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Misc 2 - If A = [1 1 1 1 1 1], prove An = [3n-1 3n-1 3n-1 - Proof using mathematical induction

  1. Chapter 3 Class 12 Matrices
  2. Serial order wise
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Misc. 2 If A = [โ– 8(1&1&1@1&1&1@1&1&1)] , prove that An = [โ– 8(3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1) )], n โˆˆ N We shall prove the result by using mathematical induction Step 1: P(n) : If A = [โ– 8(1&1&1@1&1&1@1&1&1)] , An = [โ– 8(3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1) )] Step 2: Prove for n = 1 For n = 1 L.H.S = A1 = A = [โ– 8(1&1&1@1&1&1@1&1&1)] R.H.S = [โ– 8(3^(1โˆ’1)&3^(1โˆ’1)&3^(1โˆ’1)@3^(1โˆ’1)&3^(1โˆ’1)&3^(1โˆ’1)@3^(1โˆ’1)&3^(1โˆ’1)&3^(1โˆ’1) )] = [โ– 8(30&30&30@30&30&30@30&30&30)] = [โ– 8(1&1&1@1&1&1@1&1&1)] So, L.H.S = R.H.S โˆด P(n) is true for n = 1 Step 3: Assume P(k) to be true and then prove P(k+1) is true Assuming P(k) is true P(k): If A = [โ– 8(1&1&1@1&1&1@1&1&1)] , Ak = [โ– 8(3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1)@3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1)@3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1) )] We will prove that P (k + 1)is true P (k + 1): If A = [โ– 8(1&1&1@1&1&1@1&1&1)] , then Ak+1 = [โ– 8(3^((k+1)โˆ’1)&3^((k+1)โˆ’1)&3^((k+1)โˆ’1)@3^((k+1)โˆ’1)&3^((k+1)โˆ’1)&3^((k+1)โˆ’1)@3^((k+1)โˆ’1)&3^((k+1)โˆ’1)&3^((k+1)โˆ’1) )] Ak+1 = [โ– 8(3^(k+1โˆ’1)&3^(k+1โˆ’1)&3^(k+1โˆ’1)@3^(k+1โˆ’1)&3^(k+1โˆ’1)&3^(k+1โˆ’1)@3^(k+1โˆ’1)&3^(k+1โˆ’1)&3^(k+1โˆ’1) )] Consider L.H.S Ak +1 = Ak . A1 =[โ– 8(3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1)@3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1)@3^(kโˆ’1)&3^(kโˆ’1)&3^(kโˆ’1) )][โ– 8(1&1&1@1&1&1@1&1&1)] = [โ– 8(3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)@3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)@3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1)&3^(kโˆ’1) (1)+3^(kโˆ’1) (1)+3^(kโˆ’1) (1) )] = [โ– 8(3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)@3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)@3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1)&3^(kโˆ’1) ใ€–+3ใ€—^(kโˆ’1)+3^(kโˆ’1) )] = [โ– 8(ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1)@ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1)@ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1)&ใ€–3.3ใ€—^(๐‘˜โˆ’1) )]= [โ– 8(3^k&3^k&3^k@3^k&3^k&3^k@3^k&3^k&3^k )] = R.H.S Hence P (k+1) is true โˆด By the mathematical induction P(n) is true for all n where n is natural number Thus if A = = [โ– 8(1&1&1@1&1&1@1&1&1)] then An = [โ– 8(3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)@3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1)&3^(๐‘›โˆ’1) )] for n โˆˆ N

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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