Inverse of matrix using elementary transformation

Chapter 3 Class 12 Matrices
Concept wise

We have learned about elementary operations

Letβs learn how to find inverse of a matrix using it.

We will find inverse of a 2 Γ 2 & a 3 Γ 3 matrix

Β

Note:- While doing elementary operations, we use

Only rows

Β  Β  OR

Only columns

Not both

Β

Let's take some examples

Β

Β

Β

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### Transcript

Thus, We can use either π_1β π_3 π_1β π_2 + π_1 π_3β 2π_3 OR πΆ_1β" " πΆ_2 + 9πΆ_2 πΆ_1β 5/2 πΆ_1 πΆ_2β πΆ_3β3πΆ_1 But not π_1β π_3 π_1β π_2 + γ2πγ_1 πͺ_πβ πͺ_π + γππͺγ_π This is wrong Find inverse of [β 8(π&π@π&π)] Let A = [β 8(3&[email protected]&4)] We know that A = A I This becomes Aβ1 Convert to I using elementary transformation A = A I [β 8(3&[email protected]&4)] = A [β 8(1&[email protected]&1)] Making 3 to 1 π_1 βπ_1β 2π_2 [β 8(3β2 (1) &2β2(4)@1&4)] = A [β 8(1β2 (0) &0β2(1)@0&1)] [β 8(3β2 &2β[email protected]&4)] = A [β 8(1&β[email protected]&1)] [β 8(1&β[email protected]&4)] = A [β 8(1&β[email protected]&1)] Making 1 to 0 π_2 βπ_2β π_1 [β 8(1&β[email protected]β1&4β(β6))] = A [β 8(1 &β[email protected]β1&1β(β2))] [β 8(1&β[email protected]&4+6)] = A [β 8(1&β[email protected]β1&1+2)] [β 8(1&β[email protected]&10)] = A [β 8(1&β[email protected]β1&3)] Making 10 to 1 π_2 βπ_2/10 [β 8(1&β[email protected]/10&10/10)] = A [β 8(1 &β[email protected](β1)/10&3/10)] [β 8(1&β[email protected]&1)] = A [β 8(1 &β[email protected](β1)/10&3/10)] Making β6 to 0 π_1 βπ_1+ 6π_2 [β 8(1+6(0)&β6+6(1)@0&1)] = A [β 8(1+6((β1)/10) &β2+6(3/10)@(β1)/10&3/10)] [β 8(1&β[email protected]&1)] = A [β 8(1β6/10&β2+18/[email protected](β1)/10&3/10)] [β 8(1&[email protected]&1)] = A [β 8(4/10&(β2)/[email protected](β1)/10&3/10)] This is similar to I = AAβ1 Therefore, Aβ1 = [β 8(4/10&(β2)/[email protected](β1)/10&3/10)] Find inverse of [β 8(9&2&[email protected]&β1&[email protected]&0&β2)] Let A = [β 8(9&2&[email protected]&β1&[email protected]&0&β2)] We know that A = AI [β 8(9&2&[email protected]&β1&[email protected]&0&β2)] = A [β 8(1&0&[email protected]&1&[email protected]&0&1)] Making 9 to 1 π_1 βπ_1β 2π_3 [β 8(9β2(4)&2β2(0)&1β2(β2)@5&β1&[email protected]&0&β2)] = A [β 8(1β2(0)&0β2(0)&0β2(1)@0&1&[email protected]&0&1)] [β 8(9β8&2β0&[email protected]&β1&[email protected]&0&β2)] = A [β 8(1&0&β[email protected]&1&[email protected]&0&1)] [β 8(1&2&[email protected]&β1&[email protected]&0&β2)] = A [β 8(1&0&β[email protected]&1&[email protected]&0&1)] Making 5 to 0 π_2 βπ_2β 5π_1 [β 8(1&2&[email protected]β5(1)&β1β5(2)&6β5(5)@4&0&β2)] = A [β 8(1&0&β[email protected]β5(1)&1β5(0)&0β5(β2)@0&0&1)] [β 8(1&2&[email protected]β5&β1β10&6β[email protected]&0&β2)] = A [β 8(1&0&β[email protected]β5&1&[email protected]&0&1)] [β 8(1&2&[email protected]&β11&β[email protected]&0&β2)] = A [β 8(1&0&β[email protected]β5&1&[email protected]&0&1)] Making 4 to 0 π_3 βπ_3β 4π_1 [β 8(1&2&[email protected]&β11&β[email protected]β4(1)&0β4(2)&β2β4(5))] = A [β 8(1&0&β[email protected]β5&1&[email protected]β4(1)&0β4(0)&1β4(β2))] [β 8(1&2&[email protected]&β11&β[email protected]β4&β8&β2β20)] = A [β 8(1&0&β[email protected]β5&1&[email protected]β4&0&1+8)] [β 8(1&2&[email protected]&βππ&β[email protected]&β8&β22)] = A [β 8(1&0&β[email protected]β5&1&[email protected]β4&0&9)] Making β11 to 1 π_2 βπ_2/(β11) [β 8(1&2&[email protected]/(β11)&(β11)/(β11)&(β19)/(β11)@0&β8&β22)] = A [β 8(1&0&β[email protected](β5)/(β11)&1/(β11)&10/(β11)@β4&0&9)] [β 8(1&π&[email protected]&1&19/[email protected]&β8&β22)] = A [β 8(1&0&β[email protected]/11&(β1)/11&(β10)/[email protected]β4&0&9)] Making 2 as 0 π_1 βπ_1β 2π_2 [β 8(1β2(0)&2β2(1)&5β2(19/11)@0&1&19/[email protected]&β8&β22)] = A [β 8(1β2(5/11)&0β2((β1)/11)&β2β2((β10)/11)@5/11&(β1)/11&(β10)/[email protected]β4&0&9)] [β 8(1&0&5β38/[email protected]&1&19/[email protected]&β8&β22)] = A [β 8(1β10/11&2/11&β2+20/[email protected]/11&(β1)/11&(β10)/[email protected]β4&0&9)] [β 8(1&0&17/[email protected]&1&19/[email protected]&βπ&β22)] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected]β4&0&9)] Making β8 as 0 π_3 βπ_3 + 8π_2 [β 8(1&0&17/[email protected]&1&19/[email protected]+8(0)&β8+8(1)&β22+8(19/11) )] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected]β4+8(5/11)&0+8((β1)/11)&9+8((β10)/11) )] [β 8(1&0&17/[email protected]&1&19/[email protected]&β8+8&β22+152/11)] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected]β4+40/11&(β8)/11&9β80/11)] [β 8(1&0&17/[email protected]&1&19/[email protected]&0&(βππ)/ππ)] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected](β4)/11&(β8)/11&19/11)] We Make (β90)/11 to 1 π_3 βπ_3 Γ (β11)/90 [β 8(1&0&17/[email protected]&1&19/[email protected]Γ(β11)/90&0Γ(β11)/90&(β90)/11Γ(β11)/90)] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected]β4Γ(β11)/90&(β8)/11Γ(β11)/90&19/11Γ(β11)/90)] [β 8(1&0&17/[email protected]&1&19/[email protected]&0&1)] = A [β 8(1/11&2/11&(β2)/[email protected]/11&(β1)/11&(β10)/[email protected]/45&4/45&(β19)/90)] We Make 17/11 to 0 π_1 βπ_1 β 17/11 π_3 [β 8(1β17/11(0)&0β17/11(0)&17/11β17/11(1)@0&1&19/[email protected]&0&1)] = A [β 8(1/11β17/11 (2/45)&2/11β17/11 (4/45)&(β2)/11β17/11 ((β19)/90)@5/11&(β1)/11&(β10)/[email protected]/45&4/45&(β19)/90)] [β 8(1β17/11(0)&0β17/11(0)&17/11β17/11(1)@0&1&19/[email protected]&0&1)] = A [β 8(1/11 (1β34/45) &2/11 (1β34/35)&1/11 (β2+323/90)@5/11&(β1)/11&(β10)/[email protected]/45&4/45&(β19)/90)] [β 8(1&0&[email protected]&1&19/[email protected]&0&1)] = A [β 8(1/11Γ11/45&2/11Γ11/45&1/11Γ143/[email protected]/11&(β1)/11&(β10)/[email protected]/45&4/45&(β19)/90)] [β 8(1&0&[email protected]&1&19/[email protected]&0&1)] = A [β 8(1/45&2/45&13/[email protected]/11&(β1)/11&(β10)/[email protected]/45&4/45&(β19)/90)] We Make 19/11 as 1 π_2 βπ_2 β 19/11 π_3 [β 8(1&0&[email protected]β19/11(0)&1β19/11(0)&19/11β19/11(1)@0&0&1)] = A [β 8(1/45&2/11&13/[email protected]/11β19/11 (2/45)&(β1)/11β19/11 (4/45)&(β10)/11β19/11 ((β19)/11)@2/45&4/45&(β19)/90)] [β 8(1&0&[email protected]&1&[email protected]&0&1)] = A [β 8(1/45&2/45&13/[email protected]/11 (5β38/45)&(β1)/11Γ121/45&1/11Γ((β539)/90)@2/45&4/45&(β19)/90)] [β 8(1&0&[email protected]&1&[email protected]&0&1)] = A [β 8(1/45&2/45&13/[email protected]/45&(β11)/45&(β49)/[email protected]/45&4/45&(β19)/90)] I = AAβ1 Therefore, Aβ1 = [β 8(1/45&2/45&13/[email protected]/45&(β11)/45&(β49)/[email protected]/45&4/45&(β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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.