Question

The system of equation

3x + 4y + 5z = μ

x + 2y + 3z = 1

4x + 4y + 4z = δ

is inconsistent, then (δ, μ) can be

(1) (4, 6)

(2) (3, 4)

(3) (4, 3)

(4) (1, 0)

Answer 1

Answer is (3) (4, 3)

Now let P3 = 4x + 4y + 4z - δ = 0. If the system has solutions it will have infinite solution,

so P3 = αP1 + βP2

Hence 3α + β = 4 & 4α + 2β = 4 ⇒ α = 2 & β = - 2

So for infinite solution 2μ - 2 = δ ⇒ for 2μ ≠ δ + 2

system inconsistent