GATE CS 2006 Q23 — Linear Algebra: Matrices, Determinants, System of Linear Equations

We are given that $F$ is an $n \times n$ real matrix, and there exist two distinct vectors $u$ and $v$ ($u \neq v$) such that:

Subtracting the second equation from the first, we get:

Let $x = u - v$. Since $u \neq v$, we have $x \neq 0$. Thus, there exists a non-zero vector $x$ such that $Fx = 0$.

Let's analyze the given statements:

• Statement C is True: As derived above, $x = u - v \neq 0$ satisfies $Fx = 0$.
• Statement A is True: Since there exists a non-zero vector $x$ in the null space of $F$ (i.e., the system $Fx = 0$ has a non-trivial solution), the matrix $F$ is singular. Therefore, the determinant of $F$ is zero ($\det(F) = 0$).
• Statement B is True: For any scalar $k \in \mathbb{R}$, the vector $y = u + k(u - v)$ is a solution because: $$Fy = F(u + k(u - v)) = Fu + kF(u - v) = b + k(0) = b$$ Since there are infinitely many real values for $k$, there are infinitely many solutions to $Fx = b$.
• Statement D is False: A matrix with a determinant of zero must have linearly dependent rows, but they do not need to be identical. For example, consider the matrix: $$F = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$$ Here, the rows are not identical, yet $\det(F) = 0$, and it satisfies all the given conditions.

Thus, the false statement is D.