सर्वप्रथम A=IA लेने पर,
`{:[(0,1,2),(1,2,3),(3,1,1)]=[(1,0,0),(0,1,0),(0,0,1)]:}A`
`{:[(1,3,5),(1,2,3),(3,1,1)]=[(1,1,0),(0,1,0),(0,0,1)]:}A,R_(1)toR_(1)+R_(2)" के प्रयोग से " `
`{:[(1,3,5),(0, 1,2),(0,8,14)]=[(1,1,0),(1,0,0),(3,3,-1)]:}A,R_(2)toR_(1)-R_(2)" तथा "R_(3)to3R_(1)-R_(3)` के प्रयोग से
`{:[(1,3,5),(0, 1,2),(0,0,2)]=[(1,1,0),(1,0,0),(5,-3,1)]:}A,R_(3)to8R_(2)-R_(3)` के प्रयोग से
`{:[(1,3,5),(0, 1,0),(0,0,2)]=[(1,1,0),(-4,3,-1),(5,-3,1)]:}A,R_(2)toR_(2)-R_(3)` के प्रयोग से
`{:[(1,3,1),(0, 1,0),(0,0,2)]=[(-9,7,-2),(-4,3,-1),(5,-3,1)]:}A,R_(1)toR_(1)-2R_(3)` के प्रयोग से
`{:[(1,0,1),(0, 1,0),(0,0,1)]=[(3,-2,1),(-4,3,-1),((5)/(2),-(3)/(5),(1)/(2))]:}A,R_(1)toR_(1)-3R_(2)" तथा "R_(3)to(1)/(2)R_(3)` के प्रयोग से
`{:[(1,0,1),(0, 1,0),(0,0,1)]=[((1)/(2),-(1)/(2),(1)/(2)),(-4,3,-1),((5)/(2),-(3)/(2),(1)/(2))]:}A,R_(1)toR_(1)-3R_(2)" तथा "R_(3)to(1)/(2)R_(3)` के प्रयोग से
`{:[(1,0,0),(0, 1,0),(0,0,1)]=[((1)/(2),-(1)/(2),(1)/(2)),(-4,3,-1),((5)/(2),-(3)/(2),(1)/(2))]:}A,R_(1)toR_(1)-R_(3)` के प्रयोग से
अतः `A^(-1)={:[((1)/(2),-(1)/(2),(1)/(2)),(-4,3,-1),((5)/(2),-(3)/(2),(1)/(2))]:}`