Consider the following statements: (A) The system 3x - 2y = 4,\ 6x -…

The correct answer is Option (1) → (A), (B) and (C) only

For AX = B: if \(|A| \ne 0\) there is a unique solution; if \(|A| = 0\) the system has either infinitely many solutions or none, depending on whether the equations agree.

The coefficient determinants of the four systems:

    \( \text{(A)}:\ 3 \times (-4) - (-2) \times 6 = 0 \)
    \( \text{(B)}:\ 1 \times (-2) - (-1) \times 2 = 0 \)
    \( \text{(C)}:\ 3 \times (-2) - 1 \times 1 = -7 \)
    \( \text{(D)}:\ 1 \times 5 - 2 \times 3 = -1 \)

So (A) and (B) have determinant 0, while (C) and (D) have non-zero determinants and hence exactly one solution each.

Checking the statements

(A) The system \(3x - 2y = 4,\ 6x - 4y = 8\) has infinitely many solutions. — Correct. The second equation is twice the first, so both are the same line.

(B) The system \(x - y = 2,\ 2x - 2y = 3\) has no solution. — Correct. The second equation says \(x - y = \frac{3}{2}\), which contradicts \(x - y = 2\).

(C) The system \(3x + y = 4,\ x - 2y = -1\) has the unique solution \(x = 1,\ y = 1\). — Correct. Determinant −7 ≠ 0, and \(3 + 1 = 4\), \(1 - 2 = -1\).

(D) The system \(x + 2y = 0,\ 3x + 5y = 0\) has infinitely many solutions. — Incorrect. Determinant −1 ≠ 0, so x = 0, y = 0 is the only solution.