State with reason, whether the following are true of false. A, B and C are matrices of order `2 xx 2`.
`A^2-B^2=(A+B) (A-B)`

Answer

To determine whether the statement \( A^2 - B^2 = (A + B)(A - B) \) is true or false for matrices \( A \), \( B \), and \( C \) of order \( 2 \times 2 \), let's analyze the expression step by step.

### Step-by-Step Solution

1. **Understanding the Expression**:
The expression \( A^2 - B^2 \) is a difference of squares. In algebra, the difference of squares can be factored as:
\[ A^2 - B^2 = (A + B)(A - B) \]
However, we need to check if this property holds for matrices.

2. **Matrix Multiplication is Not Commutative**:
One of the key properties of matrices is that matrix multiplication is not commutative. This means that \( AB \neq BA \) in general for matrices \( A \) and \( B \).

3. **Counterexample**:
To illustrate that the equation does not hold for matrices, we can consider specific \( 2 \times 2 \) matrices. Let:
\[ A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, \quad B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} \]
Now, calculate \( A^2 \) and \( B^2 \):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 2 \cdot 3 & 1 \cdot 2 + 2 \cdot 4 \\ 3 \cdot 1 + 4 \cdot 3 & 3 \cdot 2 + 4 \cdot 4 \end{pmatrix} = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} \]
\[ B^2 = B \cdot B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} \cdot \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} = \begin{pmatrix} 5 \cdot 5 + 6 \cdot 7 & 5 \cdot 6 + 6 \cdot 8 \\ 7 \cdot 5 + 8 \cdot 7 & 7 \cdot 6 + 8 \cdot 8 \end{pmatrix} = \begin{pmatrix} 67 & 78 \\ 91 & 110 \end{pmatrix} \]

4. **Calculating \( A^2 - B^2 \)**:
Now, compute \( A^2 - B^2 \):
\[ A^2 - B^2 = \begin{pmatrix} 7 & 10 \\ 15 & 22 \end{pmatrix} - \begin{pmatrix} 67 & 78 \\ 91 & 110 \end{pmatrix} = \begin{pmatrix} 7 - 67 & 10 - 78 \\ 15 - 91 & 22 - 110 \end{pmatrix} = \begin{pmatrix} -60 & -68 \\ -76 & -88 \end{pmatrix} \]

5. **Calculating \( (A + B)(A - B) \)**:
Next, calculate \( (A + B)(A - B) \):
\[ A + B = \begin{pmatrix} 1 + 5 & 2 + 6 \\ 3 + 7 & 4 + 8 \end{pmatrix} = \begin{pmatrix} 6 & 8 \\ 10 & 12 \end{pmatrix} \]
\[ A - B = \begin{pmatrix} 1 - 5 & 2 - 6 \\ 3 - 7 & 4 - 8 \end{pmatrix} = \begin{pmatrix} -4 & -4 \\ -4 & -4 \end{pmatrix} \]
Now, compute \( (A + B)(A - B) \):
\[ (A + B)(A - B) = \begin{pmatrix} 6 & 8 \\ 10 & 12 \end{pmatrix} \cdot \begin{pmatrix} -4 & -4 \\ -4 & -4 \end{pmatrix} = \begin{pmatrix} 6 \cdot -4 + 8 \cdot -4 & 6 \cdot -4 + 8 \cdot -4 \\ 10 \cdot -4 + 12 \cdot -4 & 10 \cdot -4 + 12 \cdot -4 \end{pmatrix} = \begin{pmatrix} -48 & -48 \\ -88 & -88 \end{pmatrix} \]

6. **Conclusion**:
Since \( A^2 - B^2 \neq (A + B)(A - B) \), we conclude that the statement is **false**.

### Final Answer:
The statement \( A^2 - B^2 = (A + B)(A - B) \) is **false** for matrices.