`A` and `B` are two square matrices such that `A^(2)B=BA` and if `(AB)^(10)=A^(k)B^(10)` then the value of `k-1020` is.

Answer

To solve the problem, we need to analyze the given equations and derive the necessary values step by step.

### Step-by-Step Solution:

1. **Given Information**:
We have two square matrices \( A \) and \( B \) such that:
\[ A^2 B = BA \]
and it is also given that:
\[ (AB)^{10} = A^k B^{10} \]

2. **Finding \( (AB)^2 \)**:
We start by calculating \( (AB)^2 \):
\[ (AB)^2 = AB \cdot AB = A(BA)B \]
From the given condition \( A^2 B = BA \), we can substitute \( BA \) with \( A^2 B \):
\[ (AB)^2 = A(A^2 B)B = A^3 B^2 \]

3. **Finding \( (AB)^3 \)**:
Next, we calculate \( (AB)^3 \):
\[ (AB)^3 = (AB)^2 \cdot AB = (A^3 B^2)(AB) = A^3 (B^2 A) B \]
Again, using \( BA = A^2 B \):
\[ (AB)^3 = A^3 (A^2 B) B = A^5 B^3 \]

4. **Generalizing \( (AB)^n \)**:
From the pattern observed, we can generalize:
\[ (AB)^n = A^{2n-1} B^n \]
This can be derived from the previous calculations, where we see that each time we multiply by \( AB \), the power of \( A \) increases by 2 and the power of \( B \) increases by 1.

5. **Finding \( (AB)^{10} \)**:
Using our generalized formula:
\[ (AB)^{10} = A^{2 \cdot 10 - 1} B^{10} = A^{19} B^{10} \]

6. **Equating to Given Condition**:
We know from the problem statement that:
\[ (AB)^{10} = A^k B^{10} \]
Therefore, we can equate:
\[ A^{19} B^{10} = A^k B^{10} \]
This implies:
\[ k = 19 \]

7. **Finding \( k - 1020 \)**:
Finally, we need to find:
\[ k - 1020 = 19 - 1020 = -1001 \]

### Final Answer:
The value of \( k - 1020 \) is:
\[ \boxed{-1001} \]