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Mathematics Question on Matrices

If A=[111 111 111]A=\begin{bmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\end{bmatrix},Prove that An=[3n−13n−13n−1 3n−13n−13n−1 3n−13n−13n−1],n∈NA^n=\begin{bmatrix}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{bmatrix},n∈N

Answer

It is given that A=[111 111 111]A=\begin{bmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\end{bmatrix}
To show P(n):An=[3n−13n−13n−1 3n−13n−13n−1 3n−13n−13n−1]P(n):A^n=\begin{bmatrix}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{bmatrix}
We shall prove the result by using the principle of mathematical induction.
For n = 1, we have:
P(1)=[31−131−131−1 31−131−131−1 31−131−131−1]P(1)=\begin{bmatrix}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{bmatrix}
An=[303030 303030 303030]A^n=\begin{bmatrix}3^0& 3^0& 3^0\\\ 3^0& 3^0& 3^0\\\ 3^0& 3^0& 3^0\end{bmatrix}
A=[111 111 111]A=\begin{bmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\end{bmatrix}
Therefore, the result is true for n = 1.
Let the result be true for n = k.
That is P(k)=Ak=[3k−13k−13k−1 3k−13k−13k−1 3k−13k−13k−1]P(k)=A^k=\begin{bmatrix}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{bmatrix}
Now, we prove that the result is true for n = k + 1.
Ak+1=A.AkA^{k+1}=A.A^k
=[111 111 111][3k−13k−13k−1 3k−13k−13k−1 3k−13k−13k−1]=\begin{bmatrix}1&1&1\\\ 1&1&1\\\ 1&1&1\end{bmatrix}\begin{bmatrix}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{bmatrix}
[3.3k−13.3k−13.3k−1 3.3k−13.3k−13.3k−1 3.3k−13.3k−13.3k−1]\begin{bmatrix}3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\\\ 3.3^{k-1}& 3.3^{k-1}& 3.3^{k-1}\end{bmatrix}
[3k+1−13k+1−13k+1−1 3k+1−13k+1−13k+1−1 3k+1−13k+1−13k+1−1]\begin{bmatrix}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{bmatrix}
Therefore, the result is true for n=k+1.n = k + 1.
Thus by the principle of mathematical induction, we have:
An=[3n−13n−13n−1 3n−13n−13n−1 3n−13n−13n−1],n∈NA^n=\begin{bmatrix}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{bmatrix},n∈N