If A and B are square matrices of same order, then (A+B) (A-B) is equal to

Options

A. `A^(2)B^(2)`
B. `A^(2)=BA-AB+B^(2)`
C. `A^(2)-AB+BA-B^(2)`
D. `A^(2)+AB-BA-B^(2)`

Answer

To solve the problem, we need to find the expression for \((A + B)(A - B)\) where \(A\) and \(B\) are square matrices of the same order.

### Step-by-Step Solution:

1. **Write the Expression**:
We start with the expression \((A + B)(A - B)\).

2. **Apply the Distributive Property**:
We will use the distributive property (also known as the FOIL method for binomials) to expand the expression:
\[ (A + B)(A - B) = A(A - B) + B(A - B) \]

3. **Distribute Each Term**:
Now, we distribute \(A\) and \(B\) across \((A - B)\):
\[ = A^2 - AB + BA - B^2 \]

4. **Combine the Terms**:
The expression can be rewritten as:
\[ = A^2 - B^2 + BA - AB \]

5. **Final Result**:
Therefore, the final result of \((A + B)(A - B)\) is:
\[ A^2 - AB + BA - B^2 \]

### Final Answer:
\[ (A + B)(A - B) = A^2 - AB + BA - B^2 \]