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Ex 3.3, 2 - Verify (A + B)' = A' + B' if A = [-1 2 3 5 7 9

Ex 3.3, 2 - Chapter 3 Class 12 Matrices - Part 2
Ex 3.3, 2 - Chapter 3 Class 12 Matrices - Part 3 Ex 3.3, 2 - Chapter 3 Class 12 Matrices - Part 4 Ex 3.3, 2 - Chapter 3 Class 12 Matrices - Part 5 Ex 3.3, 2 - Chapter 3 Class 12 Matrices - Part 6

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Ex 3.3, 2 If A = [■8(−1&2&[email protected]&7&9@−2&1&1)] and B= [■8(−4&1&−[email protected]&2&[email protected]&3&1)] , then verify that (i) (A + B)’ = A’ + B’ Solving L.H.S (A + B)’ First we will calculate A + B A + B = [■8(−1&2&[email protected]&7&9@−2&1&1)] + [■8(−4&1&−[email protected]&2&[email protected]&3&1)] = [■8(−1+(−4)&2+1&3+(−5)@5+1&7+2&9+0@−2+1&1+3&1+1)] = [■8(−5&3&−[email protected]&9&9@−1&4&2)] Thus, A + B = [■8(−5&3&−[email protected]&9&9@−1&4&2)] (A + B)’ = [■8(−5&6&−[email protected]&9&4@−2&9&2)] Solving R.H.S A’ + B’ First we will calculate A’ and B’ A = [■8(−1&2&[email protected]&7&9@−2&1&1)] A’ =[■8(−1&5&−[email protected]&7&[email protected]&9&1)] B = [■8(−4&1&−[email protected]&2&[email protected]&3&1)] B’ = [■8(−4&1&[email protected]&2&3@−5&0&1)] Now, A’ + B’ = [■8(−1&5&−[email protected]&7&[email protected]&9&1)]+[■8(−4&1&[email protected]&2&3@−5&0&1)] = [■8(−1+(−4)&5+1&−[email protected]+1&7+2&[email protected]+(−5)&9+0&1+1)] =[■8(−5&6&−[email protected]&9&4@−2&9&2)] = L.H.S Hence Proved Ex 3.3, 2 If A = [■8(−1&2&[email protected]&7&9@−2&1&1)] and B= [■8(−4&1&−[email protected]&2&[email protected]&3&1)] , then verify that (ii) (A – B)’ = A’ – B’ Solving L.H.S (A – B)’ First we will calculate A – B A – B = [■8(−1&2&[email protected]&7&9@−2&1&1)] – [■8(−4&1&−[email protected]&2&[email protected]&3&1)] = [■8(−1−(−4)&2−1&3−(−5)@5−1&7−2&9−0@−2−1&1−3&1−1)] = [■8(−1+4&1&[email protected]&5&9@−3&−2&0)] = [■8(3&1&[email protected]&5&9@−3&−2&0)] Thus, A – B = [■8(3&1&[email protected]&5&9@−3&−2&0)] Now, (A – B)’ = [■8(3&4&−[email protected]&5&−[email protected]&9&0)] Solving R.H.S A’ – B’ First we will calculate A’ and B’ A = [■8(−1&2&[email protected]&7&9@−2&1&1)] A’ = [■8(−1&5&−[email protected]&7&[email protected]&9&1)] B = [■8(−4&1&−[email protected]&2&[email protected]&3&1)] B’ = [■8(−4&1&[email protected]&2&3@−5&0&1)] Now, A’ – B’ = [■8(−1&5&−[email protected]&7&[email protected]&9&1)]−[■8(−4&1&[email protected]&2&3@−5&0&1)] = [■8(−1−(−4)&5−1&−2−[email protected]−1&7−2&1−[email protected]−(−5)&9−0&1−1)] = [■8(−1+4&4&−[email protected]&5&−[email protected]+5&9&0)] = [■8(3&4&−[email protected]&5&−[email protected]&9&0)] = 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.