प्रारम्भिक संक्रियाओं का उपयोग कर निम्न आव्यूह का व्यत्क्रम ज्ञात कीजिए :
`A={:[(2,-3,3),(2,2,3),(3,-2,2)]:}`

Explanation

A=IA लेने पर, `{:[(2,-3,3),(2,2,3),(3,-2,2)]=[(1,0,0),(0,1,0),(0,0,1)]:}A`
`{:[(3,-2,2),(2,2,3),(2,-3,3)]=[(0,0,1),(0,1,0),(1,0,0)]:}A," "R_(1)harrR_(3)`
`{:[(1,-4,-1),(2,2,3),(2,-3,3)]=[(0,-1,1),(0,1,0),(1,0,0)]:}A," "R_(1)toR_(1)-R_(2)`
`{:[(1,-4,-1),(0,10,5),(0,5,5)]=[(0,-1,1),(1,3,-2),(1,2,-2)]:}A," "R_(2)toR_(2)-2R_(1)" तथा "R_(3)toR_(3)-2R_(1)`
`{:[(1,-4,-1),(0,5,0),(0,5,5)]=[(0,-1,1),(-1,1,0),(1,2,-2)]:}A," "R_(2)toR_(2)-R_(3)`
`{:[(1,-4,-1),(0,1,0),(0,5,5)]=[(0,-1,1),(-1//5,1//5,0),(1,2,-2)]:}A," "R_(2)to(1)/(5)R_(2)`
`{:[(1,-4,-1),(0,1,0),(0,1,1)]=[(0,-1,1),(-1//5,1//5,0),(1//5,2//5,-2//5)]:}A," "R_(3)to(1)/(5)R_(3)`
`{:[(1,0,-1),(0,1,0),(0,1,1)]=[(-4//5,-1//5,1),(-1//5,1//5,0),(1//5,2//5,-2//5)]:}A,=(1)/(5){:[(-4,-1,5),(-1,1,0),(1,2,-2)]:}A" "R_(1)toR_(1)+4R_(2)`
`{:[(1,0,-1),(0,1,0),(0,0,1)]=(1)/(5)[(-4,-1,5),(-1,1,0),(2,1,-2)]:}A," "R_(3)toR_(3)-R_(2)`
`{:[(1,0,0),(0,1,0),(0,0,1)]=(1)/(5)[(-2,0,3),(-1,1,0),(2,1,-2)]:}A={:[(-2//5,0,3//5),(-1//5,1//5,0),(2//5,1//5,-2//5)]:}A`
अतः `A^(-1)={:[(-2//5,0,3//5),(-1//5,1//5,0),(2//5,1//5,-2//5)]:}`