Question

Solve the following by inversion method 2x + y = 5, 3x + 5y = −3

Answer 1

Matrix form of the given system of equations is

`[(2, 1),(3, 5)] [(x), (y)] = [(5),(-3)]`

This is of the form AX = B,

where A = `[(2, 1),(3, 5)]`, X = `[(x),(y)]` and B = `[(5),(-3)]`

To determine X, we have to find A⁻¹ 

|A| = `[(2, 1),(3, 5)]`

= 10 − 3

= 7 ≠ 0

∴ A⁻¹ exists.

Consider AA⁻¹ = I

∴ `[(2, 1),(3, 5)]` A⁻¹ = `[(1, 0),(0, 1)]`

Applying R₂ → 2R₂ − 3R₁, we get

`[(2, 1),(0, 7)]` A⁻¹ = `[(1, 0),(-3, 2)]`

Applying R₁ → 7R₁ – R₂, we get

`[(14, 0),(0, 7)]` A⁻¹ = `[(10, -2),(-3, 2)]`

Applying R₁ → `(1/14)` R₁ and R₂ → `(1/7)` R₂, we get

`[(1, 0),(0, 1)]` A⁻¹ = `[(10/14, (-2)/14),((-3)/7, 2/7)]`

∴ A⁻¹ = `1/7[(5, -1),(-3, 2)]`

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 = `1/7[(5, -1),(-3, 2)] [(5),(-3)]`

∴ `[(x),(y)] = 1/7[(25 + 3),(-15 - 6)]`

= `1/7[(28),(-21)]`

= `[(4),(-3)]`

∴ By equality of matrices, we get

x = 4, y = – 3