Consider the matrix `A=[(3,1),(-6,-2)]`, then `(I+A)^(40)` is equal to

Options

A. `I+2^(38)A`
B. `I+2^(39)A`
C. `I+(2^(40)+1)A`
D. `I+(2^(40)-1)A`

Answer

To solve the problem, we need to calculate \((I + A)^{40}\) where \(A = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix}\) and \(I\) is the identity matrix of the same size.

### Step 1: Calculate \(A^2\)
First, we will find \(A^2\) to check if \(A\) has any special properties.

\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} \]

Calculating the elements:

- First row, first column: \(3 \cdot 3 + 1 \cdot (-6) = 9 - 6 = 3\)
- First row, second column: \(3 \cdot 1 + 1 \cdot (-2) = 3 - 2 = 1\)
- Second row, first column: \(-6 \cdot 3 + (-2) \cdot (-6) = -18 + 12 = -6\)
- Second row, second column: \(-6 \cdot 1 + (-2) \cdot (-2) = -6 + 4 = -2\)

Thus, we have:

\[ A^2 = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} = A \]

### Step 2: Determine \(A^n\)
Since \(A^2 = A\), we can conclude that for any \(n \geq 1\):

\[ A^n = A \]

### Step 3: Calculate \((I + A)^{40}\)
Now, we can express \((I + A)^{40}\) using the binomial theorem:

\[ (I + A)^{40} = I^{40} + \binom{40}{1} I^{39} A + \binom{40}{2} I^{38} A^2 + \ldots + \binom{40}{40} A^{40} \]

Since \(I^n = I\) for any \(n\) and \(A^n = A\):

\[ (I + A)^{40} = I + 40A + \binom{40}{2} A + \ldots + A \]

### Step 4: Factor out \(A\)
Now we can factor out \(A\):

\[ (I + A)^{40} = I + A \left(40 + \binom{40}{2} + \ldots + \binom{40}{40}\right) \]

### Step 5: Sum of Binomial Coefficients
The sum of the binomial coefficients from \(k=1\) to \(k=40\) is given by:

\[ \sum_{k=0}^{40} \binom{40}{k} = 2^{40} \]

Thus, the sum from \(k=1\) to \(k=40\) is:

\[ 2^{40} - 1 \]

### Step 6: Final Expression
Therefore, we can write:

\[ (I + A)^{40} = I + A(2^{40} - 1) \]

### Conclusion
The final result is:

\[ (I + A)^{40} = I + (2^{40} - 1)A \]