The given system of equations is:

2x + 3y + 3z = 5

x – 2y + z = -4

3x - y - 2z = 3

The given system of equations can be written in the form of AX = B, where

Now |A| = 2(4+1)-3(-5) +3(5) = 40 ≠ 0

∴ A is a non-singular matrix and hence A^-1 exists.

Now A₁₁ = 5, A₁₂ = 5, A₁₃ = 5, A₂₁ = 3, A₂₂ = -13, A₂₃ = 11, A₃₁ = 9, A₃₂ = 1, A₃₃ = -7

So AdjA =

∴

And hence X = A^-1B

So

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