Question

Solve the following equations by method of inversion.
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 deterrmine X, we have to find A^–1.

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

= 10 – 3
= 7 ≠ 0
∴ A^–1 exissts.
Consider AA^–1 = I

∴ `[(2, 1),(3, 5)] "A"^-1 = [(1, 0),(0, 1)]`

Applying R₁ ↔ R₂, we get

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

Applying R₁ → R₁ – R₂, we get

`[(1, 4),(2, 1)] "A"^-1 = [(-1, 1),(1, 0)]`

Applying R₂ → R₂ – 2R₁, we get

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

Applying R₂ → `(-1/7)` R₂, we get

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

Applying R₁ → R₁ – 4R₂, we get

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

∴ A^–1 =`[(5/7, -1/7),(-3/7, 2/7)]`

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

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

∴ `[(x),(y)] = [(25/7, (+3)/7),(-15/7, -6/7)] = [(28/7),(-21/7)] = [(4),(-3)]`

∴ By equality of matrices, we get
x = 4 and y = – 3.