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

    OR

Only columns

Not both

 

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Let's take some examples
Finding inverse of a matrix using Elementary Operations - Part 2

Finding inverse of a matrix using Elementary Operations - Part 3

 

Finding inverse of a matrix using Elementary Operations - Part 4

Finding inverse of a matrix using Elementary Operations - Part 5

 

 


Transcript

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)]

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.