




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
Ex 3.4, 16 Deleted for CBSE Board 2022 Exams You are here
Ex 3.4, 17 Important Deleted for CBSE Board 2022 Exams
Ex 3.4, 18 (MCQ) Deleted for CBSE Board 2022 Exams
Ex 3.4, 16 Find the inverse of each of the matrices, if it exists. [■8(1&3&−2@−3&0&−5@2&5&0)] Let A = [■8(1&3&−2@−3&0&−5@2&5&0)] A = IA [■8(1&3&−2@−3&0&−5@2&5&0)] = [■8(1&0&0@0&1&0@0&0&1)] A R2 → R2 + 3R1 [■8(1&3&−2@−𝟑+𝟑(𝟏)&0+3(3)&−5+3(−2)@2&5&0)] = [■8(1&0&0@0+3(1)&1+3(0)&0+3(0)@0&0&1)] A [■8(1&3&−2@𝟎&9&−11@2&5&0)] = [■8(1&0&0@3&1&0@0&0&1)] A R3 → R3 − 2R1 [■8(1&3&−2@0&9&−11@𝟐−𝟐(𝟏)&5−2(3)&0−2(−2))] = [■8(1&0&0@3&1&0@0−2(1)&0−2(0)&1−2(0))] A [■8(1&3&−2@0&9&−11@𝟎&−1&4)] = [■8(1&0&0@3&1&0@−2&0&1)] A R2 → 1/9 R2 [■8(1&3&−2@0/9&𝟗/𝟗&(−11)/9@0&−1&4)]= [■8(1&0&0@3/9&1/9&0/9@−2&0&1)]A [■8(1&3&−2@0&𝟏&(−11)/9@0&−1&4)] = [■8(1&0&0@1/3&1/9&0@−2&0&1)] A R3 → R3 + R2 [■8(1&3&−2@0&1&(−11)/9@0+0&−𝟏+𝟏&4+((−11)/9) )] = [■8(1&0&0@1/3&1/9&0@−2+1/3&0+1/9&1+0)] A [■8(1&3&−2@0&1&(−11)/9@0&𝟎&25/9)] = [■8(1&0&0@1/3&1/9&0@(−5)/3&1/9&1)] A R1 → R1 − 3R2 [■8(1−3(0)&𝟑−𝟑(𝟏)&−2−3((−11)/9)@0&1&(−11)/9@0&0&25/3)] =[■8(1−3(1/3)&0−3(1/9)&0−3(0)@1/3&1/9&0@(−5)/3&1/9&1)] A ` [■8(1&𝟎&5/3@0&1&(−11)/9@0&0&25/9)] = [■8(0&(−1)/3&0@1/3&1/9&0@(−5)/3&1/9&1)] A R3 → 9/25 R3 [■8(1&0&5/3@0&1&(−11)/9@0&0&𝟏)] = [■8(0&(−1)/3&0@1/3&1/9&0@(−3)/5&1/25&9/25)] A R2 → R2 + 11/9 R3 [■8(1&0&5/3@0+11/9 (0)&1+11/9(0)&(−𝟏𝟏)/𝟗+𝟏𝟏/𝟗(𝟏)@0&0&1)] =[■8(0&(−1)/3&0@1/3+11/9 ((−3)/5)&1/9+11/9 (1/25)&0+11/9 (9/25)@(−3)/5&1/25&9/25)] A [■8(1&0&5/3@0&1&𝟎@0&0&1)] = [■8(0&(−1)/3&0@(−2)/5&4/25&11/25@(−3)/5&1/25&9/25)] A R1 → R1 – 5/3 R3 [■8(1− 5/3 (0)&0− 5/3 (0)&𝟓/𝟑− 𝟓/𝟑 (𝟏)@0&1&0@0&0&1)] =[■8(0−5/3 ((−3)/5)&(−1)/3− 5/3 (1/25)&0−5/3 (9/25)@(−2)/5&4/25&11/25@(−3)/5&1/25&9/25)] A [■8(1&0&𝟎@0&1&0@0&0&1)] = [■8(5/5&(−6)/15 &(−3)/5@(−2)/5&4/25&11/25@(−3)/5&1/25&9/25)] A I = [■8(1&(−2)/5&(−3)/5@(−2)/5&4/25&11/25@(−3)/5&1/25&9/25)] A This is similar to I = A-1 A Hence, A-1 = [■8(𝟏&(−𝟐)/𝟓&(−𝟑)/𝟓@(−𝟐)/𝟓&𝟒/𝟐𝟓&𝟏𝟏/𝟐𝟓@(−𝟑)/𝟓&𝟏/𝟐𝟓&𝟗/𝟐𝟓)]