Suppose `A` is a matrix such that `A^2 =A and (I + A)^6 =I+ KA`, then `K` is________.

Answer

To solve the problem, we need to find the value of \( K \) given that \( A \) is a matrix such that \( A^2 = A \) and \( (I + A)^6 = I + KA \).

### Step 1: Understand the properties of matrix \( A \)
Since \( A^2 = A \), we can conclude that \( A \) is an idempotent matrix. This means that \( A \) can be thought of as a projection matrix.

### Step 2: Expand \( (I + A)^6 \) using the Binomial Theorem
We can use the Binomial Theorem to expand \( (I + A)^6 \):

\[ (I + A)^6 = \sum_{k=0}^{6} \binom{6}{k} I^{6-k} A^k \]

Since \( I^n = I \) for any integer \( n \), we can simplify this to:

\[ (I + A)^6 = \sum_{k=0}^{6} \binom{6}{k} I A^k = I + \binom{6}{1} I A + \binom{6}{2} I A^2 + \binom{6}{3} I A^3 + \binom{6}{4} I A^4 + \binom{6}{5} I A^5 + \binom{6}{6} I A^6 \]

### Step 3: Substitute \( A^k \) for \( k \geq 1 \)
Since \( A^2 = A \), we have \( A^k = A \) for all \( k \geq 1 \). Therefore, we can replace \( A^k \) in the expansion:

\[ (I + A)^6 = I + 6IA + 15IA + 20IA + 15IA + 6IA + A \]

### Step 4: Calculate the coefficients
Now we can calculate the coefficients:

- For \( k = 1 \): \( \binom{6}{1} = 6 \)
- For \( k = 2 \): \( \binom{6}{2} = 15 \)
- For \( k = 3 \): \( \binom{6}{3} = 20 \)
- For \( k = 4 \): \( \binom{6}{4} = 15 \)
- For \( k = 5 \): \( \binom{6}{5} = 6 \)
- For \( k = 6 \): \( \binom{6}{6} = 1 \)

### Step 5: Combine the terms
Combining all the terms gives:

\[ (I + A)^6 = I + (6 + 15 + 20 + 15 + 6 + 1)A = I + 63A \]

### Step 6: Set equal to \( I + KA \)
According to the problem, we have:

\[ I + KA = I + 63A \]

### Step 7: Compare coefficients
By comparing coefficients of \( A \) on both sides, we find:

\[ K = 63 \]

### Final Answer
Thus, the value of \( K \) is \( \boxed{63} \).