The image contains a mathematical problem that reads:

Question 10 (16 pts): For matrices $A$ and $B$, where $AB = BA$ (i.e., $A$ and $B$ commute), prove or disprove the following statement:

$(AB)^{10} = A^{10}B^{10}.$

Let's go through the proof or disproof of the given statement.

Proof/Disproof:

Given that $A$ and $B$ are matrices and $AB = BA$, the goal is to determine whether $(AB)^{10} = A^{10}B^{10}$.

First, consider the general property for any two commuting matrices $A$ and $B$: $(AB)^n = A^nB^n \quad \text{if } AB = BA.$

We can prove this by induction on $n$.

Base Case ($n = 1$): $(AB)^1 = AB = A^1B^1.$ The statement holds true for $n = 1$.

Inductive Step:

Assume the statement is true for some $n = k$, i.e., $(AB)^k = A^kB^k.$ We need to show it holds for $n = k + 1$.

Consider: $(AB)^{k+1} = (AB)^k \cdot AB.$ Using the inductive hypothesis: $(AB)^{k+1} = (A^kB^k) \cdot AB.$ Since $A$ and $B$ commute, we have: $(A^kB^k) \cdot AB = A^k \cdot (B^k \cdot A) \cdot B = A^k \cdot (A \cdot B^k) \cdot B = A^k \cdot A \cdot B^k \cdot B = A^{k+1} \cdot B^{k+1}.$

Thus: $(AB)^{k+1} = A^{k+1}B^{k+1}.$

By induction, the statement $(AB)^n = A^nB^n$ holds for all positive integers $n$.

Conclusion:

Since the above holds true for any integer $n$, specifically for $n = 10$: $(AB)^{10} = A^{10}B^{10}.$ Thus, the statement $(AB)^{10} = A^{10}B^{10}$ is proven to be true.

Would you like further details or have any specific questions?

Here are some related questions:

1. What other properties of commuting matrices can be derived?
2. How does this proof change if the matrices $A$ and $B$ do not commute?
3. What are some examples of matrices where $AB \neq BA$?
4. How would this proof look in the case of $(AB)^n = A^nB^n$ for $n = 2$?
5. Can we generalize this result to more than two matrices?

Tip: When working with matrices, always check if they commute, as many matrix identities and simplifications depend on this property.