If `A=[(1,2,-3),(2,3,2),(3,-3,-4)]` then find `A^-1` and hence solve the follwoing equations: `x+2y-3z=4, 2x+3y+2z=2 and 3x-3y-4z=11`

Explanation

The given equation are `x+2y-3z=-4. " "…(i)`
`2x+3y+2z=2,"     "…(ii)`
`3x-3y-4z=11."     "…(iii)`
`Let A=[{:(1 " "2 " "-3),(2 " "3 " "          2),(3 "  -3 "  -4):}],X=[{:(x),(y),(z):}] and B=[{:(-4),(2),(11):}]`
So , the given system in matrix form is AX=B.
`Now, |A|=[{:(1 " "2 " "-3),(2 " "3 " "          2),(3 "  -3 "  -4):}]=[{:(1 " "2 " "-3),(0 ""-1 " "          8),(0 "  -9 "             5):}]` `[R_2rarrR_2-2R_1 and R_3rarrR_3-3R_1]=1.(-5+72)=67 ne 0`
Thus , A is invertible
So the has a unique solution , `X=A^(-1)B`
Now , the confactors of the elements of |A| are
`A_(11)=-6, A_(12)=14, A_(13)=-15`
 `A_(21)=17, A_(22)=5, A_(23)=9`
 `A_(31)=13, A_(32)=-8, A_(33)=-1`
`therefore adjA=[{:(-6 " "14 " "-15),(17 "     "5 "        "9),(13 " "-8 " "-1):}]=[{:(-6 " "17 "     "13),(14 "     "5 " "-8),(-15 " "9 " "-1):}]`
So,`A^(-1)=1/|A|.adjA=1/(67).[{:(-6 " "17 "     "13),(14 "     "5 " "-8),(-15 " "9 " "-1):}]`
`thereforeX=A^(-1)B`
or `[{:(x),(y),(z):}]=1/(67).[{:(-6 " "17 "     "13),(14 "     "5 " "-8),(-15 " "9 " "-1):}].[{:(-4),(2),(11):}]`
`=1/(67).[{:(201),(-134),(67):}]=[{:(3),(-2),(1):}]`
`therefore` x=3 , y=-2 and z=1