



Inverse of matrix using elementary transformation
Inverse of matrix using elementary transformation
Last updated at Dec. 16, 2024 by Teachoo
Ex3.4, 17 Find the inverse of each of the matrices, if it exists. [■8(2&0&−1@5&1&0@0&1&3)] Let A =[■8(2&0&−1@5&1&0@0&1&3)] We know that A = IA [■8(2&0&−1@5&1&0@0&1&3)]= [■8(1&0&0@0&1&0@0&0&1)] A R1 →1/2 R1 , [■8(𝟐/𝟐&0/2&(−1)/2@5&1&0@0&1&3)]= [■8(1/2&0/2&0/2@0&1&0@0&0&1)] A [■8(𝟏&0&−1/2@5&1&0@0&1&3)] = [■8(1/2&0&0@0&1&0@0&0&1)] A R2 → R2 – 5R1 [■8(1&0&−1/2@𝟓−𝟓(𝟏)&1−5(0)&0−5((−1)/2)@0&1&3)] = [■8(1/2&0&0@0−5(1/2)&1−5(0)&0−5(0)@0&0&1)] A [■8(1&0&−1/2@𝟎&1&5/2@0&0&3)]= [■8(1/2&0&0@−5/2&1&0@0&0&1)] A R3 → R3 – R2 [■8(1&0&−1/2@0&1&5/2@𝟎&0&3−5/2)]= [■8(1/2&0&0@−5/2&1&0@0−(−5/2)&0−1&1−0)] A [■8(1&0&−1/2@0&1&5/2@𝟎&0&1/2)]= [■8(1/2&0&0@−5/2&1&0@5/2&−1&1)] A R3 → 2R3 [■8(1&0&−1/2@0&1&5/2@2 × 0&2 × 0&𝟐 × 𝟏/𝟐)] = [■8(1/2&0&0@−5/2&1&0@2 × 5/2&2 × (−1)&2 × 1)] A [■8(1&0&−1/2@0&1&5/2@0&0&𝟏)]= [■8(1/2&0&0@−5/2&1&0@5&−2&2)] A R1 → R1 + 1/2 R3 [■8(1+1/2(0)&0+1/2(0)&(−𝟏)/𝟐+𝟏/𝟐(𝟏)@0&1&5/2@0&0&1)]= [■8(1/2+1/2(5)&0+1/2(−2)&0+1/2(2)@(−5)/2&1&0@5&−2&2)] A [■8(1&0&𝟎@0&1&5/2@0&0&1)]= [■8(3&−1&1@(−5)/2&1&0@5&−2&2)] A R2 → R2 − 5/2 R3 [■8(1&0&0@0−5/2(0)&1−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)@0&0&1)] = [■8(3&−1&1@(−5)/2−5/2 (5)&1−5/2 (−2)&0−5/2@5&−2&2) (2)] A [■8(1&0&0@0&1&𝟎@0&0&1)]= [■8(3&−1&1@(−30)/2&6&−5@5&−2&2)] A "I" = [■8(3&−1&1@−15&6&−5@5&−2&2)] A This is similar to I = A-1A Thus, A-1 = [■8(𝟑&−𝟏&𝟏@−𝟏𝟓&𝟔&−𝟓@𝟓&−𝟐&𝟐)]