# Ex 3.4, 17 - Chapter 3 Class 12 Matrices

Last updated at Jan. 17, 2020 by Teachoo

Last updated at Jan. 17, 2020 by Teachoo

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)]

Ex 3.4

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Ex 3.4, 17 Important Deleted for CBSE Board 2021 Exams only You are here

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.