Let a be a matrix of order `2xx2` such that `A^(2)=O`.
`(I+A)^(100) =`

Options

A. 100 A
B. `100 (I+A)`
C. `100 I+A`
D. `I+100 A`

Explanation

To solve the problem, we need to evaluate \((I + A)^{100}\) given that \(A\) is a \(2 \times 2\) matrix such that \(A^2 = O\) (the zero matrix).

### Step-by-Step Solution:

1. **Understand the properties of the matrix \(A\)**:
Since \(A^2 = O\), we know that any higher power of \(A\) (i.e., \(A^3, A^4, \ldots\)) will also be the zero matrix. This is because:
\[ A^3 = A \cdot A^2 = A \cdot O = O \]
Similarly, \(A^4 = O\), \(A^5 = O\), and so on.

**Hint**: Remember that if \(A^2 = O\), then \(A^n = O\) for all \(n \geq 2\).

2. **Use the Binomial Theorem**:
We can expand \((I + A)^{100}\) using the binomial theorem:
\[ (I + A)^{100} = \sum_{k=0}^{100} \binom{100}{k} I^{100-k} A^k \]
This expansion consists of terms where \(I^{100-k} = I\) for any \(k\) (since the identity matrix raised to any power is still the identity matrix).

**Hint**: The binomial theorem allows us to expand expressions of the form \((x + y)^n\).

3. **Evaluate the terms in the expansion**:
Since \(A^2 = O\), any term where \(k \geq 2\) will be zero:
- For \(k = 0\): \(\binom{100}{0} I^{100} A^0 = I\)
- For \(k = 1\): \(\binom{100}{1} I^{99} A^1 = 100A\)
- For \(k \geq 2\): \(\binom{100}{k} I^{100-k} A^k = 0\)

Therefore, we only need to consider the first two terms of the expansion:
\[ (I + A)^{100} = I + 100A \]

**Hint**: Identify which terms contribute to the final result based on the properties of \(A\).

4. **Final Result**:
Thus, we conclude that:
\[ (I + A)^{100} = I + 100A \]

### Final Answer:
\[ (I + A)^{100} = I + 100A \]