Question

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

x + y + z = 1, 2x + 3y + 2z = 2 and x + y + 2z = 4

Answer 1

Matrix form of the given system of equations is

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

This is of the form AX = B,

Where A = `[(1, 1, 1),(2, 3, 2),(1, 1, 2)]`, X = `[(x), (y), (z)]` and B = `[(1), (2), (4)]`

Applying R₂ → R₂ – 2R₁

`[(1, 1, 1),(0, 1, 0),(1, 1, 2)][(x), (y), (z)] = [(1), (0), (4)]`

Applying R₃ → R₃ – R₁

We get, `[(1, 1, 1),(0, 1, 0),(0, 0, 1)][(x), (y), (z)] = [(1), (0), (3)]`

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

∴ `[(x + y + z),(0 + y + 0),(0 + 0 + z)] = [(1), (0), (3)]`

∴ By equality of matrices, we get

x + y + z = 1  ...(i)

y = 0

z = 3

Substituting y = 0 and z = 3 in equation (i), we get

x + 0 + 3 = 1

∴ x = 1 – 3 = –2

∴ x = –2, y = 0 and z = 3 is the required solution.