Check sibling questions

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

 

 

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Thus, We can use either 𝑅_1↔ 𝑅_3 𝑅_1→ 𝑅_2 + 𝑅_1 𝑅_3→ 2𝑅_3 OR 𝐶_1→" " 𝐶_2 + 9𝐶_2 𝐶_1→ 5/2 𝐶_1 𝐶_2→ 𝐶_3−3𝐶_1 But not 𝑅_1↔ 𝑅_3 𝑅_1→ 𝑅_2 + 〖2𝑅〗_1 𝑪_𝟐→ 𝑪_𝟑 + 〖𝟑𝑪〗_𝟏 This is wrong Find inverse of [■8(𝟑&𝟐@𝟏&𝟒)] Let A = [■8(3&2@1&4)] We know that A = A I This becomes A−1 Convert to I using elementary transformation A = A I [■8(3&2@1&4)] = A [■8(1&0@0&1)] Making 3 to 1 𝑅_1 →𝑅_1− 2𝑅_2 [■8(3−2 (1) &2−2(4)@1&4)] = A [■8(1−2 (0) &0−2(1)@0&1)] [■8(3−2 &2−8@1&4)] = A [■8(1&−2@0&1)] [■8(1&−6@1&4)] = A [■8(1&−2@0&1)] Making 1 to 0 𝑅_2 →𝑅_2− 𝑅_1 [■8(1&−6@1−1&4−(−6))] = A [■8(1 &−2@0−1&1−(−2))] [■8(1&−6@0&4+6)] = A [■8(1&−2@−1&1+2)] [■8(1&−6@0&10)] = A [■8(1&−2@−1&3)] Making 10 to 1 𝑅_2 →𝑅_2/10 [■8(1&−6@0/10&10/10)] = A [■8(1 &−2@(−1)/10&3/10)] [■8(1&−6@0&1)] = A [■8(1 &−2@(−1)/10&3/10)] Making −6 to 0 𝑅_1 →𝑅_1+ 6𝑅_2 [■8(1+6(0)&−6+6(1)@0&1)] = A [■8(1+6((−1)/10) &−2+6(3/10)@(−1)/10&3/10)] [■8(1&−6+6@0&1)] = A [■8(1−6/10&−2+18/10@(−1)/10&3/10)] [■8(1&0@0&1)] = A [■8(4/10&(−2)/10@(−1)/10&3/10)] This is similar to I = AA−1 Therefore, A−1 = [■8(4/10&(−2)/10@(−1)/10&3/10)] Find inverse of [■8(9&2&1@5&−1&6@4&0&−2)] Let A = [■8(9&2&1@5&−1&6@4&0&−2)] We know that A = AI [■8(9&2&1@5&−1&6@4&0&−2)] = A [■8(1&0&0@0&1&0@0&0&1)] Making 9 to 1 𝑅_1 →𝑅_1− 2𝑅_3 [■8(9−2(4)&2−2(0)&1−2(−2)@5&−1&6@4&0&−2)] = A [■8(1−2(0)&0−2(0)&0−2(1)@0&1&0@0&0&1)] [■8(9−8&2−0&1+4@5&−1&6@4&0&−2)] = A [■8(1&0&−2@0&1&0@0&0&1)] [■8(1&2&5@5&−1&6@4&0&−2)] = A [■8(1&0&−2@0&1&0@0&0&1)] Making 5 to 0 𝑅_2 →𝑅_2− 5𝑅_1 [■8(1&2&5@5−5(1)&−1−5(2)&6−5(5)@4&0&−2)] = A [■8(1&0&−2@0−5(1)&1−5(0)&0−5(−2)@0&0&1)] [■8(1&2&5@5−5&−1−10&6−25@4&0&−2)] = A [■8(1&0&−2@−5&1&10@0&0&1)] [■8(1&2&5@0&−11&−19@4&0&−2)] = A [■8(1&0&−2@−5&1&10@0&0&1)] Making 4 to 0 𝑅_3 →𝑅_3− 4𝑅_1 [■8(1&2&5@0&−11&−19@4−4(1)&0−4(2)&−2−4(5))] = A [■8(1&0&−2@−5&1&10@0−4(1)&0−4(0)&1−4(−2))] [■8(1&2&5@0&−11&−19@4−4&−8&−2−20)] = A [■8(1&0&−2@−5&1&10@−4&0&1+8)] [■8(1&2&5@0&−𝟏𝟏&−19@4&−8&−22)] = A [■8(1&0&−2@−5&1&10@−4&0&9)] Making −11 to 1 𝑅_2 →𝑅_2/(−11) [■8(1&2&5@0/(−11)&(−11)/(−11)&(−19)/(−11)@0&−8&−22)] = A [■8(1&0&−2@(−5)/(−11)&1/(−11)&10/(−11)@−4&0&9)] [■8(1&𝟐&5@0&1&19/11@0&−8&−22)] = A [■8(1&0&−2@5/11&(−1)/11&(−10)/11@−4&0&9)] Making 2 as 0 𝑅_1 →𝑅_1− 2𝑅_2 [■8(1−2(0)&2−2(1)&5−2(19/11)@0&1&19/11@0&−8&−22)] = A [■8(1−2(5/11)&0−2((−1)/11)&−2−2((−10)/11)@5/11&(−1)/11&(−10)/11@−4&0&9)] [■8(1&0&5−38/11@0&1&19/11@0&−8&−22)] = A [■8(1−10/11&2/11&−2+20/11@5/11&(−1)/11&(−10)/11@−4&0&9)] [■8(1&0&17/11@0&1&19/11@0&−𝟖&−22)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4&0&9)] Making −8 as 0 𝑅_3 →𝑅_3 + 8𝑅_2 [■8(1&0&17/11@0&1&19/11@0+8(0)&−8+8(1)&−22+8(19/11) )] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4+8(5/11)&0+8((−1)/11)&9+8((−10)/11) )] [■8(1&0&17/11@0&1&19/11@0&−8+8&−22+152/11)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4+40/11&(−8)/11&9−80/11)] [■8(1&0&17/11@0&1&19/11@0&0&(−𝟗𝟎)/𝟏𝟏)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@(−4)/11&(−8)/11&19/11)] We Make (−90)/11 to 1 𝑅_3 →𝑅_3 × (−11)/90 [■8(1&0&17/11@0&1&19/11@0×(−11)/90&0×(−11)/90&(−90)/11×(−11)/90)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@−4×(−11)/90&(−8)/11×(−11)/90&19/11×(−11)/90)] [■8(1&0&17/11@0&1&19/11@0&0&1)] = A [■8(1/11&2/11&(−2)/11@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] We Make 17/11 to 0 𝑅_1 →𝑅_1 – 17/11 𝑅_3 [■8(1−17/11(0)&0−17/11(0)&17/11−17/11(1)@0&1&19/11@0&0&1)] = A [■8(1/11−17/11 (2/45)&2/11−17/11 (4/45)&(−2)/11−17/11 ((−19)/90)@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(1−17/11(0)&0−17/11(0)&17/11−17/11(1)@0&1&19/11@0&0&1)] = A [■8(1/11 (1−34/45) &2/11 (1−34/35)&1/11 (−2+323/90)@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(1&0&0@0&1&19/11@0&0&1)] = A [■8(1/11×11/45&2/11×11/45&1/11×143/90@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] [■8(1&0&0@0&1&19/11@0&0&1)] = A [■8(1/45&2/45&13/90@5/11&(−1)/11&(−10)/11@2/45&4/45&(−19)/90)] We Make 19/11 as 1 𝑅_2 →𝑅_2 − 19/11 𝑅_3 [■8(1&0&0@0−19/11(0)&1−19/11(0)&19/11−19/11(1)@0&0&1)] = A [■8(1/45&2/11&13/90@5/11−19/11 (2/45)&(−1)/11−19/11 (4/45)&(−10)/11−19/11 ((−19)/11)@2/45&4/45&(−19)/90)] [■8(1&0&0@0&1&0@0&0&1)] = A [■8(1/45&2/45&13/90@1/11 (5−38/45)&(−1)/11×121/45&1/11×((−539)/90)@2/45&4/45&(−19)/90)] [■8(1&0&0@0&1&0@0&0&1)] = A [■8(1/45&2/45&13/90@17/45&(−11)/45&(−49)/90@2/45&4/45&(−19)/90)] I = AA−1 Therefore, A−1 = [■8(1/45&2/45&13/90@17/45&(−11)/45&(−49)/90@2/45&4/45&(−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.