Ex 3.3, 2 - Chapter 3 Class 12 Matrices
Last updated at April 16, 2024 by Teachoo
Ex 3.3
Ex 3.3, 2 You are here
Ex 3.3, 3
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 April 16, 2024 by Teachoo
Ex 3.3, 2 If A = [■8(−1&2&3@5&7&9@−2&1&1)] and B= [■8(−4&1&−5@1&2&0@1&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&3@5&7&9@−2&1&1)] + [■8(−4&1&−5@1&2&0@1&3&1)] = [■8(−1+(−4)&2+1&3+(−5)@5+1&7+2&9+0@−2+1&1+3&1+1)] = [■8(−5&3&−2@6&9&9@−1&4&2)] Thus, A + B = [■8(−5&3&−2@6&9&9@−1&4&2)] (A + B)’ = [■8(−𝟓&𝟔&−𝟏@𝟑&𝟗&𝟒@−𝟐&𝟗&𝟐)] Solving R.H.S A’ + B’ First we will calculate A’ and B’ A = [■8(−1&2&3@5&7&9@−2&1&1)] A’ =[■8(−1&5&−2@2&7&1@3&9&1)] B = [■8(−4&1&−5@1&2&0@1&3&1)] B’ = [■8(−4&1&1@1&2&3@−5&0&1)] Now, A’ + B’ = [■8(−1&5&−2@2&7&1@3&9&1)]+[■8(−4&1&1@1&2&3@−5&0&1)] = [■8(−1+(−4)&5+1&−2+1@2+1&7+2&1+3@3+(−5)&9+0&1+1)] =[■8(−𝟓&𝟔&−𝟏@𝟑&𝟗&𝟒@−𝟐&𝟗&𝟐)] = L.H.S Hence Proved Ex 3.3, 2 If A = [■8(−1&2&3@5&7&9@−2&1&1)] and B= [■8(−4&1&−5@1&2&0@1&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&3@5&7&9@−2&1&1)] – [■8(−4&1&−5@1&2&0@1&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&3+5@4&5&9@−3&−2&0)] = [■8(𝟑&𝟏&𝟖@𝟒&𝟓&𝟗@−𝟑&−𝟐&𝟎)] Thus, A – B = [■8(3&1&8@4&5&9@−3&−2&0)] Now, (A – B)’ = [■8(𝟑&𝟒&−𝟑@𝟏&𝟓&−𝟐@𝟖&𝟗&𝟎)] Solving R.H.S A’ – B’ First we will calculate A’ and B’ A = [■8(−1&2&3@5&7&9@−2&1&1)] A’ = [■8(−1&5&−2@2&7&1@3&9&1)] Now, A’ – B’ = [■8(−1&5&−2@2&7&1@3&9&1)]−[■8(−4&1&1@1&2&3@−5&0&1)] = [■8(−1−(−4)&5−1&−2−1@2−1&7−2&1−3@3−(−5)&9−0&1−1)] = [■8(−1+4&4&−3@1&5&−2@3+5&9&0)] = [■8(𝟑&𝟒&−𝟑@𝟏&𝟓&−𝟐@𝟖&𝟗&𝟎)] = L.H.S Hence L.H.S = R.H.S Hence Proved