Question

If $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$, prove that

$A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right],n\in N$.

Answer 1

Given $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$

To prove :

$A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right],n\in N$.

Proof :

Let P(n) be the given statement.

For $n=1$,

$P\left(1\right)=A=\left[\begin{array}{ccc}{3}^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\end{array}\right]=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]=A$

Therefore, statement is true for $n=1$.

Now, let us assume that statement is true for $n=k$.

$P\left(k\right):A^k=\left[\begin{array}{ccc}{3}^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{array}\right]$

Now, we shall prove the statement for $n=k+1$, we have to show,

$P(k+1)=A^{k+1}=\left[\begin{array}{ccc}{3}^k& 3^k& 3^k\\ 3^k& 3^k& 3^k\\ 3^k& 3^k& 3^k\end{array}\right]$

LHS = $A^{k+1}=A^k.A$

$=\left[\begin{array}{ccc}{3}^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{array}\right]\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$

$=\left[\begin{array}{ccc}{3}^k& 3^k& 3^k\\ 3^k& 3^k& 3^k\\ 3^k& 3^k& 3^k\end{array}\right]$ = RHS

Statement is true for $n=k+1$.

Hence, by principal of mathematical induction, statemebt is true for all n, where $n\in N$.