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Byju's Answer
Standard XII
Mathematics
Proof by mathematical induction
If A = begin ...
Question

If A=⎡⎢⎣111111111⎤⎥⎦, prove that An=⎡⎢⎣3n−13n−13n−13n−13n−13n−13n−13n−13n−1⎤⎥⎦,n∈N.

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Solution

Here, A=⎡⎢⎣111111111⎤⎥⎦

We shall prove the result by principle of mathematical induction.

Let P(n)An=⎡⎢⎣3n−13n−13n−13n−13n−13n−13n−13n−13n−1⎤⎥⎦
Putting n =1
P(1):A1=⎡⎢⎣31−131−131−131−131−131−131−131−131−1⎤⎥⎦=⎡⎢⎣303030303030303030⎤⎥⎦=⎡⎢⎣111111111⎤⎥⎦........(i)
which is true for n=1. Let the result be true for n =k.
∴P(k):Ak=⎡⎢ ⎢⎣3k−13k−13k−13k−13k−13k−13k−13k−13k−1⎤⎥ ⎥⎦..........(ii)

Putting n=k+1∴P(k+1):Ak+1=⎡⎢ ⎢⎣3k3k3k3k3k3k3k3k3k⎤⎥ ⎥⎦
Now, LHS=Ak+1=Ak.A1Ak.A1=⎡⎢ ⎢⎣3k−13k−13k−13k−13k−13k−13k−13k−13k−1⎤⎥ ⎥⎦⎡⎢⎣111111111⎤⎥⎦ [Using Eqs.(i)and (ii)]

=⎡⎢ ⎢⎣3k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−13k−1+3k−1+3k−1⎤⎥ ⎥⎦
Using multiplication of matrix.
=⎡⎢ ⎢⎣3×3k−13×3k−13×3k−13×3k−13×3k−13×3k−13×3k−13×3k−13×3k−1⎤⎥ ⎥⎦=⎡⎢ ⎢⎣3k3k3k3k3k3k3k3k3k⎤⎥ ⎥⎦=RHS
Therefore, the result is true for n=k+1 whenever it is true for n=k. So, by principle of mathematical induction it is true for all n∈N.


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