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Example 20 If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] Verify that (i) (A’)’ = A, A = [■8(3&√3&2@4&2&0)] A’ = [■8(3&√3&2@4&2&0)]^′= [■8(𝟑&𝟒@√𝟑&𝟐@𝟐&𝟎)] (A’)’ = [■8(3&4@√3&2@2&0)]^′= [■8(3&√3&2@4&2&0)] = A Thus (A’)’ = A Example 20 If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] Verify that (ii) (A + B)’ = A’ + B’, Solving L.H.S First finding (A + B) (A + B) = [■8(3&√3&2@4&2&0)] + [■8(2&−1&2@1&2&4)] = [■8(3+2 &√3+(−1)&2+2@4+1&2+2&0+4)] = [■8(5&√3−1&4@5&4&4)] Thus, (A + B)’ = [■8(𝟓&𝟓@√𝟑−𝟏&𝟒@𝟒&𝟒)] Solving R.H.S A’ + B’ Finding A’ A = [■8(3&√3&2@4&2&0)] A’ = [■8(𝟑&𝟒@√𝟑&𝟐@𝟐&𝟎)] Also, B = [■8(2&−1&2@1&2&4)] B‘ = [■8(𝟐&𝟏@−𝟏&𝟐@𝟐&𝟒)] Now, A’ + B’ =[■8(3&4@√3&2@2&0)] +[■8(2&1@−1&2@2&4)] = [■8(3+2&4+1@√3+(−1)&2+2@2+0&0+4)] = [■8(𝟓&𝟓@√𝟑−𝟏&𝟒@𝟒&𝟒)] = L.H.S Since L.H.S = R.H.S Hence Proved Example 20 If A = [■8(3&√3&2@4&2&0)] and B = [■8(2&−1&2@1&2&4)] .Verify that (iii) (kB)’ = kB’, where k is any constant. Solving L.H.S (kB)’ Finding kB first kB = k [■8(2&−1&2@1&2&4)] = [■8(2𝑘&−𝑘&2𝑘@𝑘&2𝑘&4𝑘)] (kB)’ = [■8(𝟐𝒌&𝒌@−𝒌&𝟐𝒌@𝟐𝒌&𝟒𝒌)] Solving R.H.S kB’ Finding B’ first B = [■8(2&−1&2@1&2&4)] B’ = [■8(𝟐&𝟏@−𝟏&𝟐@𝟐&𝟒)] kB’ = k[■8(2&1@−1&2@2&4)] = [■8(2𝑘&𝑘@−𝑘&2𝑘@2𝑘&4𝑘)] = L.H.S Since 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.