For which of the following ordered pairs (μ, δ), the system of linear equations
x + 2y + 3z = 1
3x + 4y + 5z = μ
4x + 4y + 4z = δ
is inconsistent ?

Solution

Maths sir · 3 steps

Once D = 0, find how row 3 is built from rows 1 and 2, then check whether the right sides follow.

1. Check the determinant. The coefficient determinant is 1(16-20) - 2(12-20) + 3(12-16) = -4 + 16 - 12 = 0. So the system has either no solution or infinitely many, depending on the right-hand sides.
2. Find the row relation. Write (4, 4, 4) = a(1, 2, 3) + b(3, 4, 5). The first two entries give a + 3b = 4 and 2a + 4b = 4, so b = 2 and a = -2. The third entry checks: -6 + 10 = 4.

   Eq₃ = 2 Eq₂ - 2 Eq₁ ⇒ consistent only if δ = 2μ - 2

   Option	2μ - 2	δ	System
   (a) (1, 0)	0	0	consistent
   (b) (4, 3)	6	3	inconsistent
   (c) (4, 6)	6	6	consistent
   (d) (3, 4)	4	4	consistent

3. So. Only (μ, δ) = (4, 3) breaks δ = 2μ - 2, because the left sides demand 2(4) - 2(1) = 6 but the third equation says 3. The answer is (b).

The idea to keep. When D = 0, find the combination of equations that makes the left sides match. The system is consistent only if the same combination holds for the right sides too.

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