If A is an idempotent matrix and I is an identify matrix of the Same order, then the value of n, such that `(A+I)^n =I+127A` is

Explanation

To solve the problem, we need to find the value of \( n \) such that \( (A + I)^n = I + 127A \), given that \( A \) is an idempotent matrix and \( I \) is the identity matrix of the same order.

### Step-by-Step Solution:

1. **Understanding Idempotent Matrix**:
An idempotent matrix \( A \) satisfies the property \( A^2 = A \). This means that any power of \( A \) greater than or equal to 1 will equal \( A \).

2. **Expanding \( (A + I)^n \)**:
We can use the binomial theorem to expand \( (A + I)^n \):
\[ (A + I)^n = \sum_{k=0}^{n} \binom{n}{k} A^k I^{n-k} \]
Since \( I^{n-k} = I \) for any \( n-k \), we can simplify this to:
\[ (A + I)^n = \sum_{k=0}^{n} \binom{n}{k} A^k \]

3. **Substituting Idempotency**:
Using the property of idempotency \( A^k = A \) for \( k \geq 1 \):
- For \( k = 0 \): \( \binom{n}{0} A^0 = 1 \)
- For \( k = 1 \): \( \binom{n}{1} A^1 = nA \)
- For \( k \geq 2 \): \( \binom{n}{k} A^k = \binom{n}{k} A \)

Therefore, the expansion becomes:
\[ (A + I)^n = I + nA + (nC2 + nC3 + \ldots + nCn)A \]

4. **Simplifying the Expression**:
The sum of the binomial coefficients from \( k = 2 \) to \( n \) is:
\[ \sum_{k=2}^{n} \binom{n}{k} = 2^n - 1 - n \]
Thus, we can rewrite:
\[ (A + I)^n = I + nA + (2^n - 1 - n)A \]
This simplifies to:
\[ (A + I)^n = I + (2^n - 1)A \]

5. **Setting the Equation**:
We set this equal to \( I + 127A \):
\[ I + (2^n - 1)A = I + 127A \]

6. **Equating Coefficients**:
By equating the coefficients of \( A \):
\[ 2^n - 1 = 127 \]
Therefore:
\[ 2^n = 128 \]

7. **Finding \( n \)**:
Since \( 128 = 2^7 \), we have:
\[ n = 7 \]

### Final Answer:
Thus, the value of \( n \) is \( \boxed{7} \).