If `A=[[2, 0, 0], [0, 2, 0], [0, 0, 2]]`, then `A^(6)=`

Options

A. `6A`
B. `12A`
C. `16A`
D. `32A`

Answer

To solve the problem of finding \( A^6 \) for the matrix \( A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \), we can follow these steps:

### Step 1: Factor out the scalar from the matrix
We can factor out the scalar \( 2 \) from the matrix \( A \):
\[ A = 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = 2I \]
where \( I \) is the identity matrix.

**Hint:** Remember that you can factor out a scalar from a diagonal matrix by taking it out of each diagonal entry.

### Step 2: Raise the matrix to the power of 6
Now, we can express \( A^6 \) as:
\[ A^6 = (2I)^6 \]

**Hint:** When raising a product to a power, remember to apply the power to both the scalar and the matrix.

### Step 3: Apply the power to the scalar and the identity matrix
Using the property of exponents, we can separate the scalar and the identity matrix:
\[ A^6 = 2^6 I^6 \]

**Hint:** The identity matrix raised to any power remains the identity matrix.

### Step 4: Simplify the expression
Since \( I^6 = I \), we have:
\[ A^6 = 2^6 I \]
Calculating \( 2^6 \):
\[ 2^6 = 64 \]
Thus, we can write:
\[ A^6 = 64I \]

**Hint:** Always calculate the power of the scalar carefully, as it contributes to the final result.

### Final Result
The final result is:
\[ A^6 = 64 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix} \]

### Summary
The answer is:
\[ A^6 = \begin{bmatrix} 64 & 0 & 0 \\ 0 & 64 & 0 \\ 0 & 0 & 64 \end{bmatrix} \]