Question

Express the following equations in matrix form and solve them by the method of reduction:

2x - y + z = 1, x + 2y + 3z = 8, 3x + y - 4z = 1.

Answer 1

The given equations can be written in the matrix form as:

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

By R₁ ↔ R₂, we get,

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

By R₂ - 2R₁ and R₃ - 3R₁, we get,

`[(1,2,3),(0,-5,-5),(0,-5,-13)] [("x"),("y"),("z")] = [(8),(-15),(-23)]`

By R₃ - R₂, we get,

`[(1,2,3),(0,-5,-5),(0,0,-8)] [("x"),("y"),("z")] = [(8),(-15),(-8)]`

∴ `[("x"+"2y" + 3"z"),(0 - "5y" - "5z"),(0 + 0 - "8z")] = [(8),(-15),(-8)]`

By equality of matrices,

x + 2y + 3z = 8     ...(1)

- 5y - 5z = - 15      ....(2)

- 8z = - 8              ....(3)

From (3), z = 1

Substituting z = 1 in (2), we get,

- 5y - 5 = - 15

∴ - 5y = - 10

∴ y = 2

Substituting y = 2, z = 1 in (1), we get,

x + 4 + 3 = 8

∴ x = 1

Hence, x = 1, y = 2, z = 1 is the required solution.