Question

Solve by using matrix inversion method:

x - y + z = 2, 2x - y = 0, 2y - z = 1

Answer 1

The given equations can be written in matrix form as

`[(1,-1,1),(2,-1,0),(0,2,-1)] [(x),(y),(z)] = [(2),(0),(1)] => "AX" = "B"`

Where A = `[(1,-1,1),(2,-1,0),(0,2,-1)]`, X = `[(x),(y),(z)]`, B = `[(2),(0),(1)]`

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

`= 1|(-1,0),(2,-1)| + 1|(2,0),(0,-1)| + 1|(2,-1),(0,2)|`

= 1(1 - 0) + 1(- 2 - 0) + 1(4 - 0)

= 1 - 2 + 4 = 3 ≠ 0

∴ A^-1 exists.

adj A = `[(+|(-1,0),(2,-1)|, -|(2,0),(0,-1)|, +|(2,-1),(0,2)|),(-|(-1,1),(2,-1)|, +|(1,1),(0,-1)|, -|(1,-1),(0,2)|),(+|(-1,1),(-1,0)|, -|(1,1),(2,0)|, +|(1,-1),(2,-1)|)]^"T"`

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

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

∴ A^-1 = `1/|"A"|`adj A

`= 1/3 [(1,1,1),(2,-1,2),(4,-2,1)]`

Now X = `"A"^-1"B"`

`[(x),(y),(z)] = 1/3[(1,1,1),(2,-1,2),(4,-2,1)] [(2),(0),(1)]`

`[(x),(y),(z)] = 1/3[(2+0+1),(4+0+2),(8+0+1)]`

`[(x),(y),(z)] = 1/3[(3),(6),(9)]`

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

∴ x = 1, y = 2, z = 3