Ex 3.3

Chapter 3 Class 12 Matrices
Serial order wise

### Transcript

Ex 3.3,3 If A’ = [■8(3&4@−1&2@0&1)] and B =[■8(−1&2&1@1&2&3)] , then verify that (i) (A + B)’ = A’ + B’ Solving L.H.S (A + B)’ First finding A + B Given A’ = [■8(3&4@−1&2@0&1)] A = (A’)’ = [■8(3&4@−1&2@0&1)]^′ = [■8(3&−1&0@4&2&1)] Now, A + B = [■8(𝟑&−𝟏&𝟎@𝟒&𝟐&𝟏)] + [■8(−𝟏&𝟐&𝟏@𝟏&𝟐&𝟑)] = [■8(3+(−1)&−1+2&0+1@4+1&2+2&1+3)] = [■8(𝟐&𝟏&𝟏@𝟓&𝟒&𝟒)] So, (A + B)’ = [■8(2&5@1&4@1&4)] Solving R.H.S. (A’ + B’) Given A’ = [■8(𝟑&𝟒@−𝟏&𝟐@𝟎&𝟏)] Also B = [■8(−1&2&1@1&2&3)] B’ = [■8(−𝟏&𝟏@𝟐&𝟐@𝟏&𝟑)] A’ + B’ = [■8(3&4@−1&2@0&1)] + [■8(−1&1@2&2@1&3)] = [■8(3+(−1)&4+1@−1+2&2+2@0+1&1+3)] = [■8(𝟐&𝟓@𝟏&𝟒@𝟏&𝟒)] = L.H.S Hence, L.H.S = R.H.S Hence proved Ex 3.3, 3 If A’ = [■8(3&4@−1&2@0&1)] and B =[■8(−1&2&1@1&2&3)] , then verify that (ii) (A – B)’ = A’ – B’ Solving L.H.S (A – B)’ First finding A – B A – B = [■8(3&−1&0@4&2&1)] – [■8(−1&2&1@1&2&3)] = [■8(3−(−1)&−1−2&0−1@4−1&2−2&1−3)] = [■8(𝟑+𝟏&−𝟑&−𝟏@𝟑&𝟎&−𝟐)] = [■8(4&−3&−1@3&0&−2)] (A – B)’ = [■8(𝟒&𝟑@−𝟑&𝟎@−𝟏&−𝟐)] Solving R.H.S A’ – B’ Given A’ = [■8(3&4@−1&2@0&1)] A’ – B’ = [■8(3&4@−1&2@0&1)] – [■8(−1&1@2&2@1&3)] = [■8(3−(−1)&4−1@−1−2&2−2@0−1&1−3)] = [■8(3+1&3@−3&0@−1&−2)] = [■8(𝟒&𝟑@−𝟑&𝟎@−𝟏&−𝟐)] = 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.