Question

Solve the following system of equations by the method of reduction:

x + y + z = 6, y + 3z = 11, x + z = 2y.

Answer 1

Given:

x + y + z = 6

y + 3z = 11

x + z = 2y

`[(1, 1, 1), (0, 1, 3), (1, -2, 1)] [(x), (y), (z)] = [(6), (11), (0)]`

using R₃ → R₃ – R₁

we get `[(1, 1, 1), (0, 1, 3),(0, -3, 0)] [(x), (y), (z)] = [(6), (11), (-6)]`

using R₃ → R₃ + 3R₂

we get `[(1, 1, 1), (0, 1, 3),(0, 0, 9)] [(x), (y), (z)] = [(6), (11), (27)]`

∴ `[(x + y + z),(0 + y + 3z),(0 + 0 + 9z)] = [(6), (11), (27)]`

Equating, we get

x + y + z = 6   ...(1)

y + 3z = 11  ...(2)

9z = 27     ...(3)

z = `27/9`    ...[from eq. (3)]

z = 3

put this value of z in equation (2), we get

y + 3(3) = 11

y + 9 = 11

y = 11 − 9

y = 2

put the values of y and z in equation (1), we get

x + y + z = 6

x + 2 + 3 = 6

x + 5 = 6

x = 6 − 5

x = 1