Misc 3 - If A = [3 -4 1 -1], prove An = [1 + 2n -4n n 1 - 2n]

Misc. 3 - Chapter 3 Class 12 Matrices - Part 2
Misc. 3 - Chapter 3 Class 12 Matrices - Part 3 Misc. 3 - Chapter 3 Class 12 Matrices - Part 4

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Question 3 If A = [■8(3&−4@1&−1)] , then prove An = [■8(1+2n&−4n@n&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&−4@1&−1)] , then An = [■8(1+2n&−4n@n&1−2n)] , n ∈ N Step 2: Prove for n = 1 For n = 1 L.H.S = A1 = A = [■8(3&−4@1&−1)] R.H.S = [■8(1+2(1)&−4(1)@1&1−2(1))]=[■8(1+2&−4@1&1−2)]" = " [■8(3&−4@1&−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&−4@1&−1)] , then Ak = [■8(1+2k&−4k@k&1−2k)] We will have to prove that P( k +1) is true P(k + 1) : If A= [■8(3&−4@1&−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&−4k@k&1−2k)] [■8(3&−4@1&−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&−4−8k+4k@3k+1−2k&−4k−1+2k)] = [■8(3+2k&−4−4k@1+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&−4@1&−1)] , then An = [■8(1+2n&−4n@n&1−2n)] , n ∈ N

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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.