Q.
If A=1 1 11 1 11 1 1 then An for every positive integer n is
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a
3n+1 3n+1 3n+13n+1 3n+1 3n+13n+1 3n+1 3n+1
b
3n−1 3n−1 3n−13n−1 3n−1 3n−13n−1 3n−1 3n−1
c
3n 3n 3n3n 3n 3n3n 3n 3n
d
3n−2 3n−2 3n−23n−2 3n−2 3n−23n−2 3n−2 3n−2
answer is B.
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Detailed Solution
the result by the principle of mathematical induction on n.Step1 When n=1, by the definition of integral powers of a matrix, we have A1=A=1 1 11 1 11 1 1=31−1 31−1 31−131−1 31−1 31−131−1 31−1 31−1So, the result is true forn=1.Step2 Let the result be true forn=m .Then Am=3m−1 3m−1 3m−13m−1 3m−1 3m−13m−1 3m−1 3m−1 ...(i)Now we shall show that the result is true for n=m+1, i.e.n=m+1Am+1=3m 3m 3m3m 3m 3m3m 3m 3mBy the definition of integral powers of a matrix, we have Am+1=Am·A=3m-13m-13m-13m-13m-13m-13m-13m-13m-1111111111=3m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-13m-1+3m-1+3m-1=3.3m-13.3m-13.3m-13.3m-13.3m-13.3m-13.3m-13.3m-13.3m-1=3m3m3m3m3m3m3m3m3mThis shows that the result is true for, whenever it is true forn=m Hence, by the principle of mathematical induction the result is valid for any positive integer n.
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