Multiple choice

$F$ is an $n\times n$ real matrix. $b$ is an $n\times 1$ real vector. Suppose there are two $n\times 1$ vectors, $u$ and $v$ such that, $u ≠ v$ and $Fu = b, Fv = b$. Which one of the following statements is false?

  1. Determinant of F is zero

  2. There are an infinite number of solutions to Fx = b
  3. There is an $x≠0$ such that $Fx = 0$

  4. F must have two identical rows
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(A) : Correct. We are given

$$Fu = b$$

$$Fv = b$$

So $F(u-v) = 0$

Since $u \neq v$, so we have a non-zero solution $w = (u-v)$ to homogeneous equation $Fx=0$. Now any vector $\lambda w$ is also a solution of $Fx=0$, and so we have infinitely many solutions of $Fx=0$, and so determinant of F is zero.

(B) : Correct. Consider a vector $u+\lambda w$.

$$F(u+\lambda w) = Fu+F(\lambda w) = b + 0 = b$$

So there are infinitely many vectors of the form $u+\lambda w$, which are solutions to equation $Fx=b$.

(C) : Correct. In option (a), we proved that vector $(u-v) \neq 0$ satisfies equation $Fx=0$.

(D) : False. This is not necessary.

So option (D) is the answer.

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