If `A^(20) = [(a,b),(c,d)]` where A`=[(1,1),(0,2)]` then a+b+c+d is equal to

Options

A. `2^(19)`
B. `2^(20)`
C. `2^(21)`
D. `2^(22)`

Answer

To solve the problem, we need to find the sum \( a + b + c + d \) where \( A^{20} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) and \( A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)
First, we calculate \( A^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 1 \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot 1 + 1 \cdot 2 = 3 \)
- Second row, first column: \( 0 \cdot 1 + 2 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 1 + 2 \cdot 2 = 4 \)

Thus,
\[ A^2 = \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix} \]

### Step 2: Calculate \( A^3 \)
Next, we calculate \( A^3 = A^2 \cdot A \):
\[ A^3 = \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \]
Calculating the elements:
- First row, first column: \( 1 \cdot 1 + 3 \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot 1 + 3 \cdot 2 = 7 \)
- Second row, first column: \( 0 \cdot 1 + 4 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot 1 + 4 \cdot 2 = 8 \)

Thus,
\[ A^3 = \begin{pmatrix} 1 & 7 \\ 0 & 8 \end{pmatrix} \]

### Step 3: Identify the pattern
From the calculations, we observe a pattern:
- \( A^1 = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix} \)
- \( A^2 = \begin{pmatrix} 1 & 3 \\ 0 & 4 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 1 & 7 \\ 0 & 8 \end{pmatrix} \)

We can see that:
- The first element remains 1.
- The second element in the first row appears to follow the pattern \( 2^n - 1 \).
- The second element in the second row is \( 2^n \).

### Step 4: Generalize the pattern for \( A^n \)
Based on the observations, we can generalize:
\[ A^n = \begin{pmatrix} 1 & 2^n - 1 \\ 0 & 2^n \end{pmatrix} \]

### Step 5: Calculate \( A^{20} \)
Using the generalized formula:
\[ A^{20} = \begin{pmatrix} 1 & 2^{20} - 1 \\ 0 & 2^{20} \end{pmatrix} \]

### Step 6: Identify \( a, b, c, d \)
From \( A^{20} \):
- \( a = 1 \)
- \( b = 2^{20} - 1 \)
- \( c = 0 \)
- \( d = 2^{20} \)

### Step 7: Calculate \( a + b + c + d \)
Now, we calculate:
\[ a + b + c + d = 1 + (2^{20} - 1) + 0 + 2^{20} \]
This simplifies to:
\[ = 1 + 2^{20} - 1 + 2^{20} = 2 \cdot 2^{20} = 2^{21} \]

### Final Answer
Thus, the value of \( a + b + c + d \) is \( 2^{21} \).