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$F$ 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

Ans is D. I am getting option A, B  be true and D  be false. I'm not understanding option C. Pls explain it

If D would have been “ F must have two linearly dependent rows” then it would be correct as well.

(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.

Having two identical rows means, that F determinant must be 0 which is same as (A). Then why (D) is considered as false?
Your statement is true, but its converse is not true i.e. if F determinant is 0, then it is not necessary that there must be two identical rows.
$u ≠ v$ and $Fu = b, Fv = b$ this tells that $Fx=b$  has infinite solutions, (either system of equation has unique solution or infinite solutions)

this is possible only when $\text{rank} \lt n$ that means determinant of $F$ is zero.

if $\text{rank} \lt n$ then of course $Fx=0$ also has infinite solutions.
How you assumed it's homogeneous equation?
Does having more than one solution ==> Infinitely many solutions Always ??

Does having more than one solution ==> Infinitely many solutions

Yes.

@Sachin Mittal 1 sir
If the determinant of a matrix is 0. Then is it always possible to write one row as a linear combination of another row?

@Verma Ashish

I don't think so. Is it?

@`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)

@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.

option A is true but question is asking which option is false.
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

1 comment

AB= 0 if B is non singular then A must be singular or it can be non singular also??
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

option c is wrong?
here equation has infinite solutions so there exists vectors like y,z...etc others than given u v which will be satisfy Fx=0

"A matrix with two identical rows has a determinant of zero."
https://en.wikibooks.org/wiki/Linear_Algebra/Properties_of_Determinants

edited by
can someone explain option B and C in a simpler way

A nxn homogeneous system of linear equations has a unique solution (the trivial solution) if and only if its determinant is non-zero.

If this determinant is zero, then the system has an infinite number of solutions.

A nxn non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero.

If 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.