Question

Solve the following equations by method of inversion :

4x – 3y – 2 = 0, 3x – 4y + 6 = 0

Answer 1

Matrix form of the given system of equations is

`[(4, -3),(3, -4)][(x),(y)] = [(2),(-6)]`

This of the form AX = B, where

A = `[(4, -3),(3, -4)], "X" = [(x),(y)] "and B" = [(2),(-6)]`

To determine X, we have to find A^–1.

|A| = `|(4, -3),(3, -4)|`

= – 16 + 9

= – 7 ≠ 0

∴ A^–1 exists

Consider AA^–1 = I

`[(4, -3),(3, -4)] "A"^-1 = [(1, 0),(0, 1)]`

Applying R₁ → R₁ – R₂, we get

`[(1, 1),(3, -4)] "A"^-1 = [(1, -1),(0, 1)]`

Applying R₂ → R₂ – 3R₁, we get

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

Applying R₂ → `(-1/7)`R₂, we get

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

Applying R₁ → R₁ – R₂, we get

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

∴ A^–1 = `(1)/(7)[(4, -3),(3, -4)]`

Pre-multiplyig AX =  by A^–1, we get

A^–1(AX) = A^–1B

∴ (A^–1A) X = A^–1B

∴ IX = A^–1B

∴ X = A^–1B

∴ X = `(1)/(7)[(4, -3),(3, -4)][(2),(-6)]`

∴ `[(x),(y)] = (1)/(7)[(8 + 18),(6 + 24)]`

= `(1)/(7)[(26),(30)]`

= `[(26/7),(30/7)]`

∴ By equality of martices, we get

x = `(26)/(7)`  and y = `(30)/(7)`.