in Linear Algebra
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38 votes
38 votes

Let $Ax = b$ be a system of linear equations where $A$ is an $m \times n$ matrix and $b$ is a $m \times 1$ column vector and $X$ is an $n \times1$ column vector of unknowns. Which of the following is false?

  1. The system has a solution if and only if, both $A$ and the augmented matrix $[Ab]$ have the same rank.

  2. If $m < n$ and $b$ is the zero vector, then the system has infinitely many solutions.

  3. If $m=n$ and $b$ is a non-zero vector, then the system has a unique solution.

  4. The system will have only a trivial solution when $m=n$, $b$ is the zero vector and $\text{rank}(A) =n$.

in Linear Algebra
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4 Comments

Corrected.
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Ans:C


A) True

B) The homogeneous (with all constant terms equal to zero) underdetermined linear system always has non-trivial solutions.

https://en.wikipedia.org/wiki/Underdetermined_system

C) Can't claim anything about the nature of solution.

D) True

$A_{n*n}X_{n*1}$=0 & |A|$\neq$ 0 & $\rho (A)=n$, then trivial solution

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In option b they are talking about homogenous system of equation but how it can be infinite no of solution when nothing is given about rank .solution is infinite when rank<no of unknown.no such thing is given in option so how we can decide it is correct??
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4 Answers

29 votes
29 votes
Best answer

Answer is $C$ because it is a case of linear non-homogeneous equations. By having m = n, we can't say that it will have unique solution. Solution depends on rank of matrix A and matrix [A B].

If rank[ A ] = rank[A B], then it will have solution, otherwise no solution

edited by

4 Comments

option d is right because , rank of the matrix = n which means the the number of linearly independent vectors is n. If we have only linearly independent vectors in the matrix  then the only values of X=( x1,x2,x3) that is going to satisfy AX=0 will be 0,0,0. No other values of x1,x2,x3 is going to satisfy AX=0

so why they have given that when the rank = n ,  only trivial solution exist for AX=0

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AX = 0 , here X has trivial solution only if |A| !=0.
Now m=n means let's say the order of the matrix is n*n. Now rank(A) = n means smallest submatrix which have n*n order have determinant not equals to 0 which is the matrix A itself as m=n. i.e |A| != 0.

So option d is right.
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how to check option B?
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14 votes
14 votes

A. It is true, either system will have unique or infinitely many solutions, but system is consistent.
B. Rank of matrix will be less than the number of unknowns and it's an homogenous system of linear equations, So, this is also true.
C. It may not be true, homogenous system may have infinite number of solutions.
D. It's an homogenous system, number of unknowns are equal to the rank, hence the system will have unique solution.

Hence, (C) is false!

1 comment

Hello manu

in 'c' he is talking about non-homo system , see b is non-zero vector.
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3 votes
3 votes

Consider a matrix $A_{3X4}$ = $\begin{bmatrix} 1 & 3& 3 &2 \\ 2& 6 & 9&7 \\ -1 &-3 &3 & 4 \end{bmatrix}$

The Echelon form of this matrix after Row operations

$R_2=R_2-2R_1$

$R_3=R_3+R_1$

$R_3=R_3-2R_2$

would be

$\begin{bmatrix} 1 & 3& 3 &2 \\ 0& 0 & 3&3\\ 0 &0 &0 & 0 \end{bmatrix}$

Now, we have our system $Ax=b$

which can be represented as

$\begin{bmatrix} 1 & 3& 3 &2 \\ 2& 6 & 9&7 \\ -1 &-3 &3 & 4 \end{bmatrix}$.$\begin{bmatrix} u\\ v\\ w\\ y \end{bmatrix}=\begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}$

Perform the same Row operations as listed above and then it becomes

$\begin{bmatrix} 1 & 3 & 3 & 2\\ 0&0 & 3 & 3\\ 0&0 & 0 & 0 \end{bmatrix}.\begin{bmatrix} u\\ v\\ w\\ y \end{bmatrix}=\begin{bmatrix} b_1\\ b_2-2b_1\\ b_3-2b_2+5b_1 \end{bmatrix}$

Let this system be $Ux=c$

Now, your system would be consistent, if $b_3-2b_2+5b_1=0$ which validates option (A).

Option (B), says if $Ax=0$ then this system has infinitely many solutions and yes, because you take any constant $c \not=0$,

$Acx=0$

Option(D) says if A is an nxn matrix and rank(A)=n, means all the rows and columns of A are linearly independent, then there is no such non-zero vector that can take some combinations of columns of A and produce a zero vector. And, thus the only solution to $Ax=0$ is the vector $X=0$ which is a trivial solution, so this option is true.

Option(C) asserts that if m=n and b is a non-zero vector, then system has unique solution which is clearly false.

Why?

For the system, $Ax=b$ to have a solution for a non-zero b, the vector b must be such that it can be produced by some linear combinations of columns of A.This means, b must be in the Column Space of A.

In the above system $Ux=c$ it has a solution  when  $c=$$\begin{bmatrix} 1(b_1))\\ 5(b_2)\\ 5(b_3) \end{bmatrix}$ and it is inconsistent, means does not have a solution when $c=$$\begin{bmatrix} 1(b_1))\\ 5(b_2)\\ 6(b_3) \end{bmatrix}$

Hence, Option (C) is the answer.

3 Comments

Appreciated.

Thanks
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@ Is there any good reference for this concept??

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@MRINMOY_HALDER-Gilbert strang lectures plus his book

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0 votes
0 votes
A)the system has a solution if and only if ,both A and augmented matrix[AB] have the same rank because [0 0 0 0 0 nonzero] does not exist .it is true.

B)it is true because if m<n there is possibility free variable exist , then there is infinite many solutions exist.

c)it is false because if m==n it does not mean that there exist a unique solution soltion depend on rank if rank[A]=rank[AB] then only unque solution exist.

d)it is true because rank=n and does not exist any free variable
Answer:

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