If `A=((2,-3,5),(3,2,-4),(1,1,-2))` find `A^(-1)`. Use it to solve the system of equations `2x-3y+5z=11` , `3x+2y-4z=-5` and `x+y-2z=-3`

Explanation

` A=[{:(2,-3,5),(3,2,-4),(1,1,-2):}]`
`rArr" " |A|=[{:(2,-3,5),(3,2,-4),(1,1,-2):}]`
=2(-4+4)-(-23)(-6+4)+5(3-2)
`=-6+5=-1ne0`
`therefore` A is invertible
`"Now," A_(11)=0, A_(12)=2, A_(13)=1`
`A_(21)=-1, A_(22)=-9, A_(23)=-5`
`A_(31)=2, A_(32)=23, A_(33)=13`
`therefore" adjA"=[{:(0,2,1),(-1,-9,-5),(2,23,13):}]=[{:(0,-1,2),(2,-9,23),(1,-5,13):}]`
`"and A"^(-1)=1/|A|"adj A"=1/-1"=[{:(0,-1,2),(2,-9,23),(1,-5,13):}]=[{:(0,1,-2),(2,-9,23),(-1,5,-13):}]`
Given system of equations,
2x-3y+5y=11
3x+2y-4z=-5
x+y-2z=-3
`"=[{:(2,-3,5),(3,2,-4),(1,1,12):}][{:(x),(y),(z):}]=[{:(11),(-5),(-3):}]`
`rArr" "AX=B=rArrX=A^(-1)B`
`[{:(x),(y),(z):}]=[{:(0,1,-2),(-2,9,-23),(-1,5,-13):}][{:(11),(-5),(-3):}]`
`=[{:(0-5+6),(-22-45+69),(-11-25+39):}][{:(1),(2),(3):}]`
`rArr " "x=1, y=2, z=3`