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 compute \((I + A)^{40}\) where \(A = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix}\).

### Step 1: Calculate the identity matrix \(I\)
The identity matrix \(I\) for a \(2 \times 2\) matrix is:
\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Compute \(I + A\)
Now, we add the matrix \(A\) to the identity matrix \(I\):
\[ I + A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} = \begin{pmatrix} 1 + 3 & 0 + 1 \\ 0 - 6 & 1 - 2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ -6 & -1 \end{pmatrix} \]

### Step 3: Calculate the determinant of \(A\)
The determinant of matrix \(A\) is calculated as follows:
\[ \text{det}(A) = (3)(-2) - (-6)(1) = -6 + 6 = 0 \]

### Step 4: Analyze the implications of the determinant
Since the determinant of \(A\) is \(0\), this means that \(A\) is a singular matrix. For singular matrices, we have \(A^2 = A\) (idempotent property).

### Step 5: Compute \(A^2\)
We can verify this:
\[ A^2 = A \cdot A = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} \cdot \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} = \begin{pmatrix} 3 \cdot 3 + 1 \cdot (-6) & 3 \cdot 1 + 1 \cdot (-2) \\ -6 \cdot 3 + (-2) \cdot (-6) & -6 \cdot 1 + (-2) \cdot (-2) \end{pmatrix} \]
Calculating this gives:
\[ = \begin{pmatrix} 9 - 6 & 3 - 2 \\ -18 + 12 & -6 + 4 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} = A \]

### Step 6: Generalize the powers of \(A\)
Since \(A^2 = A\), it follows that:
\[ A^n = A \quad \text{for all } n \geq 1 \]

### Step 7: Compute \((I + A)^{40}\)
Using the binomial expansion:
\[ (I + A)^{40} = I^{40} + \binom{40}{1} I^{39} A + \binom{40}{2} I^{38} A^2 + \ldots + A^{40} \]
Since \(I^n = I\) and \(A^n = A\) for \(n \geq 1\), we can simplify:
\[ (I + A)^{40} = I + 40A + 0 + 0 + \ldots + 0 = I + 40A \]

### Step 8: Substitute \(A\) back into the equation
Now substituting \(A\):
\[ I + 40A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + 40 \begin{pmatrix} 3 & 1 \\ -6 & -2 \end{pmatrix} = \begin{pmatrix} 1 + 120 & 0 + 40 \\ 0 - 240 & 1 - 80 \end{pmatrix} = \begin{pmatrix} 121 & 40 \\ -240 & -79 \end{pmatrix} \]

### Final Answer
Thus, \((I + A)^{40} = \begin{pmatrix} 121 & 40 \\ -240 & -79 \end{pmatrix}\).

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