If , find A^–1. Using A^–1 solve the system of equations

2x – 3y + 5z = 11

3x + 2y – 4z = – 5

x + y – 2z = – 3

∴ |A| = 2(-4+4) + 3(-6+4) + 5(3-2) = -1 ≠ 0

So inverse of A exists,

Now A₁₁ = 0, A₁₂ = 2, A₁₃ = 1, A₂₁ = -1, A₂₂ = -9, A₂₃ = -5, A₃₁ = 2, A₃₂ = 23, A₃₃ = 13

So AdjA =

∴

So .

Hence x = 1, y = 2 and z = 3