If `A and B` are two square matrices such that they commute, then which of the following is true?

Options

A. `AB^(2013)=B^(2013)A`
B. `(AB)^(2013)=A^(2013)B^(2013)`
C. `(A+B)^(n)=.^(n)C_(0)A^(n)+.^(n)C_(1)A^(n-1)B+.^(n)C_(2)A^(n-2)B^(2)+. . . . +.^(n)C_(n)B^(n)`
D. `A^(2)-B^(2)=(A-B)(A+B)`

Answer

To solve the problem, we need to analyze the properties of two square matrices \( A \) and \( B \) that commute, meaning \( AB = BA \). We will evaluate the truth of each of the given options based on this property.

### Step-by-Step Solution:

1. **Understanding Commutative Property**:
Since \( A \) and \( B \) commute, we have:
\[ AB = BA \]

2. **Option A: Proving \( AB^{2013} = B^{2013}A \)**:
- Start with \( AB = BA \).
- Post-multiply both sides by \( B \):
\[ AB \cdot B = BA \cdot B \implies AB^2 = B^2A \]
- Continue this process:
\[ AB^3 = B^3A, \quad AB^4 = B^4A, \quad \ldots, \quad AB^{2013} = B^{2013}A \]
- Thus, option A is true.

3. **Option B: Proving \( (AB)^{2013} = A^{2013}B^{2013} \)**:
- Start with \( (AB)^2 = AB \cdot AB \).
- Using commutativity:
\[ (AB)^2 = A(BA)B = A(AB)B = A^2B^2 \]
- By induction, we can show that:
\[ (AB)^n = A^nB^n \]
- Therefore, \( (AB)^{2013} = A^{2013}B^{2013} \) is true.

4. **Option C: Binomial Expansion**:
- We need to show that \( (A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k}B^k \).
- Using the principle of mathematical induction:
- Base case \( n=1 \):
\[ A + B = 1 \cdot A + 1 \cdot B \]
- Assume true for \( n=k \):
\[ (A + B)^k = \sum_{j=0}^{k} \binom{k}{j} A^{k-j}B^j \]
- For \( n=k+1 \):
\[ (A + B)^{k+1} = (A + B)(A + B)^k \]
- Expanding and using commutativity leads to the required form.
- Thus, option C is true.

5. **Option D: Proving \( (A - B)(A + B) = A^2 - B^2 \)**:
- Expand the left-hand side:
\[ (A - B)(A + B) = A^2 + AB - BA - B^2 \]
- Since \( AB = BA \), we have:
\[ A^2 - B^2 \]
- Therefore, option D is true.

### Conclusion:
All options A, B, C, and D are true for two commuting square matrices \( A \) and \( B \).