Consider a system of linear equations $PX = Q$ where $P \in \mathbb{R}^{3\times 3}$ and $Q \in \mathbb{R}^{3\times 1}$. Suppose $P$ has an LU decomposition, $P = LU$, where $$L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \text{ and } U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}.$$ Which of the following statement(s) is/are TRUE?

Let us analyze each statement based on the properties of LU decomposition:

• Statement (A): Substituting $P = LU$ into $PX = Q$ gives $L(UX) = Q$. Letting $Y = UX$, we first solve $LY = Q$ for $Y$ by forward substitution, and subsequently solve $UX = Y$ for $X$ by back substitution. Thus, statement (A) is TRUE.

• Statement (B): The determinant of the product of two matrices is the product of their determinants:

  $$\det(P) = \det(LU) = \det(L) \cdot \det(U)$$

  Since $L$ is a unit lower triangular matrix, $\det(L) = 1 \neq 0$, meaning $L$ is always invertible. If $P$ is invertible, then $\det(P) \neq 0$, which implies:

  $$\det(U) = \frac{\det(P)}{\det(L)} = \det(P) \neq 0$$

  Thus, $U$ is also invertible. Therefore, statement (B) is TRUE.

• Statement (C): If $P$ is singular, then $\det(P) = 0$. Since $\det(L) = 1$, we have:

  $$\det(P) = \det(L) \cdot \det(U) = 1 \cdot (u_{11} u_{22} u_{33}) = 0$$

  This requires at least one of the diagonal elements of $U$ ($u_{11}, u_{22}, \text{ or } u_{33}$) to be zero. Thus, statement (C) is TRUE.

• Statement (D): A triangular matrix (upper or lower) can only be symmetric if it is a diagonal matrix. For a general symmetric matrix $P$, $L$ has non-zero off-diagonal entries below the diagonal and $U$ has non-zero off-diagonal entries above the diagonal, so neither $L$ nor $U$ is symmetric in general (instead, $P = LDL^T$). Thus, statement (D) is FALSE.

Hence, the correct options are A, B, and C.