By using elementary row operations, find the inverse of the matrix
`A=[{:(3,-1,-2),(2,0,-1),(3,-5," "0):}].`

Explanation

We have
`[{:(3,-1,-2),(2,0,-1),(3,-5," "0):}]=[{:(1,0,0),(0,1,0),(0,0,1):}].A`
`implies" "[{:(1,-1,-1),(2,0,-1),(3,-5," "0):}]=[{:(1,-1,0),(0,1,0),(0,0,1):}].A" "[R_(1)toR_(2)-2R_(1)]`
`implies" "[{:(1,-1,-1),(0," "2," "1),(0,-2," "3):}]=[{:(1,-1,0),(-2," "3,0),(-3," "3,1):}].A" "[{:(R_(2)toR_(2)-2R_(1)),(R_(3)toR_(3)-3R_(1)):}]`
`implies" "[{:(1,-1,-1),(0," "1,(1)/(2)),(0,-2,3):}]=[{:(1,-1,0),(-1,(3)/(2),0),(-3,3,1):}].A" "{R_(2)to(1)/(2)R_(2)}`
`implies" "[{:(1,0,(-1)/(2)),(0,1,(1)/(2)),(0,0,4):}]=[{:(0,(1)/(2),0),(-1,(3)/(2),0),(-5,6,1):}].A" "{{:(R_(1)toR_(1)+R_(2)),(R_(3)toR_(3)+2R_(2)):}}`
`=[{:(1,0,(-1)/(2)),(0,1," "(1)/(2)),(0,0," "1):}]=[{:(0,(1)/(2),0),(-1,(3)/(2),0),((-5)/(4),(3)/(2),(1)/(4)):}].A" "{R_(3)to(1)/(4)R_(3)}`
`implies" "[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:((-5)/(8),(5)/(4),(1)/(8)),((-3)/(8),(3)/(4),(-1)/(8)),((-5)/(4),(3)/(2),(1)/(4)):}].A." "{{:(R_(1)toR_(1)+(1)/(2)R_(3)),(R_(2)toR_(2)-(1)/(2)R^(3)):}}`
Hence, `A^(-1)=[{:((-5)/(8),(5)/(4),(1)/(8)),((-3)/(8),(3)/(4),(-1)/(8)),((-5)/(4),(3)/(2),(1)/(4)):}].`