



Introducing your new favourite teacher - Teachoo Black, at only ₹83 per month
Inverse of matrix using elementary transformation
Inverse of a matrix
Finding inverse of a matrix using Elementary Operations
Ex 3.4, 18 (MCQ)
Example 23
Ex 3.4, 1
Ex 3.4, 2
Ex 3.4, 3
Ex 3.4, 4
Ex 3.4, 5
Ex 3.4, 6
Ex 3.4, 7
Ex 3.4, 8 Important
Ex 3.4, 9
Ex 3.4, 10
Ex 3.4, 11
Ex 3.4, 13
Example 25 Important
Ex 3.4, 12
Ex 3.4, 14
Example 24 Important You are here
Ex 3.4, 15 Important
Ex 3.4, 16
Ex 3.4, 17 Important
Inverse of matrix using elementary transformation
Last updated at Jan. 17, 2020 by Teachoo
Example 24 Obtain the inverse of the following matrix using elementary operations A = [■8(0&1&2@1&2&3@3&1&1)] Given A = [■8(0&1&2@1&2&3@3&1&1)] We know that A = IA [■8(0&1&2@1&2&3@3&1&1)] = [■8(1&0&0@0&1&0@0&0&1)] A R1↔R2 [■8(𝟏&2&3@0&1&2@3&1&1)] = [■8(0&1&0@1&0&0@0&0&1)] A R3 → R3 – 3R1 [■8(1&2&3@0&1&2@𝟑−𝟑(𝟏)&1−3(2)&1−3(3))] = [■8(0&1&0@1&0&0@0−3(0)&0−3(1)&1−3(0))]A [■8(1&2&3@0&1&2@𝟎&−5&−8)] = [■8(0&1&0@1&0&0@0&−3&1)] R1 → R1 – 2R2 [■8(1−2(0)&𝟐−𝟐(𝟏)&3−2(2)@0&1&2@0&−5&−8)] = [■8(0−2(1)&1−2(0)&0−2(0)@1&0&0@0&−3&1)]A [■8(1&𝟎&−1@0&1&2@0&−5&−8)] = [■8(−2&1&0@1&0&0@0&−3&1)] A R3 → R3 + 5R2 [■8(1&0&−1@0&1&2@0+5(0)&−𝟓+𝟓(𝟏)&−8+5(2))] = [■8(−2&1&0@1&0&0@0+5(1)&−3+5(0)&1+5(0))] A [■8(1&0&−1@0&1&2@0&𝟎&2)] = [■8(−2&1&0@1&0&0@5&−3&1)] A R3 → 1/2 R3 [■8(1&0&−1@0&1&2@0/2&0/2&𝟐/𝟐)] = [■8(−2&1&0@1&0&0@5/2&(−3)/2&1/2)] A R1 → R1 + R3 [■8(1+0&0+0&−𝟏+𝟏@0&1&2@0&0&1)]=[■8(−2+5/2&1+((−3)/2)&0+1/2@1&0&0@5/2&(−3)/2&1/2)] A [■8(1&0&𝟎@0&1&2@0&0&1)] = [■8(1/2&(−1)/2&1/2@1&0&0@5/2&(−3)/2&1/2)] A R2 → R2 – 2R3 [■8(1&0&0@0−2(0)&1−2(0)&𝟐−𝟐(𝟏)@0&0&1)] = [■8(1/2&(−1)/2&1/2@1−2(5/2)&0−2((−3)/2)&0−2(1/2)@5/2&(−3)/2&1/2)]A [■8(1&0&0@0&1&𝟎@0&0&1)] = [■8(1/2&(−1)/2&1/2@−4&3&−1@5/2&(−3)/2&1/2)] A I= [■8(1/2&(−1)/2&1/2@−4&3&−1@5/2&(−3)/2&1/2)] A This is similar to I = A-1 A Hence, A-1 = [■8(𝟏/𝟐&(−𝟏)/𝟐&𝟏/𝟐@−𝟒&𝟑&−𝟏@𝟓/𝟐&(−𝟑)/𝟐&𝟏/𝟐)]