For a matrix A, if `A^(2)=A` and `B=I-A` then `AB+BA +I-(I-A)^(2)` is equal to (where, I is the identity matrix of the same order of matrix A)

Options

A. B
B. A
C. AB
D. I

Answer

To solve the problem, we need to evaluate the expression \( AB + BA + I - (I - A)^2 \) given that \( A^2 = A \) and \( B = I - A \).

### Step 1: Substitute \( B \)
We start by substituting \( B \) in the expression:
\[ AB + BA + I - (I - A)^2 \]
Substituting \( B = I - A \):
\[ A(I - A) + (I - A)A + I - (I - A)^2 \]

### Step 2: Expand the terms
Now we expand the terms:
1. \( A(I - A) = AI - A^2 = A - A^2 \) (since \( A^2 = A \))
2. \( (I - A)A = IA - A^2 = A - A^2 \)
3. \( (I - A)^2 = I^2 - 2IA + A^2 = I - 2A + A \) (since \( A^2 = A \))

Now, substituting these back into the expression:
\[ (A - A^2) + (A - A^2) + I - (I - 2A + A) \]

### Step 3: Simplify the expression
Now we simplify:
\[ (A - A^2) + (A - A^2) + I - (I - A) \]
Combine like terms:
\[ 2A - 2A^2 + I - I + A \]
This simplifies to:
\[ 2A - 2A^2 + A \]

### Step 4: Use the property \( A^2 = A \)
Since \( A^2 = A \), we replace \( A^2 \) in the expression:
\[ 2A - 2A + A = A \]

### Final Result
Thus, the expression simplifies to:
\[ \boxed{A} \]