1. Problem statement: (a) Given a matrix $$A=\begin{bmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1 \\ 1 & 1 & a & 1 \\ 1 & 1 & 1 & a \end{bmatrix}$$ We want to express the determinant of $$A$$ in terms of $$a$$ and find when $$A$$ is singular or nonsingular. 2. Expressing determinant: Note the structure of $$A$$: diagonal entries are $$a$$, off-diagonal entries are $$1$$. We can write $$A = aI + J - I$$ where $$I$$ is the identity matrix and $$J$$ is the matrix of all ones. Since $$J$$ has eigenvalues $$4$$ (because it's a 4x4 matrix) and $$0$$ (multiplicity 3), determinant can be computed using eigenvalues. The eigenvalues of $$A$$ are: $$\lambda_1 = a - 1 + 4 = a + 3$$ $$\lambda_2 = a - 1$$ (with multiplicity 3) 3. Therefore, $$\det(A) = (a + 3)(a - 1)^3$$ 4. Singular and nonsingular: - $$A$$ is singular if $$\det(A) = 0$$, which occurs when $$a = -3$$ or $$a = 1$$. - $$A$$ is nonsingular if $$a \neq -3$$ and $$a \neq 1$$. --- 5. Problem statement (b): Theorem 1 states if $$B$$ is a square matrix, then $$B + B^T$$ is symmetric. 6. To prove: - Recall a matrix $$M$$ is symmetric if $$M = M^T$$. 7. Derivation: Consider $$M = B + B^T$$. Then, $$M^T = (B + B^T)^T = B^T + (B^T)^T = B^T + B = M$$ Since $$M^T = M$$, $$B + B^T$$ is symmetric. 8. Explanation: Adding a matrix to its transpose ensures symmetry because the operation "averages" entries with their transposed positions. --- Final answers: (a) $$\det(A) = (a + 3)(a - 1)^3$$; singular when $$a=-3$$ or $$a=1$$; nonsingular otherwise. (b) The matrix $$B + B^T$$ is symmetric for any square matrix $$B$$, since $$(B + B^T)^T = B + B^T$$.