F is an n x n real matrix. b is an n x 1 real vector. Suppose there are two n x 1 vectors, u and v such that u != v, Fu = b and Fv = b. Which one of the following statements is FALSE? A. Determinant of F is zero. B. There are an infinite number of solutions to Fx = b. C. There is an x != 0 such that Fx = 0. D. F must have two identical rows.
GATE 2006 · Engineering Mathematics · Matrix · medium
Answer: Statement D is FALSE. Answer: D.
- Establish key consequence: non-trivial null space: Let w = u - v. Since u != v, w != 0. And Fw = 0. So F has a non-zero null space vector. Therefore F is singular: det(F) = 0. Statement A is TRUE.
- Verify statements B and C: Statement B: Take x* = u. Then for any scalar t, F(u + tw) = Fu + tFw = b + 0 = b. So infinitely many solutions exist. TRUE. Statement C: w = u-v != 0 and Fw = 0. So x = w satisfies Fx = 0 with x != 0. TRUE.
- Evaluate statement D: Statement D claims F must have two identical rows. This is FALSE. A matrix can be singular without having identical rows. Example: [[1,1],[2,2]] is singular (det=0) and has proportional (not identical) rows. More generally, [[1,0,1],[0,1,1],[1,1,2]] is singular (det=0) with all rows distinct.