Question

Solve the following equations by method of reduction:

x + 2y + z = 3, 3x – y + 2z = 1 and 2x – 3y + 3z = 2

Answer 1

The matrix form of the given system of equations is

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

This is of the form AX = B, where

A = `[(1, 2,1),(3, -1, 2),(2, -3, 3)], "X" =  [(x),(y),(z)] "and B"= [(3),(1),(2)]`

Applying R₂ → R₂ – 3R₁ and R₃ → R₃ – 2R₁, we get

`[(1, 2, 1),(0, -7, -1),(0, -7, 1)] [(x),(y),(z)] = [(3),(-8),(-4)]`

Applying R₃ → R₃ – R₂, we get

`[(1, 2, 1),(0, -7, -1),(0, 0, 2)] [(x),(y),(z)] = [(3),(-8),(4)]`

Hence, the original matrix A is reduced to an upper triangular matrix.

∴ `[(x + 2y + z),(0 - 7y - z),(0 + 0 + 2z)] = [(3),(-8),(4)]`

By equality of matrices, we get
x + 2y + z = 3         ...(i)
– 7y –z = – 8
i.e., 7y + z = 8        ...(ii)
2z = 4
∴ z = 2
Substituting z = 2 in equation (ii), we get
7y + 2 = 8
∴ 7y = 6
∴ y = `(6)/(7)`
Substituting y = `6/(7)` and z = 2 in equation (i), we get

`x + 2(6/7) + 2` = 3

∴ x = `3 - 2 - 12/7 = (-5)/(7)`

∴ x = `-5/(7), y = (6)/(7)` and z = 2 is the required solution.