For the system of linear equations
2 x+4 y+2 a z=b
x+2 y+3 z=4
2 x-5 y+2 z=8
which of the following is NOT correct?

Solution

Maths sir · 4 steps

Find where the determinant is zero, then check whether the equations agree or clash there.

1. Coefficient determinant. A unique solution exists exactly when the determinant of the coefficients is non-zero. Expand along the first row.

   Δ = |2, 4, 2a; 1, 2, 3; 2, -5, 2| = 2(19) - 4(-4) + 2a(-9) = 54 - 18a

2. Non-zero determinant. Δ = 0 only when a = 3. For a = 8 or a = 6, Δ ≠ 0, so the solution is unique whatever b is. Options (c) and (d) are correct.
3. When a equals 3. The first equation becomes 2x + 4y + 6z = b, whose left side is exactly twice the second equation's. If b = 8, it is just the second equation doubled, so two independent equations remain and there are infinitely many solutions. If b = 6, it says x + 2y + 3z = 3 while the second says x + 2y + 3z = 4, a contradiction, so there is no solution.
4. So. With a = 3, b = 6 the system has no solution, not infinitely many. The statement that is NOT correct is (b).

The idea to keep. When Δ = 0, do not stop there. Compare the dependent equations: the same right side (in the same ratio) gives infinitely many solutions, and a different one gives no solution.

Answer from the NTA final answer key. Solution by PrepWiser.