Question 1164620
To show that the matrix $M = (A^T + k B^T)(A + k B)$ is symmetric, we must prove that the transpose of the entire expression is equal to the expression itself. That is, we must show that $M^T = M$.

### **Definitions and Properties**
We will use the following fundamental properties of matrix transposes:
1.  **Sum/Difference Rule:** $(X + Y)^T = X^T + Y^T$
2.  **Product Rule:** $(XY)^T = Y^T X^T$
3.  **Scalar Rule:** $(kX)^T = k(X^T)$
4.  **Double Transpose Rule:** $(X^T)^T = X$

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### **Proof**

Let $M = (A^T + k B^T)(A + k B)$.

**Step 1: Apply the transpose to the entire product.**
Using the **Product Rule** $(XY)^T = Y^T X^T$, we swap the order of the two main factors and transpose them:
$$M^T = \left[ (A^T + k B^T) (A + k B) \right]^T = (A + k B)^T (A^T + k B^T)^T$$

**Step 2: Distribute the transpose into the sums.**
Using the **Sum Rule** and the **Scalar Rule** on both factors:
$$(A + k B)^T = A^T + (kB)^T = A^T + k B^T$$
$$(A^T + k B^T)^T = (A^T)^T + (kB^T)^T = A + k(B^T)^T = A + kB$$

**Step 3: Substitute the results back into the expression.**
Now substitute the simplified factors back into the equation from Step 1:
$$M^T = (A^T + k B^T) (A + k B)$$

**Step 4: Conclusion.**
We observe that:
$$M^T = (A^T + k B^T) (A + k B) = M$$

Since the transpose of the matrix is equal to the original matrix, the expression $(A^T + k B^T)(A + k B)$ is **symmetric**. $\blacksquare$