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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
asked in Linear Algebra by Active (3.7k points)
edited by | 1.4k views
0
Ans is D. I am getting option A, B  be true and D  be false. I'm not understanding option C. Pls explain it

3 Answers

+23 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.

answered by Boss (11.3k points)
edited by
0
Having two identical rows means, that F determinant must be 0 which is same as (A). Then why (D) is considered as false?
+11
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.
+17
$u ≠ v$ 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.
0
How you assumed it's homogeneous equation?
+10 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
answered by Boss (31.7k points)
0
option c is wrong?
+1
here equation has infinite solutions so there exists vectors like y,z...etc others than given u v which will be satisfy Fx=0
0

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

0
can someone explain option B and C in a simpler way
+7 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
answered by Loyal (9k points)


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