InterviewSolution
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InterviewSolution
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If A=`[[3,-4] , [1,-1]]` then by the method of mathematical induction prove that `A^n=[[1+2n,-4n] , [n,1-2n]]` |
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Answer» We shall prove the result by using the principle of mathematical induction. When `n=1`, we have `A^(1)=[{:(1+2.1,-4.1),(1,1-2.1):}]=[{:(1+2,-4),(1,1-2):}]=[{:(3,-4),(1,-1):}].` Thus, the result it true for n=1. Let the result be true for n=k. Then, `A^(k)=[{:(1+2k,-4k),(k,1-2k):}].` `:." "A^(k+1)=A.A^(k)` `=[{:(3,-4),(1,-1):}][{:(1+2k,-4k),(k,1-2k):}]` `=[{:(3+6k-4k,-12k-4+8k),(1+2k-k,-4k-1+2k):}]=[{:(3+2k,-4k-4),(k+1,1-2k):}]` `=[{:(1+2(k+1),-4(k+1)),(k+1,1-2(k+1)):}].` Thus, the result is true for `n=(k+1)`, whenever it is true for `n=k.` So, the result is true for all `n inN.` Hence, `A^(n)=[{:(1+2n,-4n),(n,1-2n):}]` for all values of `n inN.` |
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