Use product `[1-1 2 0 2-3 3-2 4]\ \ [-2 0 1 9 2-3 6 1-2]` to solve the system of equation: `x-y+2z=1` `2y-3z=1` `3x-2y+4z=2`

Explanation

The given equations are
`x-2y+2z=1, " " (i)`
`2y-3z=1, " " (ii)`
`3x-2y+4z=2 " " (iii)`
`Let A=[{:(1 " "-1 " "2),(0" "2 " "-3),(3 " "-2 " "4):}], X=[{:(x),(y),(z):}] and B=[{:(1),(1),(2):}]`
Then , the given system in matrix form is AX=B
`Now,[{:(1 " "-1 " "2),(0" "2 " "-3),(3 " "-2 " "4):}][{:(-2 " "0 " "1),(9" "2 " "-3),(6 " "1 " "-2):}]`
`=[{:(-2-9+12 " "0-2+2 " "1+3-4),(0+18-18 " "0+4-3 " "0-6+6),(-6-18+24 " "0-4+4 " "3+6-8):}]=[{:(1 " "0 " "0),(0 " "1 " "0),(0 " "0 " "1):}]`
`rArr A.[{:(-2 " "0 " "1),(9 " "2 " "-3),(6 " "1 " "-2):}]=I`
`rArrA^(-1)=[{:(-2 " "0 " "1),(9 " "2 " "-3),(6 " "1 " "-2):}]`
Now,AX=B
`rArrx=A^(-1)B`
`rArr[{:(x),(y),(z):}]=[{:(-2 " "0 " "1),(9 " "2 " "-3),(6 " "1 " "-2):}][{:(1),(1),(2):}]`
`=[{:(-2+0+2),(9+2-6),(6+1-4):}]=[{:(0),(5),(3):}]`
`rArr` x=0,y=5 and z=3
Hence x=0,y=5 and z=3