Consider a system of linear equations \(𝑃𝑋 = 𝑄\) where \(𝑃 ∈ ℝ^{3×3}\) and…

Summary of which statements are true and why:

• The statement that the system can be solved by first solving the lower triangular system and then the upper triangular system is true. Given P = L U and P X = Q, set Y = U X. Then L Y = Q, which is solved by forward substitution, and then U X = Y is solved by backward substitution.

• The statement that if P is invertible then both L and U are invertible is true. Because L is unit lower triangular, det(L)=1 so L is invertible. Since det(P)=det(L)·det(U)=det(U), det(P)≠0 implies det(U)≠0, so U is invertible.

• The statement that if P is singular then at least one diagonal element of U is zero is true. If P is singular then det(P)=0. With det(L)=1, det(P)=det(U), so det(U)=0. For a triangular matrix, det(U) is the product of its diagonal entries, hence at least one diagonal entry must be zero.

• The statement that if P is symmetric then both L and U are symmetric is false. Symmetry of P does not force L and U from a general LU factorization to be symmetric. Only in special factorizations for symmetric positive definite matrices (Cholesky) do we get a lower triangular factor whose transpose is the upper factor. A general symmetric matrix can have an LU factorization where L and U are not symmetric.

Final conclusion: The first, second, and third statements are true; the fourth statement is false.