




Ex 3.4
Ex 3.4, 2 Deleted for CBSE Board 2022 Exams
Ex 3.4, 3 Deleted for CBSE Board 2022 Exams
Ex 3.4, 4 Deleted for CBSE Board 2022 Exams
Ex 3.4, 5 Deleted for CBSE Board 2022 Exams
Ex 3.4, 6 Deleted for CBSE Board 2022 Exams
Ex 3.4, 7 Deleted for CBSE Board 2022 Exams
Ex 3.4, 8 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 9 Deleted for CBSE Board 2022 Exams
Ex 3.4, 10 Deleted for CBSE Board 2022 Exams
Ex 3.4, 11 Deleted for CBSE Board 2022 Exams
Ex 3.4, 12 Deleted for CBSE Board 2022 Exams
Ex 3.4, 13 Deleted for CBSE Board 2022 Exams
Ex 3.4, 14 Deleted for CBSE Board 2022 Exams
Ex 3.4, 15 Important Deleted for CBSE Board 2022 Exams You are here
Ex 3.4, 16 Deleted for CBSE Board 2022 Exams
Ex 3.4, 17 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 18 (MCQ) Deleted for CBSE Board 2022 Exams
Last updated at Jan. 17, 2020 by Teachoo
Ex 3.4, 15 Find the inverse of each of the matrices, if it exists. [■8(2&−3&3@2&2&3@3&−2&2)] Let A =[■8(2&−3&3@2&2&3@3&−2&2)] We know that A = IA [■8(2&−3&3@2&2&3@3&−2&2)]= [■8(1&0&0@0&1&0@0&0&1)] A R1 →R_1/2 [■8(𝟐/𝟐&(−3)/2&3/2@2&2&3@3&−2&2)]= [■8(1/2&0/2&0/2@0&1&0@0&0&1)] A [■8(𝟏&(−3)/2&3/2@2&2&3@3&−2&2)]= [■8(1/2&0&0@0&1&0@0&0&1)] A R2 → R2 – 2R1 [■8(1&(−3)/2&3/2@𝟐−𝟐(𝟏)&2−2((−3)/2)&3−2(3/2)@3&−2&2)] = [■8(1/2&0&0@0−2(1/2)&1−2(0)&0−2(0)@0&0&1)] A [■8(1&(−3)/2&3/2@𝟎&5&0@3&−2&2)] = [■8(1/2&0&0@−1&1&0@0&0&1)] A R3 → R3 – 3R1 [■8(1&(−3)/2&3/2@0&5&0@𝟑−𝟑(𝟏)&−2−3((−3)/2)&2−3(3/2) )]= [■8(1/2&0&0@−1&1&0@0−3(1/2)&0−3(0)&1−3(0))] A [■8(1&(−3)/2&3/2@0&5&0@𝟎&5/2&(−5)/2)]= [■8(1/2&0&0@−1&1&0@(−3)/2&0&1)] A R2 →R_2/5 , [■8(1&(−3)/2&3/2@0&𝟓/𝟓&0@0&5/2&(−5)/2)] = [■8(1/2&0&0@(−1)/5&1/5&0@(−3)/2&0&1)] A [■8(1&(−3)/2&3/2@0&𝟏&0@0&5/2&(−5)/2)] = [■8(1/2&0&0@(−1)/5&1/5&0@(−3)/2&0&1)] A R1 → R1 + 3/2 R2 [■8(1+3/2(0)&(−𝟑)/𝟐+𝟏(𝟑/𝟐)&3/2+3/2(0)@0&1&0@0&5/2&(−5)/2)]= [■8(1/2+3/2 ((−1)/5)&0+3/2 (1/5)&0+3/2(0)@(−1)/5&1/5&0@(−3)/2&0&1)] A [■8(1&𝟎&3/2@0&1&0@0&5/2&(−5)/2)]= [■8(2/10&3/10&0@(−1)/5&1/5&0@(−3)/2&0&1)] A R3 → R3 – 5/2 R2 [■8(1&0&3/2@0&1&0@0−5/2(0)&𝟓/𝟐−𝟓/𝟐(𝟏)&(−5)/2−(−5)/2(0))]= [■8(1/5&3/10&0@(−1)/5&1/5&0@(−3)/2−5/2 ((−1)/5)&0−5/2 (1/5)&1−5/2(0))] A [■8(1&0&3/2@0&1&0@0&𝟎&(−5)/2)] = [■8(1/5&3/10&0@(−1)/5&1/5&0@−1&(−1)/2&1)] A R3 → (−2)/5 R3 [■8(1&0&3/2@0&1&0@0((−2)/5)&0(−2/5)&(−𝟓)/𝟐 ((−𝟐)/𝟓) )]= [■8(1/5&3/10&0@(−1)/5&1/5&0@−1((−2)/5)&(−1)/2 ((−2)/5)&1((−2)/5) )] A [■8(1&0&3/2@0&1&0@0&0&𝟏)] = [■8(1/5&3/10&0@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A R1 → R1 −3/2 R3 [■8(1−3/2(0)&0−3/2(0)&𝟑/𝟐−𝟑/𝟐(𝟏)@0&1&0@0&0&1)]= [■8(1/5−3/2 (2/5)&3/10−3/2 (1/5)&0−3/2 ((−2)/5)@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A [■8(1&0&𝟎@0&1&0@0&0&1)] = [■8((−2)/5&0&3/5@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A I = [■8((−2)/5&0&3/5@(−1)/5&1/5&0@2/5&1/5&(−2)/5)] A This is similar to I = A-1A Thus, A-1 = [■8((−𝟐)/𝟓&𝟎&𝟑/𝟓@(−𝟏)/𝟓&𝟏/𝟓&𝟎@𝟐/𝟓&𝟏/𝟓&(−𝟐)/𝟓)]