We have learned about elementary operations

Let’s learn how to find inverse of a matrix using it.

We will find inverse of a 2 × 2 & a 3 × 3 matrix


Note:- While doing elementary operations, we use

Only rows


Only columns

Not both



Let's take some examples







  1. Chapter 3 Class 12 Matrices
  2. Concept wise


Thus, We can use either 𝑅_1↔ 𝑅_3 𝑅_1β†’ 𝑅_2 + 𝑅_1 𝑅_3β†’ 2𝑅_3 OR 𝐢_1β†’" " 𝐢_2 + 9𝐢_2 𝐢_1β†’ 5/2 𝐢_1 𝐢_2β†’ 𝐢_3βˆ’3𝐢_1 But not 𝑅_1↔ 𝑅_3 𝑅_1β†’ 𝑅_2 + γ€–2𝑅〗_1 π‘ͺ_πŸβ†’ π‘ͺ_πŸ‘ + γ€–πŸ‘π‘ͺγ€—_𝟏 This is wrong Find inverse of [β– 8(πŸ‘&𝟐@𝟏&πŸ’)] Let A = [β– 8(3&2@1&4)] We know that A = A I This becomes Aβˆ’1 Convert to I using elementary transformation A = A I [β– 8(3&2@1&4)] = A [β– 8(1&0@0&1)] Making 3 to 1 𝑅_1 →𝑅_1βˆ’ 2𝑅_2 [β– 8(3βˆ’2 (1) &2βˆ’2(4)@1&4)] = A [β– 8(1βˆ’2 (0) &0βˆ’2(1)@0&1)] [β– 8(3βˆ’2 &2βˆ’8@1&4)] = A [β– 8(1&βˆ’2@0&1)] [β– 8(1&βˆ’6@1&4)] = A [β– 8(1&βˆ’2@0&1)] Making 1 to 0 𝑅_2 →𝑅_2βˆ’ 𝑅_1 [β– 8(1&βˆ’6@1βˆ’1&4βˆ’(βˆ’6))] = A [β– 8(1 &βˆ’2@0βˆ’1&1βˆ’(βˆ’2))] [β– 8(1&βˆ’6@0&4+6)] = A [β– 8(1&βˆ’2@βˆ’1&1+2)] [β– 8(1&βˆ’6@0&10)] = A [β– 8(1&βˆ’2@βˆ’1&3)] Making 10 to 1 𝑅_2 →𝑅_2/10 [β– 8(1&βˆ’6@0/10&10/10)] = A [β– 8(1 &βˆ’2@(βˆ’1)/10&3/10)] [β– 8(1&βˆ’6@0&1)] = A [β– 8(1 &βˆ’2@(βˆ’1)/10&3/10)] Making βˆ’6 to 0 𝑅_1 →𝑅_1+ 6𝑅_2 [β– 8(1+6(0)&βˆ’6+6(1)@0&1)] = A [β– 8(1+6((βˆ’1)/10) &βˆ’2+6(3/10)@(βˆ’1)/10&3/10)] [β– 8(1&βˆ’6+6@0&1)] = A [β– 8(1βˆ’6/10&βˆ’2+18/10@(βˆ’1)/10&3/10)] [β– 8(1&0@0&1)] = A [β– 8(4/10&(βˆ’2)/10@(βˆ’1)/10&3/10)] This is similar to I = AAβˆ’1 Therefore, Aβˆ’1 = [β– 8(4/10&(βˆ’2)/10@(βˆ’1)/10&3/10)] Find inverse of [β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)] Let A = [β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)] We know that A = AI [β– 8(9&2&1@5&βˆ’1&6@4&0&βˆ’2)] = A [β– 8(1&0&0@0&1&0@0&0&1)] Making 9 to 1 𝑅_1 →𝑅_1βˆ’ 2𝑅_3 [β– 8(9βˆ’2(4)&2βˆ’2(0)&1βˆ’2(βˆ’2)@5&βˆ’1&6@4&0&βˆ’2)] = A [β– 8(1βˆ’2(0)&0βˆ’2(0)&0βˆ’2(1)@0&1&0@0&0&1)] [β– 8(9βˆ’8&2βˆ’0&1+4@5&βˆ’1&6@4&0&βˆ’2)] = A [β– 8(1&0&βˆ’2@0&1&0@0&0&1)] [β– 8(1&2&5@5&βˆ’1&6@4&0&βˆ’2)] = A [β– 8(1&0&βˆ’2@0&1&0@0&0&1)] Making 5 to 0 𝑅_2 →𝑅_2βˆ’ 5𝑅_1 [β– 8(1&2&5@5βˆ’5(1)&βˆ’1βˆ’5(2)&6βˆ’5(5)@4&0&βˆ’2)] = A [β– 8(1&0&βˆ’2@0βˆ’5(1)&1βˆ’5(0)&0βˆ’5(βˆ’2)@0&0&1)] [β– 8(1&2&5@5βˆ’5&βˆ’1βˆ’10&6βˆ’25@4&0&βˆ’2)] = A [β– 8(1&0&βˆ’2@βˆ’5&1&10@0&0&1)] [β– 8(1&2&5@0&βˆ’11&βˆ’19@4&0&βˆ’2)] = A [β– 8(1&0&βˆ’2@βˆ’5&1&10@0&0&1)] Making 4 to 0 𝑅_3 →𝑅_3βˆ’ 4𝑅_1 [β– 8(1&2&5@0&βˆ’11&βˆ’19@4βˆ’4(1)&0βˆ’4(2)&βˆ’2βˆ’4(5))] = A [β– 8(1&0&βˆ’2@βˆ’5&1&10@0βˆ’4(1)&0βˆ’4(0)&1βˆ’4(βˆ’2))] [β– 8(1&2&5@0&βˆ’11&βˆ’19@4βˆ’4&βˆ’8&βˆ’2βˆ’20)] = A [β– 8(1&0&βˆ’2@βˆ’5&1&10@βˆ’4&0&1+8)] [β– 8(1&2&5@0&βˆ’πŸπŸ&βˆ’19@4&βˆ’8&βˆ’22)] = A [β– 8(1&0&βˆ’2@βˆ’5&1&10@βˆ’4&0&9)] Making βˆ’11 to 1 𝑅_2 →𝑅_2/(βˆ’11) [β– 8(1&2&5@0/(βˆ’11)&(βˆ’11)/(βˆ’11)&(βˆ’19)/(βˆ’11)@0&βˆ’8&βˆ’22)] = A [β– 8(1&0&βˆ’2@(βˆ’5)/(βˆ’11)&1/(βˆ’11)&10/(βˆ’11)@βˆ’4&0&9)] [β– 8(1&𝟐&5@0&1&19/11@0&βˆ’8&βˆ’22)] = A [β– 8(1&0&βˆ’2@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4&0&9)] Making 2 as 0 𝑅_1 →𝑅_1βˆ’ 2𝑅_2 [β– 8(1βˆ’2(0)&2βˆ’2(1)&5βˆ’2(19/11)@0&1&19/11@0&βˆ’8&βˆ’22)] = A [β– 8(1βˆ’2(5/11)&0βˆ’2((βˆ’1)/11)&βˆ’2βˆ’2((βˆ’10)/11)@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4&0&9)] [β– 8(1&0&5βˆ’38/11@0&1&19/11@0&βˆ’8&βˆ’22)] = A [β– 8(1βˆ’10/11&2/11&βˆ’2+20/11@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4&0&9)] [β– 8(1&0&17/11@0&1&19/11@0&βˆ’πŸ–&βˆ’22)] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4&0&9)] Making βˆ’8 as 0 𝑅_3 →𝑅_3 + 8𝑅_2 [β– 8(1&0&17/11@0&1&19/11@0+8(0)&βˆ’8+8(1)&βˆ’22+8(19/11) )] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4+8(5/11)&0+8((βˆ’1)/11)&9+8((βˆ’10)/11) )] [β– 8(1&0&17/11@0&1&19/11@0&βˆ’8+8&βˆ’22+152/11)] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4+40/11&(βˆ’8)/11&9βˆ’80/11)] [β– 8(1&0&17/11@0&1&19/11@0&0&(βˆ’πŸ—πŸŽ)/𝟏𝟏)] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@(βˆ’4)/11&(βˆ’8)/11&19/11)] We Make (βˆ’90)/11 to 1 𝑅_3 →𝑅_3 Γ— (βˆ’11)/90 [β– 8(1&0&17/11@0&1&19/11@0Γ—(βˆ’11)/90&0Γ—(βˆ’11)/90&(βˆ’90)/11Γ—(βˆ’11)/90)] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@βˆ’4Γ—(βˆ’11)/90&(βˆ’8)/11Γ—(βˆ’11)/90&19/11Γ—(βˆ’11)/90)] [β– 8(1&0&17/11@0&1&19/11@0&0&1)] = A [β– 8(1/11&2/11&(βˆ’2)/11@5/11&(βˆ’1)/11&(βˆ’10)/11@2/45&4/45&(βˆ’19)/90)] We Make 17/11 to 0 𝑅_1 →𝑅_1 – 17/11 𝑅_3 [β– 8(1βˆ’17/11(0)&0βˆ’17/11(0)&17/11βˆ’17/11(1)@0&1&19/11@0&0&1)] = A [β– 8(1/11βˆ’17/11 (2/45)&2/11βˆ’17/11 (4/45)&(βˆ’2)/11βˆ’17/11 ((βˆ’19)/90)@5/11&(βˆ’1)/11&(βˆ’10)/11@2/45&4/45&(βˆ’19)/90)] [β– 8(1βˆ’17/11(0)&0βˆ’17/11(0)&17/11βˆ’17/11(1)@0&1&19/11@0&0&1)] = A [β– 8(1/11 (1βˆ’34/45) &2/11 (1βˆ’34/35)&1/11 (βˆ’2+323/90)@5/11&(βˆ’1)/11&(βˆ’10)/11@2/45&4/45&(βˆ’19)/90)] [β– 8(1&0&0@0&1&19/11@0&0&1)] = A [β– 8(1/11Γ—11/45&2/11Γ—11/45&1/11Γ—143/90@5/11&(βˆ’1)/11&(βˆ’10)/11@2/45&4/45&(βˆ’19)/90)] [β– 8(1&0&0@0&1&19/11@0&0&1)] = A [β– 8(1/45&2/45&13/90@5/11&(βˆ’1)/11&(βˆ’10)/11@2/45&4/45&(βˆ’19)/90)] We Make 19/11 as 1 𝑅_2 →𝑅_2 βˆ’ 19/11 𝑅_3 [β– 8(1&0&0@0βˆ’19/11(0)&1βˆ’19/11(0)&19/11βˆ’19/11(1)@0&0&1)] = A [β– 8(1/45&2/11&13/90@5/11βˆ’19/11 (2/45)&(βˆ’1)/11βˆ’19/11 (4/45)&(βˆ’10)/11βˆ’19/11 ((βˆ’19)/11)@2/45&4/45&(βˆ’19)/90)] [β– 8(1&0&0@0&1&0@0&0&1)] = A [β– 8(1/45&2/45&13/90@1/11 (5βˆ’38/45)&(βˆ’1)/11Γ—121/45&1/11Γ—((βˆ’539)/90)@2/45&4/45&(βˆ’19)/90)] [β– 8(1&0&0@0&1&0@0&0&1)] = A [β– 8(1/45&2/45&13/90@17/45&(βˆ’11)/45&(βˆ’49)/90@2/45&4/45&(βˆ’19)/90)] I = AAβˆ’1 Therefore, Aβˆ’1 = [β– 8(1/45&2/45&13/90@17/45&(βˆ’11)/45&(βˆ’49)/90@2/45&4/45&(βˆ’19)/90)]

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.