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Ex 3.3
Ex 3.3, 2
Ex 3.3, 3 You are here
Ex 3.3, 4 Important
Ex 3.3, 5 (i)
Ex 3.3, 5 (ii)
Ex 3.3, 6 (i)
Ex 3.3, 6 (ii) Important
Ex 3.3, 7 (i)
Ex 3.3, 7 (ii) Important
Ex 3.3, 8
Ex 3.3, 9
Ex 3.3, 10 (i) Important
Ex 3.3, 10 (ii)
Ex 3.3, 10 (iii) Important
Ex 3.3, 10 (iv)
Ex 3.3, 11 (MCQ) Important
Ex 3.3, 12 (MCQ)
Last updated at June 8, 2023 by Teachoo
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