Ex 3.2, 4 - Chapter 3 Class 12 Matrices
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 3.2, 4 If A = [■8(1&2&−3@5&0&2@1&−1&1)], B = [■8(3&−1&2@4&2&5@2&0&3)] and , C = [■8(4&1&2@0&3&2@1&−2&3)] then compute (A+B) and (B – C) . Also, verify that A + (B – C) = (A + B) – C Calculating A + B A + B = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(3&−1&2@4&2&5@2&0&3)] = [■8(1+3&2−1&−3+2@5+4&0+2&2+5@1+2&−1+0&1+3)] = [■8(𝟒&𝟏&−𝟏@𝟗&𝟐&𝟕@𝟑&−𝟏&𝟒)] Calculating B – C B – C = [■8(3&−1&2@4&2&5@2&0&3)] – [■8(4&1&2@0&3&2@1&−2&3)] = [■8(3−4&−1−1&2−2@4−0&2−3&5−2@2−1&0−(−2)&3−3)] = [■8(−𝟏&−𝟐&𝟎@𝟒&−𝟏&𝟑@𝟏&𝟐&𝟎)] We need to verify A + (B – C) = (A + B) – C Solving L.H.S A + (B – C) = [■8(1&2&−3@5&0&2@1&−1&1)]+ [■8(−1&−2&0@4&−1&3@1&2&0)] = [■8(1−1&2−2&−3+0@5+4&0−1&2+3@1+1&−1+2&1+0)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] Solving R.H.S (A + B) – C = [■8(4&1&−1@9&2&7@3&−1&4)]− [■8(4&1&2@0&3&2@1&−2&3)] = [■8(4−4&1−1&−1−2@9−0&2−3&7−2@3−1&−1+2&4−3)] = [■8(𝟎&𝟎&−𝟑@𝟗&−𝟏&𝟓@𝟐&𝟏&𝟏)] = L.H.S Hence L.H.S = R.H.S Hence proved
Ex 3.2
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 You are here
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
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo