We are tasked with evaluating $(P+Q)^2$ given that $P$ and $Q$ are square and invertible matrices, and $Q = -P^{-1}QP$.

Step 1: Expand $(P+Q)^2$

$(P+Q)^2 = P^2 + PQ + QP + Q^2$

Step 2: Substitute the given condition $Q = -P^{-1}QP$

Since $Q = -P^{-1}QP$, substitute this into the equation to simplify terms involving $Q$. First, note that this implies: $QP = (-P^{-1}QP)P = -I \cdot QP = -QP$ Thus: $QP = -QP \implies QP = 0$

Step 3: Analyze the term $PQ$

From $Q = -P^{-1}QP$, we similarly have: $P^{-1}QP = -Q \implies P^{-1}(QP) = -Q \implies PQ = 0.$

Step 4: Simplify the expression for $(P+Q)^2$

Now substitute the results into the expanded form: $(P+Q)^2 = P^2 + PQ + QP + Q^2.$ Since $PQ = 0$ and $QP = 0$, this reduces to: $(P+Q)^2 = P^2 + Q^2.$

Final Answer:

$(P+Q)^2 = P^2 + Q^2$

Would you like further clarification or additional examples?

Here are 5 related questions to consider:

1. How does the invertibility of matrices affect their properties in this problem?
2. Can this relationship $Q = -P^{-1}QP$ be extended to non-invertible matrices?
3. What would $(P-Q)^2$ evaluate to under the same conditions?
4. How does the commutative property affect operations on square matrices?
5. Can this result be generalized for other forms of matrix relationships?

Tip: Always verify matrix assumptions, such as invertibility, as they critically affect the solution.