Miscellaneous

Chapter 3 Class 12 Matrices
Serial order wise

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### Transcript

Question 3 If A = [■8(3&−[email protected]&−1)] , then prove An = [■8(1+2n&−[email protected]&1−2n)] where n is any positive integer We shall prove the result by using mathematical induction. Step 1: P(n): If A= [■8(3&−[email protected]&−1)] , then An = [■8(1+2n&−[email protected]&1−2n)] , n ∈ N Step 2: Prove for n = 1 For n = 1 L.H.S = A1 = A = [■8(3&−[email protected]&−1)] R.H.S = [■8(1+2(1)&−4(1)@1&1−2(1))]=[■8(1+2&−[email protected]&1−2)]" = " [■8(3&−[email protected]&−1)] L.H.S = R.H.S ∴ P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k+1) is true Assume that P (k) is true P(k) : If A= [■8(3&−[email protected]&−1)] , then Ak = [■8(1+2k&−[email protected]&1−2k)] We will have to prove that P( k +1) is true P(k + 1) : If A= [■8(3&−[email protected]&−1)] , then Ak+1 = [■8(1+2(k+1)&−4(k+1)@(k+1)&1−2(k+1))] Taking L.H.S Ak+1 = Ak . A = [■8(1+2k&−[email protected]&1−2k)] [■8(3&−[email protected]&−1)] = [■8((1+2k)3−4k(1)&(1+2k)(−4)−4k(−1)@k(3)+(1−2k)1&k(−4)+(1−2k)(−1))] = [■8(3+6k−4k&−4−[email protected]+1−2k&−4k−1+2k)] = [■8(3+2k&−4−[email protected]+k&−1−2k)] = [■8(1+2(k+1)&−4(k+1)@1+k&1−2(k+1))] = R.H.S Thus P (k + 1) is true ∴ By the principal of mathematical induction , P(n) is true for n ∈ N Hence, if A= [■8(3&−[email protected]&−1)] , then An = [■8(1+2n&−[email protected]&1−2n)] , n ∈ N

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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.