1. Chapter 3 Class 12 Matrices
2. Serial order wise

Transcript

Misc. 2 If A = [โ 8(1&1&1@1&1&1@1&1&1)] , prove that An = [โ 8(3^(๐โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ1) )], n โ 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โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ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^(1โ1)&3^(1โ1)&3^(1โ1)@3^(1โ1)&3^(1โ1)&3^(1โ1)@3^(1โ1)&3^(1โ1)&3^(1โ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 โด 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^(kโ1)&3^(kโ1)&3^(kโ1)@3^(kโ1)&3^(kโ1)&3^(kโ1)@3^(kโ1)&3^(kโ1)&3^(kโ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)โ1)&3^((k+1)โ1)&3^((k+1)โ1)@3^((k+1)โ1)&3^((k+1)โ1)&3^((k+1)โ1)@3^((k+1)โ1)&3^((k+1)โ1)&3^((k+1)โ1) )] Ak+1 = [โ 8(3^(k+1โ1)&3^(k+1โ1)&3^(k+1โ1)@3^(k+1โ1)&3^(k+1โ1)&3^(k+1โ1)@3^(k+1โ1)&3^(k+1โ1)&3^(k+1โ1) )] Consider L.H.S Ak +1 = Ak . A1 =[โ 8(3^(kโ1)&3^(kโ1)&3^(kโ1)@3^(kโ1)&3^(kโ1)&3^(kโ1)@3^(kโ1)&3^(kโ1)&3^(kโ1) )][โ 8(1&1&1@1&1&1@1&1&1)] = [โ 8(3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)@3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)@3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1)&3^(kโ1) (1)+3^(kโ1) (1)+3^(kโ1) (1) )] = [โ 8(3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)@3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)@3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1)&3^(kโ1) ใ+3ใ^(kโ1)+3^(kโ1) )] = [โ 8(ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ1)@ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ1)@ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ1)&ใ3.3ใ^(๐โ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 โด 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โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ1)@3^(๐โ1)&3^(๐โ1)&3^(๐โ1) )] for n โ N

Serial order wise

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Davneet Singh
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.