1. **Problem (a):** Prove that if $A$ and $B$ are symmetric $n \times n$ matrices, then $A + cB$ is symmetric for all scalars $c$. 2. **Step 1:** Recall the definition of a symmetric matrix: a matrix $M$ is symmetric if $M^T = M$. 3. **Step 2:** Since $A$ and $B$ are symmetric, we have $A^T = A$ and $B^T = B$. 4. **Step 3:** Consider the transpose of $A + cB$: $$ (A + cB)^T = A^T + (cB)^T = A + cB^T = A + cB $$ where we used the properties $(X + Y)^T = X^T + Y^T$ and $(cX)^T = cX^T$. 5. **Step 4:** Since $(A + cB)^T = A + cB$, the matrix $A + cB$ is symmetric. --- 6. **Problem (b):** Suppose $A$ and $B$ are non-symmetric matrices. Determine if there exist such $A$ and $B$ so that $A + 2B$ is symmetric, or prove no such matrices exist. 7. **Step 1:** Let $S = A + 2B$. For $S$ to be symmetric, $S^T = S$. 8. **Step 2:** Taking transpose: $$ S^T = (A + 2B)^T = A^T + 2B^T $$ 9. **Step 3:** The symmetry condition $S^T = S$ implies: $$ A^T + 2B^T = A + 2B $$ which rearranges to $$ A^T - A = -2(B^T - B) $$ 10. **Step 4:** The left side $A^T - A$ and right side $-2(B^T - B)$ are both skew-symmetric matrices (since $M^T - M$ is skew-symmetric for any matrix $M$). 11. **Step 5:** This means if we pick any skew-symmetric matrix $K$, we can set $$ A^T - A = K, \quad B^T - B = -\frac{1}{2}K $$ and choose $A$ and $B$ accordingly to satisfy these. 12. **Step 6:** For example, consider $n=2$ and $$ K = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} $$ Then $$ A^T - A = K, \quad B^T - B = -\frac{1}{2}K = \begin{bmatrix} 0 & -\frac{1}{2} \\ \frac{1}{2} & 0 \end{bmatrix} $$ 13. **Step 7:** Choose $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$ Check that $A$ and $B$ are not symmetric: $$ A^T = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \neq A, \quad B^T = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \neq B $$ 14. **Step 8:** Compute $A + 2B$: $$ A + 2B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + 2 \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 2 & 0 \end{bmatrix} $$ 15. **Step 9:** Check symmetry: $$ (A + 2B)^T = \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix} \neq A + 2B $$ So this example does not work; adjust $A$ and $B$ to satisfy the skew-symmetric difference condition. 16. **Step 10:** Instead, choose $$ A = \begin{bmatrix} a & x \\ y & d \end{bmatrix}, \quad B = \begin{bmatrix} e & p \\ q & h \end{bmatrix} $$ with the conditions $$ A^T - A = \begin{bmatrix} 0 & y - x \\ x - y & 0 \end{bmatrix} = K, \quad B^T - B = \begin{bmatrix} 0 & q - p \\ p - q & 0 \end{bmatrix} = -\frac{1}{2}K $$ which implies $$ y - x = k, \quad q - p = -\frac{k}{2} $$ for some scalar $k \neq 0$. 17. **Step 11:** For example, set $k=2$, then $$ y - x = 2, \quad q - p = -1 $$ Choose $x=0$, $y=2$, $p=0$, $q=-1$. 18. **Step 12:** Then $$ A = \begin{bmatrix} a & 0 \\ 2 & d \end{bmatrix}, \quad B = \begin{bmatrix} e & 0 \\ -1 & h \end{bmatrix} $$ 19. **Step 13:** Compute $A + 2B$: $$ A + 2B = \begin{bmatrix} a + 2e & 0 \\ 2 + 2(-1) & d + 2h \end{bmatrix} = \begin{bmatrix} a + 2e & 0 \\ 0 & d + 2h \end{bmatrix} $$ which is symmetric. 20. **Step 14:** Since $A$ and $B$ are not symmetric (because $y \neq x$ and $q \neq p$), but $A + 2B$ is symmetric, such matrices exist. **Final answers:** - (a) $A + cB$ is symmetric for all scalars $c$ if $A$ and $B$ are symmetric. - (b) There exist non-symmetric matrices $A$ and $B$ such that $A + 2B$ is symmetric, as shown by the example construction above.