If $A=\left[\begin{array}{l}1\text{ 1 1}\\ \text{1 1 1}\\ \text{1 1 1}\end{array}\right]$ then $A^n$for every positive integer n is

detailed solution

Correct option is B

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.