If, prove that

Solution:

It is given that $A=\left[\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]$

To show: $\quad \mathrm{P}(n): A^{n}=\left[\begin{array}{lll}3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1}\end{array}\right], n \in \mathbf{N}$

We shall prove the result by using the principle of mathematical induction.

For n = 1, we have:

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

Therefore, the result is true for n = 1.

 

Let the result be true for n = k.

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

 

Now, we prove that the result is true for n = k + 1.

Now, $A^{k+1}=A \cdot A^{k}$

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

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

 

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

Therefore, the result is true for n = k + 1.

Thus by the principle of mathematical induction, we have:

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