Question

Solve the following equations by method of inversion : x + y – z = 2, x – 2y + z = 3 and 2x – y – 3z = – 1

Answer 1

Matrix form of the given system of equations is

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

This is of the form AX = B, where

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

To determine X, we have to find A^–1.

|A|= `|(1, 1, -1),(1, -2, 1),(2, -1, -3)|`

= 1(6 + 1) – 1(–3 – 2) –1(–1 + 4)
= 1(7) –1(–5)–1(3)
= 7 + 5 – 3
= 9 ≠ 0
∴ A^–1 exists.
Consider AA^–1 = 

∴ `[(1, 1, -1),(1, -2, 1),(2, -1, -3)]"A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

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

`[(1, 1, -1),(0, -3, 2),(0, -3, -1)] "A"^-1 = [(1, 0, 0),(-1, 1, 0),(-2, 0, 1)]`

Applying R₂ → `((-1)/3)` R₂, we get

`[(1, 1, -1),(0, 1, -2/3),(0, -3, -1)] "A"^-1 = [(1, 0, 0),(1/3, (-1)/3, 0),(-2, 0, 1)]`

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

`[(1, 0, -1/3),(0, 1, (-2)/3),(0, 0, -3)] "A"^-1 = [(2/3, 1/3, 0),(1/3, -1/3, 0),(-1, -1, 1)]`

Applying R₃ → `(-1/3)` R₃, we get

`[(1, 0, -1/3),(0, 1, -2/3),(0, 0, 1)] "A"^-1 = [(2/3, 1/3, 0),(1/3, -1/3, 0),(1/3, 1/3, -1/3)]`

Applying R₁ → R₁ + `(-1/3)` R₃ and R₂ → R₂ + `(2/3)` R₃, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)] "A"^-1 = [(7/9, 4/9, -1/9),(5/9, -1/9, -2/9),(1/3, 1/3, -1/3)]`

∴ A^–1 = `(1)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -3)]`

Pre-multiplying AX = B by A^–1, we get
A^–1(AX) = A^–1B
∴ (A^–1A) X = A^–1B
∴ IX = A^–1B
∴ X = A^–1B

∴ X = `(1)/(9)[(7, 4, -1),(5, -1, -2),(3, 3, -3)][(2),(3),(-1)]`

∴ `[(x),(y),(z)] = (1)/(9)[(14 + 12 + 1),(10 - 3 + 2),(6 + 9 + 3)]`

= `(1)/(9)[(27),(9),(18)]`

= `[(3),(1),(2)]`

∴ By equality of martices, we get
x = 3, y = 1 and z = 2.