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38 votes

$F$ 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?

- Determinant of $F$ is zero.
- There are an infinite number of solutions to $Fx = b$
- There is an $x≠0$ such that $Fx = 0$
- $F$ must have two identical rows

48 votes

Best answer

**(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.

@`JEET why not?

For a set of **linear** non homogeneous equations, if more than one solutions exist then it means that infinite number of solutions..(other solutions can be multiple of one solution vector)

1

@Sachin Mittal 1 Sir, If the Determinant of F is zero as per your explanation of infinite solutions, then Option A must be True instead of it being false.

0

17 votes

Since Fu = b, and also Fv = b, so we have (Fu - Fb) = 0 i.e. F(u-v) = 0. Since u≠v, F is a singular matrix i.e. its determinant is 0. Now for a singular matrix F, either Fx = b has no solution or infinitely many solutions, but as we are already given two solutions u and v for x, Fx = b has to have infinitely many solutions. Moreover, by definition of singular matrix, there exists an x≠0 such that Fx = 0 . So options (A), (B), and (C) are true. Option (D) is false because it may not be necessary that two rows are identical, instead, two columns can be identical and we can get F as singular matrix then. So option (D) is correct

14 votes

there are more than two vectors satisfying equation FX=0 that means infinite solution exist...

determinant is 0 rank is less than n

but for determinant to be zero its not necessary to have two identical rows

so d is the answer

determinant is 0 rank is less than n

but for determinant to be zero its not necessary to have two identical rows

so d is the answer

1

"A matrix with two identical rows has a determinant of zero."

https://en.wikibooks.org/wiki/Linear_Algebra/Properties_of_Determinants

0

10 votes

A nxn

system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero.homogeneousIf this determinant is zero, then the system has an infinite number of solutions.

A nxn

system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero.non-homogeneousIf this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.

Now, let's look at the given options.

(a) True. Since $F x = b$ is non-homogeneous and has more than one solutions (u and v), determinant of $F$ is surely $0$.

(b) True. Infinitely many solutions are possible.

(c) This option talks about $F x = 0$ which is a homogeneous equation. Since, $\left | F \right | = 0$, thus $x \neq 0$. $\therefore$ True

(d) This option says that $\left | F \right | = 0$ only because of two identical rows in $F$. But this is not necessary.

Answer is (d).