Consider a system of linear equations PX = Q where P: R^{k-1} -> R^{k-1} and Q: R^{k-1}. Suppose P has an LU decomposition, P = LU, where L = [[1,0,0],[l_21,1,0],[l_31,l_32,1]] and U = [[u_11,u_12,u_13],[0,u_22,u_23],[0,0,u_33]] Which of the following statement(s) is/are TRUE? A. The system PX = Q can be solved by first solving LY = Q and then UX = Y. B. If P is invertible, then both L and U are invertible. C. If P is singular, then at least one of the diagonal elements of U is zero. D. If P is symmetric, then both L and U are symmetric.

GATE 2025 · Engineering Mathematics · LU Decomposition · medium

Answer: Statements A, B, and C are TRUE.

1. Verify Statement A: two-step solve: Since P = LU, we can write L(UX) = Q. Define Y = UX. Then LY = Q can be solved by forward substitution giving Y, then UX = Y by back substitution gives X.
2. Verify Statement B: invertibility: P invertible <=> det(P) != 0 <=> det(U) != 0 <=> U invertible. L is unit lower triangular so det(L) = 1, making L always invertible.
3. Verify Statement C: singular P implies zero diagonal in U: P singular => det(P) = 0 => det(U) = 0 => product of diagonals of U is zero => at least one u_ii = 0.
4. Verify Statement D: symmetry: L is lower triangular with 1s on diagonal; U is upper triangular. For both to be symmetric, L would need to equal U^T which is generally not the case. A symmetric P does not imply L or U are individually symmetric.