Question 17 - Inverse of matrix using elementary transformation - Chapter 3 Class 12 Matrices

Transcript

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)]