6y = 10\)
This matches (III) Infinitely many solutions.
\(|A| = 2 \cdot 6 - 3 \cdot 4 = 0\)
The second equation is exactly 2 times the first, so the two equations are the same line.
(B) \(2x + 3y = 5,\ 4x + 6y = 9\)
This matches (II) No solution.
Again \(|A| = 0\). Doubling the first equation gives \(4x + 6y = 10\), which contradicts \(4x + 6y = 9\), so the system is inconsistent.
(C) \(2x + 3y = 5,\ 3x - y = 2\)
This matches (I) Unique solution \(x = 1,\ y = 1\).
\(|A| = 2(-1) - 3 \cdot 3 = -11 \ne 0\)
From the second equation \(y = 3x - 2\):
\(2x + 9x - 6 = 5\)
\(x = 1,\ y = 1\)
(D) \(3x - 2y = 4,\ x + y = 3\)
This matches (IV) Unique solution \(x = 2,\ y = 1\).
\(|A| = 3 \cdot 1 - (-2) \cdot 1 = 5 \ne 0\)
From the second equation \(x = 3 - y\):
\(9 - 3y - 2y = 4\)
\(y = 1,\ x = 2\)