If we have a matrix as \[A = \left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right)\], prove that \[{A^n} = \left( {\begin{array}{*{20}{c}}
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}} \\
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}} \\
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}
\end{array}} \right)\], \[n \in N\]

Hint: This is a question of matrices, but we will apply the principle of mathematical induction. It is also known as PMI. It is used in the case of natural numbers only. According to PMI, we need to assume that the statements are true for all natural numbers and hence we will initially take \[n = 1\], then \[n = k\] and finally \[n = k + 1\].

Complete step-by-step solution:
Given,
\[{A^n} = \left( {\begin{array}{*{20}{c}}
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}} \\
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}} \\
  {{3^{n - 1}}}&{{3^{n - 1}}}&{{3^{n - 1}}}
\end{array}} \right)\]
Firstly, by assuming that the statement is true, we take
\[n = 1\]
Now, putting this value in the above equation, we get
\[ \Rightarrow {A^1} = \left( {\begin{array}{*{20}{c}}
  {{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}} \\
  {{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}} \\
  {{3^{1 - 1}}}&{{3^{1 - 1}}}&{{3^{1 - 1}}}
\end{array}} \right)\]
\[ \Rightarrow {A^1} = \left( {\begin{array}{*{20}{c}}
  {{3^0}}&{{3^0}}&{{3^0}} \\
  {{3^0}}&{{3^0}}&{{3^0}} \\
  {{3^0}}&{{3^0}}&{{3^0}}
\end{array}} \right)\]
Now, we already know that any number raise to the power 0 is 1, therefore
\[A = \left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right)\]
Thus, the above statement is true for \[n = 1\].

Now, let us assume \[n = k\]
So, by putting the value of n as k, we get
\[{A^k} = \left( {\begin{array}{*{20}{c}}
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}
\end{array}} \right)\]
Let us just assume that the statement is true when \[n = k\] as k is a natural number.
Now, by taking \[n = k + 1\]
By putting this value in the equation, we get
\[{A^{k + 1}} = \left( {\begin{array}{*{20}{c}}
  {{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}} \\
  {{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}} \\
  {{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}}&{{3^{k + 1 - 1}}}
\end{array}} \right)\]
\[ \Rightarrow {A^{k + 1}} = \left( {\begin{array}{*{20}{c}}
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}}
\end{array}} \right)\]
We can write this equation in the following way as well
\[ \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}}
\end{array}} \right)\], \[(\because {x^{m + n}} = {x^m} + {x^n})\]
Depicting \[LHS = RHS\]
By solving LHS separately, we get
\[LHS \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}} \\
  {{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&1&1 \\
  1&1&1 \\
  1&1&1
\end{array}} \right)\]
Here, we took \[{A^k}\] as a separate term as mentioned in the 2nd assumption and \[A\] separately.
Now by solving this equation by matrix multiplication method. In this method, we multiply the first row of the 1st matrix with the first column of the 2nd matrix, then the first row of the 1st matrix with the second column of the 2nd matrix and continue further with the same process. Therefore, we get
\[LHS \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}} \\
  {{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}} \\
  {{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}&{{3^{k - 1}} + {3^{k - 1}} + {3^{k - 1}}}
\end{array}} \right)\]
We can further write this as,
\[LHS \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}} \\
  {3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}} \\
  {3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}}&{3 \times {3^{k - 1}}}
\end{array}} \right)\]
\[LHS \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}} \\
  {{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}} \\
  {{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}}&{{3^{k - 1 + 1}}}
\end{array}} \right)\], \[(\because {x^m} + {x^n} = {x^{m + n}})\]
\[LHS \Rightarrow {A^k}.A = \left( {\begin{array}{*{20}{c}}
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}}
\end{array}} \right)\]
\[RHS \Rightarrow {A^{k + 1}} = \left( {\begin{array}{*{20}{c}}
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}} \\
  {{3^k}}&{{3^k}}&{{3^k}}
\end{array}} \right)\]
Hence, \[LHS = RHS\]
Therefore, our assumption is true for \[n = k\] and also for \[n = k + 1\].
Also, the statement is true for all \[n \in N\].

Note: Always keep in mind that when applying the PMI method, we'll be looking at three different cases. To prove the statement right, first prove \[n = 1\] is true, then assume \[n = k\] is true, and then solve \[n = k + 1\] with the help of \[n = k\]. If any of these cases stands invalid for the provided statement, then the statement is incorrect.