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Ex 3.2, 1 Let A = [■8(2&4@3&2)], B = [■8(1&3@−2&5)], C = [■8(−2&5@3&4)] Find each of the following (i) A + B A + B = [■8(2&4@3&2)] + [■8( 1&3@−2&5)] = [■8(2+1&4+3@3−2&2+5)] = [■8(𝟑&𝟕@𝟏&𝟕)] Ex 3.2,1 Let A = [■8(2&4@3&2)], B = [■8(1&3@−2&5)], C = [■8(−2&5@3&4)] Find each of the following (ii) A – B A – B = [■8(2&4@3&2)]− [■8( 1&3@−2&5)] = [■8(2−1&4−3@3−(−2)&2−5)] = [■8(1&1@3+2&−3)] = [■8(𝟏&𝟏@𝟓&−𝟑)] Ex 3.2, 1 Let A = [■8(2&4@3&2)], B = [■8(1&3@−2&5)], C = [■8(−2&5@3&4)] Find each of the following 3A – C Finding 3A 3A = 3[■8(2&4@3&2)] = [■8(3 × 2&3 × 4@3 × 3 &3 × 2)] = [■8(𝟔&𝟏𝟐@𝟗&𝟔)] Hence 3A – C = [■8(6&12@9&6)] ⤶7− [■8(−2&5@3&4)] = [■8(6−(−2)&12−5@9−3&6−4)] = [■8(6+2&7@6&2)] = [■8(𝟖&𝟕@𝟔&𝟐)] Ex 3.2, 1 Let A = [■8(2&4@3&2)] B = [■8(1&3@−2&5)] , C = [■8(−2&5@3&4)]. Find each of the following (iv)AB AB = [■8(2&4@3&2)] [■8(1&3@−2&5)] = [■8(2 × 1+4 × −2 &2 × 3+4 × 5@3 × 1+2 × −2&3 × 3+2 × 5)] = [■8(2−8&6+20@3−4&9+10)] = [■8(−𝟔&𝟐𝟔@−𝟏&𝟏𝟗)] Ex 3.2, 1 Let A = [■8(2&4@3&2)] B = [■8(1&3@−2&5)] , C = [■8(−2&5@3&4)]. Find each of the following (v) BA BA = [■8(1&3@−2&5)] [■8(2&4@3&2)] = [■8(1 × 2+3 × 3 &1 × 4+3 × 2@−2 × 2+5 × 3&−2 × 4+5 × 2)] = [■8(2+9&4+6@−4+15&−8+10)] = [■8(𝟏𝟏&𝟏𝟎@𝟏𝟏&𝟐)]

Ex 3.2

Ex 3.2, 1
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Ex 3.2, 2 (i)

Ex 3.2, 2 (ii) Important

Ex 3.2, 2 (iii)

Ex 3.2, 2 (iv)

Ex 3.2, 3 (i)

Ex 3.2, 3 (ii) Important

Ex 3.2, 3 (iii)

Ex 3.2, 3 (iv) Important

Ex 3.2, 3 (v)

Ex 3.2, 3 (vi) Important

Ex 3.2, 4

Ex 3.2, 5

Ex 3.2, 6

Ex 3.2, 7 (i)

Ex 3.2, 7 (ii) Important

Ex 3.2, 8

Ex 3.2, 9

Ex 3.2, 10

Ex 3.2, 11

Ex 3.2, 12 Important

Ex 3.2, 13 Important

Ex 3.2, 14

Ex 3.2, 15

Ex 3.2, 16 Important

Ex 3.2, 17 Important

Ex 3.2, 18

Ex 3.2, 19 Important

Ex 3.2, 20 Important

Ex 3.2, 21 (MCQ) Important

Ex 3.2, 22 (MCQ) Important

Chapter 3 Class 12 Matrices (Term 1)

Serial order wise

About the Author

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.