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

Answer

To determine whether the statement \((A - B)^2 = A^2 - 2AB + B^2\) is true or false for matrices \(A\), \(B\), and \(C\) of order \(2 \times 2\), we will expand the left-hand side and compare it with the right-hand side.

### Step-by-Step Solution:

1. **Expand the Left-Hand Side**:
\[ (A - B)^2 = (A - B)(A - B) \]
Using the distributive property (also known as the FOIL method for binomials), we can expand this:
\[ (A - B)(A - B) = A \cdot A - A \cdot B - B \cdot A + B \cdot B \]
This simplifies to:
\[ = A^2 - AB - BA + B^2 \]

2. **Compare with the Right-Hand Side**:
The right-hand side of the equation is:
\[ A^2 - 2AB + B^2 \]
We can rewrite this as:
\[ A^2 - AB - AB + B^2 \]

3. **Analyze the Two Expressions**:
From our expansion, we have:
\[ (A - B)^2 = A^2 - AB - BA + B^2 \]
And from the right-hand side, we have:
\[ A^2 - 2AB + B^2 = A^2 - AB - AB + B^2 \]
For the two expressions to be equal, we need:
\[ -AB - BA = -2AB \]
Rearranging gives us:
\[ -AB - BA + 2AB = 0 \implies AB - BA = 0 \]
This means that \(AB = BA\), which is only true if \(A\) and \(B\) commute.

4. **Conclusion**:
Since \(A\) and \(B\) are arbitrary \(2 \times 2\) matrices, they do not necessarily commute. Therefore, the original statement is **false**.

### Final Answer:
The statement \((A - B)^2 = A^2 - 2AB + B^2\) is **false** because the laws of algebra for factorization and expansion do not hold for matrices unless the matrices commute.