Ex 3.2, 4 - Compute (A + B), (B - C). Also verify A + (B - C) - Ex 3.2

part 2 - Ex 3.2, 4 - Ex 3.2 - Serial order wise - Chapter 3 Class 12 Matrices
part 3 - Ex 3.2, 4 - Ex 3.2 - Serial order wise - Chapter 3 Class 12 Matrices

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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

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Davneet Singh

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