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

Writing equation as `AX=B`

`[(2,-3,5),(3,2,-4),(1,1,-2)][[x],[y],[z]]=[[11],[-5],[-3]]`

Hence,
`A=[(2,-3,5),(3,2,-4),(1,1,-2)], X=[[x],[y],[z]] & B=[[11],[-5],[-3]]`

Calculating `|A|`

`|A|=|(2,-3,5),(3,2,-4),(1,1,-2)|`

` " " " " =2[-4+4]-3[6-5]+1[12-10]`

` " " " " =2[0]-3[1]+1[2]`

` " " " " =-1`

So, `|A| ne 0`

`therefore` The system of equation is consistent & has a unique solution

Now,

`AX=B`

`X=A^(-1)B`


Calculating `A^(-1)`

Now, `A^(-1)= 1/|A| adj(A)`

`adj A=[[A_11,A_21,A_31],[A_12,A_22,A_32],[A_13,A_23,A_33]]`


`A=[(2,-3,5),(3,2,-4),(1,1,-2)]`

Now,

`A_11=-4+4=0`
`A_12=-[-6-(-4)]=-(-6+4)=-(-2)=2`
`A_13=3-2=1`

`A_21=-(6-5)=-1`
`A_22=-4-5=-9`
`A_23=-[2-(-3)]=-5`

`A_31=12-10=2`
`A_32=-(-8-15)=23`
`A_33=4-(-9)=13`

Thus `adj A=[[0,-1,2],[2,-9,23],[1,-5,13]]`

Now, `A^(-1)= 1/|A| adj(A)`

`A^(-1)=1/(-1) [[0,-1,2],[2,-9,23],[1,-5,13]]=[[0,1,-2],[-2,9,-23],[-1,5,-13]]`

Solving `X=A^(-1)B`

`[[x],[y],[z]]=[[0,1,-2],[-2,9,-23],[-1,5,-13]][[11],[-5],[-3]]`

`[[x],[y],[z]]=[[0(11)+1(-5)+(-2)(-3)],[2(11)+9(-5)+(-23)(-3)],[-1(11)+5(-5)+(-13)(-3)]]`

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

`therefore` `x=1, " y=2 &z=3`