Solve the following equations by method of inversion.
x + 2y = 2, 2x + 3y = 3
Solution
Matrix form of the given system of equations is
`[(1, 2),(2, 3)] [(x),(y)] = [(2),(3)]`
This is of the form AX = B.
where A = `[(1, 2),(2, 3)], "X" = [(x),(y)] "and B" = [(2),(3)]`
To determine X, we have to find A–1.
|A| = `|(1, 2),(2, 3)|`
= 3 – 4
= – 1 ≠ 0
∴ A–1 exists.
Consider AA–1 = I
∴ `[(1, 2),(2, 3)] "A"^-1 = [(1, 0),(0, 1)]`
Applying R2 → R2 – 2R1, we get
`[(1, 2),(0, -1)] "A"^-1 = [(1, 0),(-2, 1)]`
Applying R2 → (– 1)R2, we get
`[(1, 2),(0, 1)] "A"^-1 = [(1, 0),(2, -1)]`
Applying R1 → R1 – 2R2, we get
`[(1, 0),(0, 1)] "A"^-1 = [(-3, 2),(2, -1)]`
∴ A–1 = `[(-3, 2),(2, -1)]`
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 = `[(-3, 2),(2, -1)] [(2),(3)]`
∴ `[(x),(y)] = [(-6 + 6),(4 - 3)] = [(0),(1)]`
∴ By equality of matrices, we get
x = 0 and y = 1.