F is an n x n real matrix and b is an n x 1 real vector. Suppose there are two…
2006
F is an n x n real matrix and 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.
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Correct answer: D
Given Fu = b and Fv = b with u != v.
Subtract the two equations:
Fu - Fv = b - b
F(u - v) = 0.
Since u != v, the vector u - v is nonzero. Therefore, F has a nonzero vector in its nullspace.
So F is singular, and hence det(F) = 0. Also, because one solution u exists and there is a nonzero nullspace direction v - u, infinitely many solutions exist:
u + t(v - u), for any real number t.
However, a singular matrix does not have to contain two identical rows. It only means that the rows or columns are linearly dependent. Linear dependence may occur without any two rows being identical.
Therefore, the false statement is: F must have two identical rows.