Solve system of linear equations, using matrix method,
`2x + 3y +3z = 5`
`x-2 y + z =-4`
`3x-y -2z= 3`

Explanation

Given system of equations,
2x+3y+3z=5
x-2y+z=-4
3x-y2z=3
`[{:(2,3,3),(1,-2,1),(3,-1,-2):}][{:(x),(y),(z):}]=[{:(5),(-4),(3):}]rArr AX=B`
`therefore" "A=[{:(2,3,3),(1,-2,1),(3,-1,-2):}]`
`rArr" "|A|=[{:(2,3,3),(1,-2,1),(3,-1,-2):}]`
=2(4+1)-3(-2-3)+3(-1+6)
`=10+15+15=40ne0`
`therefore `A is invertible.
`"Now" A_(11)=5, A_(12)=5, A_(13)=5`
`A_(21)=3, A_(22)=-13, A_(23)=11`
`A_(31)=9, A_(32)=1, A_(33)=-7`
`therefore adj A =[{:(5,5,5),(3,-13,11),(9,1,-7):}]=[{:(5,5,5),(3,-13,11),(9,1,-7):}]`
`"and "A^(-1)=1/|A|"adjA"=1/40[{:(5,5,5),(3,-13,11),(9,1,-7):}]`
From equation (1), AX=B `rArr X=A^(-1)`B
`rArr" "[{:(x),(y),(z):}]=1/40[{:(25,-12,+27),(25,+52,+3),(25,-44,-21):}]=[{:(1),(2),(-1):}]`
`therefore "x=1, y=2, z=-1`