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Question
Solve the following system of equations by the method of reduction:
x + y + z = 6, y + 3z = 11, x + z = 2y.
Solution
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
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