Ex 3.3

Chapter 3 Class 12 Matrices
Serial order wise

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Ex 3.3,3 If A’ = [■8(3&[email protected]−1&[email protected]&1)] and B =[■8(−1&2&[email protected]&2&3)] , then verify that (i) (A + B)’ = A’ + B’ Solving L.H.S (A + B)’ First finding A + B Given A’ = [■8(3&[email protected]−1&[email protected]&1)] A = (A’)’ = [■8(3&[email protected]−1&[email protected]&1)]^′ = [■8(3&−1&[email protected]&2&1)] Now, A + B = [■8(3&−1&[email protected]&2&1)] + [■8(−1&2&[email protected]&2&3)] = [■8(3+(−1)&−1+2&[email protected]+1&2+2&1+3)] = [■8(2&1&[email protected]&4&4)] So, (A + B)’ = [■8(2&[email protected]&[email protected]&4)] Solving R.H.S. (A’ + B’) Given A’ = [■8(3&[email protected]−1&[email protected]&1)] Also B = [■8(−1&2&[email protected]&2&3)] B’ = [■8(−1&[email protected]&[email protected]&3)] A’ + B’ = [■8(3&[email protected]−1&[email protected]&1)] + [■8(−1&[email protected]&[email protected]&3)] = [■8(3+(−1)&[email protected]−1+2&[email protected]+1&1+3)] = [■8(2&[email protected]&[email protected]&4)] = L.H.S Hence, L.H.S = R.H.S Hence proved Ex 3.3, 3 If A’ = [■8(3&[email protected]−1&[email protected]&1)] and B =[■8(−1&2&[email protected]&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&[email protected]&2&1)] – [■8(−1&2&[email protected]&2&3)] = [■8(3−(−1)&−1−2&0−[email protected]−1&2−2&1−3)] = [■8(3+1&−3&−[email protected]&0&−2)] = [■8(4&−3&−[email protected]&0&−2)] (A – B)’ = [■8(4&[email protected]−3&[email protected]−1&−2)] Solving R.H.S A’ – B’ Given A’ = [■8(3&[email protected]−1&[email protected]&1)] B = [■8(−1&2&[email protected]&2&3)] B’ = [■8(−1&[email protected]&[email protected]&3)] A’ – B’ = [■8(3&[email protected]−1&[email protected]&1)] – [■8(−1&[email protected]&[email protected]&3)] = [■8(3−(−1)&4−[email protected]−1−2&2−[email protected]−1&1−3)] = [■8(3+1&[email protected]−3&[email protected]−1&−2)] = [■8(4&[email protected]−3&[email protected]−1&−2)] = L.H.S Hence, L.H.S = R.H.S Hence proved