Question

Solve the following equations by the inversion method :
2x + 3y = - 5 and 3x + y = 3.

Answer 1

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

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

R₂ - R₁ gives

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

R₁ ↔ R₂ gives

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

R₂ - 2R₁ gives

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

`1/7`R₂ gives

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

R₁ + 2R₂
`[(1, 0), (0, 1)]A^-1 = [(-1/7, 3/7), (3/7, -2/7)]`

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

AX = B
Pre multiplying by A^-1

A^-1(AX) = A^-1B

IX = A^-1B

X = A^-1B

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

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

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

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

x = 2, y = -3