If \[A = \left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\] and \[I = \left[ {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right]\] then which one of the following holds for all \[n \ge 1\] (by the principle of mathematical induction)
A. \[{A^n} = nA + (n - 1)I\]
В. \[{A^n} = {2^{n - 1}}A + (n - 1)I\]
C. \[{A^n} = nA - (n - 1)I\]
D. \[{A^n} = {2^{n - 1}}A - (n - 1)I\]

Hint: Suppose a statement \[P\left( n \right)\] concerning the natural number n exists and is such that, for n = 1, the claim is accurate, meaning \[P\left( 1 \right)\] holds true. In our case, we are given an A matrix and are asked to determine the equations that holds the given matrix by principal of mathematical induction. For that, we have to first determine the square of the given matrix and then the cue of the given matrix and then the matrix should be verified using principle of mathematical induction to get the desired result.

Formula Used:
The concept used is that proving any statement is said to be true for every natural number N.

Complete Step-By-Step Solution:
We are given a matrix A in the question that,
\[A = \left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\]
And also given the I matrix, we have
\[I = \left[ {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right]\]
Now, we have to determine the square of matrix ‘A’, we get
\[{{\rm{A}}^2} = \left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\]
Now, we have to multiply the above two matrix.
The elements in the first matrix's column and rows are multiplied by one another\[ = \left[ {\begin{array}{*{20}{l}}{(1) \cdot (1) + (0) \cdot (1)}&{(1) \cdot (0) + (0) \cdot (1)}\\{(1) \cdot (1) + (1) \cdot (1)}&{(1) \cdot (0) + (1) \cdot (1)}\end{array}} \right]\]
Now, the results should be added, we get
\[ = \left[ {\begin{array}{*{20}{l}}1&0\\2&1\end{array}} \right]\]
Now, we gave to determine the cube of matrix ‘A’.
We have,
\[{{\rm{A}}^3}{\rm{ = }}{{\rm{A}}^2} \cdot {\rm{A}}\]
Now, we have to multiply the matrix of \[{{\rm{A}}^2}\] and matrix \[{\rm{A}}\] we get
\[{{\rm{A}}^3} = \left[ {\begin{array}{*{20}{l}}1&0\\2&1\end{array}} \right]\left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\]
Now, we have to multiply the above two matrix.
The elements in the first matrix's column and rows are multiplied by one another.
\[ = \left[ {\begin{array}{*{20}{l}}{(1) \cdot (1) + (0) \cdot (1)}&{(1) \cdot (0) + (0) \cdot (1)}\\{(2) \cdot (1) + (1) \cdot (1)}&{(2) \cdot (0) + (1) \cdot (1)}\end{array}} \right]\]
 Now, the results should be added, we get
\[ = \left[ {\begin{array}{*{20}{l}}1&0\\3&1\end{array}} \right]\]
Therefore, from the above obtained results, we can write
\[{{\rm{A}}^{\rm{n}}} = \left[ {\begin{array}{*{20}{l}}1&0\\{\rm{n}}&1\end{array}} \right]\]
And
\[{\rm{nA}} = {\rm{n}}\left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}{\rm{n}}&0\\{\rm{n}}&{\rm{n}}\end{array}} \right]\]
Now, we have to verify \[{{\rm{A}}^{\rm{n}}} = \left[ {\begin{array}{*{20}{l}}1&0\\{\rm{n}}&1\end{array}} \right]\] by principle of mathematical induction.
We have,
\[(n - 1)I = (n - 1)\left[ {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right]\]
Now, we have to multiply the term \[n - 1\] with the I matrix by multiplying the term outside with each term inside the matrix, we get
\[ = \left[ {\begin{array}{*{20}{c}}{n - 1}&0\\0&{n - 1}\end{array}} \right]\]
Now, we have to subtract the matrix value of \[nA\] and \[(n - 1)I\] we get
\[nA - (n - 1)I = \left[ {\begin{array}{*{20}{l}}n&0\\n&n\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}{n - 1}&0\\0&{n - 1}\end{array}} \right]\]
Therefore, we get
\[\left[ {\begin{array}{*{20}{c}}1&0\\n&1\end{array}} \right] = {A^n}\]
Therefore, If \[A = \left[ {\begin{array}{*{20}{l}}1&0\\1&1\end{array}} \right]\] and \[I = \left[ {\begin{array}{*{20}{l}}1&0\\0&1\end{array}} \right]\] the one that holds for all \[n \ge 1\] (by the principle of mathematical induction) is \[{A^n} = nA - (n - 1)I\]
Hence, the option C is correct

Note:
Students make mistakes in these types of problems because, it includes matrix that too matrix multiplication. We all know that matrix multiplication is tedious. So, students should keep in mind that to solve matrix multiplication, the elements in the first matrix's column and rows are multiplied by one another. And principle of induction concept should be applied correctly in order to get the desired answer.