What are w1w2w3 w1w2w31 5w175w305w15w225w305w2w30Solutionw1

What are w1,w2,w3?

w_1+w_2+w_3=1 -5w_1+.75w_3=0.5w_1-.5w_2+.25w_3=0.5w_2-w_3=0

Solution

w1 + w₂ + w₃ = 1..(1), -0.5w₁ + 0.75w₃ = 0 or, on multiplying both the sides by 4 , 2w₁ + 3w₃= 0..(2), 0.5w₁ - 0.5w₂ + 0.25w₃ = 0 or, on multiplying both the sides by 4, 2w₁ - 2w₂ + w₃ = 0 ...(3) and 0.5w₂ - w₃ = 0 or, on multiplying both the sides by 2 , w₂ - 2w₃ = 0 ...(4)

From the 2nd equation, we get w₃ = -2/3w₁ . On substituting w₃ = -2/3w₁ in the 1st, 3rd and 4th equations, we get w₁ + w₂ - 2/3 w₁ = 1 or, 1/3 w₁ + w₂= 1 or, w₁ + 3w₂ = 3 ....(5) Also,  on substituting w₃ = -2/3w₁   in the 3rd equation, we get 2w₁ - 2w₂ -2/3w₁ = 0 or, 4/3w₁ -2w₂ = 0 or, 4w₁ - 6w₂ = 0 or, 2w₁ -3w₂ = 0..(6) Further, on substituting w₃ = -2/3w₁    in the 4th equation, we get w₂ + 4/3 w₁ = 0 or, 4w₁ + 3w₂ = 0 ...(7) . From the 6th and 7th equations respectively , we get w₂= 2/3w₁ and w₂ = -4/3w₁ so that 2/3w₁= -4/4w_1. This is possible only if w₁ = 0. Then w₂ =0 and w₃ = 0. However, then from the 5th equation , we get 0 + 0 = 5 which is incorrect. Therefore, the given linear system, which is over prescribed is inconsistent and has no solution.