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Matrix exponentiation involves raising a square matrix to the power of an integer, much like you would with numbers, but using matrix multiplication rules. It is an essential tool in various practical applications such as computer graphics, solving systems of linear equations, and more.
For a given matrix \( A \), matrix exponentiation means determining \( A^n \), where \( n \) is a positive integer. Unlike scalar exponentiation, matrix exponentiation isn't just raising each element of the matrix to the power \( n \). Instead, it requires multiplying the matrix by itself repeatedly, \( n-1 \) times:

• For \( n=1 \), \( A^1 = A \)
• For \( n=2 \), \( A^2 = A \times A \)
• For \( n=3 \), \( A^3 = A \times A \times A \)

In the provided exercise, understanding matrix exponentiation is vital to determine which one of the given equations holds true for \( n \geq 1 \). Each step in the solution process applies this fundamental concept to verify the validity of the options.

An identity matrix is a special kind of matrix where all the diagonal elements are 1, and all other elements are 0. It plays a crucial role in matrix multiplication because when you multiply any matrix by the identity matrix, it retains its original form.
In mathematical terms, for any square matrix \( A \) of size \( n \times n \), there exists an identity matrix \( I \) of the same size such that:

• \( A \times I = A \)
• \( I \times A = A \)

This matrix is essentially the matrix version of the number 1 in scalar multiplication. It is crucial when performing operations like matrix exponentiation where you may need to add or subtract an identity matrix from another matrix operation to preserve certain properties of the matrix.

The inductive step is a critical component in the process of mathematical induction, a method used to prove statements for all natural numbers. To utilize mathematical induction, it requires two main parts: the base case and the inductive step.
The base case involves proving that the statement holds for the initial value, often \( n=1 \). In the provided exercise, all options were shown to hold for \( n=1 \).
The inductive step, on the other hand, requires assuming the statement is true for some arbitrary natural number \( n=k \) and then proving that it's also true for \( n=k+1 \). This part ties each mathematical truth together, forming a chain of validity that extends indefinitely.
In this exercise, the inductive step is used to demonstrate that if \( A^k = 2^{k-1}A - (k-1)I \), then this formulation holds for \( A^{k+1} \) as well. By successfully verifying both the base case and inductive step, we ensure that the statement is valid for all \( n \geq 1 \), making this option the correct one.