If `A=[(a,0,0),(0,a,0),(0,0,a)]` then `A^(n)=`

Options

A. `[(a^(n),0,0),(0,a^(n),0),(0,0,0)]`
B. `[(a^(n),0,0),(0,a,0),(0,0,a)]`
C. `[(a^(n),0,0),(0,a^(n),0),(0,0,a^(n))]`
D. `[(na,0,0),(0,na,0),(0,0,na)]`

Answer

To find \( A^n \) for the matrix \( A = \begin{pmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{pmatrix} \), we can follow these steps:

### Step 1: Understand the structure of matrix A
The matrix \( A \) is a diagonal matrix, where all the diagonal elements are equal to \( a \).

### Step 2: Recall the property of diagonal matrices
For a diagonal matrix, raising it to a power involves raising each of the diagonal elements to that power.

### Step 3: Calculate \( A^2 \)
Let's calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{pmatrix} \cdot \begin{pmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{pmatrix} \]
Calculating this gives:
\[ A^2 = \begin{pmatrix} a \cdot a & 0 & 0 \\ 0 & a \cdot a & 0 \\ 0 & 0 & a \cdot a \end{pmatrix} = \begin{pmatrix} a^2 & 0 & 0 \\ 0 & a^2 & 0 \\ 0 & 0 & a^2 \end{pmatrix} \]

### Step 4: Generalize to \( A^n \)
By applying the same logic, we can generalize this to \( A^n \):
\[ A^n = \begin{pmatrix} a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n \end{pmatrix} \]

### Final Result
Thus, the result is:
\[ A^n = \begin{pmatrix} a^n & 0 & 0 \\ 0 & a^n & 0 \\ 0 & 0 & a^n \end{pmatrix} \]