6.4k views

Let $c_{1}.....c_{n}$ be scalars, not all zero, such that $\sum_{i=1}^{n}c_{i}a_{i}$ = 0 where $a_{i}$ are column vectors in $R^{n}$.

Consider the set of linear equations

$Ax = b$

where $A=\left [ a_{1}.....a_{n} \right ]$ and $b=\sum_{i=1}^{n}a_{i}$. The set of equations has

1. a unique solution at $x=J_{n}$ where $J_{n}$ denotes a $n$-dimensional vector of all 1.
2. no solution
3. infinitely many solutions
4. finitely many solutions

edited | 6.4k views
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did not even touch it
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Representation of questions in this way, confuses me a lot.
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@Aboveallplayer

what was your gate score that year?

Just curious... :)

$\sum_i c_i a_i = 0 \text{ with } \exists i: c_i \ne 0$ indicates that column vectors of $A$ are linearly dependent. Determinant of matrix $A$ would be zero.  Therefore either $Ax=b$ has no solution or infinitely many solutions. From $\sum_i a_i = b$, it is clear that a $n$-dimensional vector of all $1$ is a solution of equation Ax=b.

Hence, $Ax=b$  will have infinitely many solutions. The correct answer is $(C)$.
by Loyal (6.1k points)
edited
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i am getting same answer option C is correct
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"column vectors of A are linearly dependent. Determinant of matrix A would be zero" This statement seems to be contradictory.  For the column vectors of A to be linearly independent, the determinant should be non-zero(Wikipedia link). Linearly independent matrix are full rank matrix and have a unique solution. The correct answer should be option A.

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I'm saying linearly dependent, not independent :)
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@suraj sir

"From ∑iai=b, it is clear that a n-dimensional vector of all 1 is a solution of equation Ax=b". how are you saying this ?? A is an n*n matrix (formed by having n column vectors) ... b is a column vector of size n*1 ... x is a n*1 vector .... Now how having x as all 1's will come a solution to Ax = b ... ??? Can you please clear this ?

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b vector is sum of all column vectors of A. That is why vector of all 1 is a solution of equation Ax = b .
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"∃i:ci≠0 " how to know this?

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@Akash Its given in question
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@suraj

what does ci represent here?
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ci is scalar (some constant).
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got it now..thanks
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@suraj sir , Thank you for the answer ! .

Sir , Please check these 2  points and correct me if I am wrong :-

1) Since , from the equation  $b = \sum_{i=1}^{n} a_{i}$  , it is clear that $n-dimensional$ $vector \; 'b'$    is the linear combination of all the linearly independent vectors of matrix $'A'$

So, vector $'b'$ should be a vector in the column space of matrix $'A'$ where column space of  $'A'$   is the subspace of $\mathbb{R}^{n}$ and it contains all linearly independent vectors of matrix $A$ and their combination.

Since , $b$ is in column space of $A$ , So, given system must have at least one solution.

2) Since , matrix $A$ does not have $'n'$ linearly independent vectors , So , dimension of column space(or rank) will be less than $'n'$ , it means system must have infinitely many solution.
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Why option A is wrong ..after getting one solution i.e (X ==> N dimensional vector of all 1's )how can we conclude that system has Infinite / finite /Unique solution.. ??
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@jatin khachane 1 i too have same doubt..did you get any explanation?

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Consider the following system of equations.

x + y + 0.z = 2

0.x + y + z = 2

x + 2y + z = 4

Here, $a_1^T = [1, 0, 1]$, $a_2^T = [1, 1, 2]$, $a_3^T = [0, 1, 1]$ .

If we take $c_1 =1$, $c_2=-1$, $c_3=1$, we can easily check that $c_1a_1 +c_2a_2 +c_3a_3=0$.

We can also check here that $b=\sum_{i=1}^{3} a_i$.

The above system of equations have solution in the form of $\begin{bmatrix} t\\ 2-t\\ t \end{bmatrix}$. Hence $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ is one of the solutions (not a unique solution) for the above system of equations.

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@suraj sir.. okay i got it how equation is having infinite solution but how you are inferring that there may be a case which have no solution

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If determinant of matrix A is zero, then  Ax=b has no solution or infinitely many solutions.

No solution case:

x+y=2

2x+2y=7

Infinitely many solution case:

x+y = 2

2x+2y=4
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Okay so it is clear that determinant of A will be zero, but how can we say it cannot have unique solution as number of unknowns is not mentioned in question. For ex, suppose A=[a1,a2,a3,a4] and X=[x,y,z]. Here determinant of A will be zero so rank(A) is not equal to 4 but it may be equal to 3 which is same as number of unknowns so unique solution is also possible. What's wrong in this?

by (427 points)
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nice one
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Suprb 👍
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Why we are considering |A| = 0 Is it given in the question? Please explain.
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A is linear dependent.....how???

Given for some scalars, not all  , means  $\exists c_i :c_i \not=0$ for $1 \leq i \leq n$

$\sum_{i=1}^{n}c_i.a_i=0$ where $a_i$ are column vectors (of size nx1) in $R_n$

so this means

$c_1\begin{bmatrix} x_{1,1}\\ x_{2,1}\\ .\\ .\\ .\\ x_{n,1} \end{bmatrix}$ +$c_2\begin{bmatrix} x_{1,2}\\ x_{2,2}\\ .\\ .\\ .\\ x_{n,2} \end{bmatrix}+....+$$c_n\begin{bmatrix} x_{1,n}\\ x_{2,n}\\ .\\ .\\ .\\ x_{n,n} \end{bmatrix}=0 so if I make a vector of all c_i for 1 \leq i \leq n, such that say vector k=$$\begin{bmatrix} c_1\\ c_2\\ .\\ .\\ .\\ c_n \end{bmatrix}$, this $k$ would be a non-zero vector.

Now, given $A=[a_1...a_n]$, so I have a vector k such that $Ak=0$

or in other way, $|A-0.I|K=0$ and 0 is an eigenvalue of this matrix and hence, this matrix A has determinant 0.

Okay, now since $Ak=0$, for any constant $h \not=0,A(hk)=0$ and I choose the domain of h as over all real numbers except 0.

Now, given that $b=\sum_{i=1}^{n}a_i$ which means you sum up all the columns of A and you would get b

so

$A.\begin{bmatrix} 1\\ 1\\ 1\\ .\\ .\\ 1 \end{bmatrix}=b$ and let this nx1 dimensional vector of all 1's be say $x$, so $Ax=b$

Now, $Ax=b$ and $A(hk)=0$ imply $A(x+hk)=b$ for any-non zero constant h.

Hence, this system of equations has infinite solutions because your h is infinite.

by Boss (29.1k points)
0
this is the correct way to answer........

A vector space can be of finite-dimension or infinite-dimension depending on the number of linearly independent basis vectors. The definition of linear dependence and the ability to determine whether a subset of vectors in a vector space is linearly dependent are central to determining a basis for a vector space

$\sum_{i=1}^{n}c_{i}a_{i}=0$

where $c_{i}$ scalar and $a_{i}$ vector

In case scalar multiplied with vector and get result 0, scalar value will be 0

means $c_{i}$ value will be 0

and it will be linearly dependent

Again given $Ax=b$

A=$\left [ a_{1},a_{2},....a_{n} \right ]$

$b=\sum_{i=1}^{n}a_{i}$$=a_{1}+a_{2}+....+a_{n}$

that means there are only one solution

And all points of solution are on that line

So, ans will be there are infinitely many solution

________________________________________________________

• Linear combination. In mathematics, a linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants
• Linearly Dependent:Intuitively vectors being linearly independent means they represent independent directions in your vector spaces, while linearly dependent vectors means they don't. So for example if you have a set of vector {x1,...,x5} and you can walk some distance in the x1 direction, then a difference distance in x2, then again in the direction of x3. If in the end you are back where you started then the vectors are linearly dependent (notice that I did not use all the vectors).
• https://math.stackexchange.com/questions/456002/what-exactly-does-linear-dependence-and-linear-independence-imply
• http://onlinemschool.com/math/library/vector/linear-independence/

Note:

by Veteran (119k points)
edited by

Take this as example

here

a1                         a2                    a3                             b

1                           3                     4                            1+3+4= 8

2                           6                     8                            2+6+8 = 16

3                           9                     12                          3+9+12 = 24

here

c1 = 1                     c2 = -1/6                      c3= -1/8

a1 *c1                   a2 *c2                           a3*c3                                 sum

1                          3 * (-1/6) = -1/2              4 * (-1/8) = -1/2                   1 + (-1/2) + (-1/2) =0

2                          6*  (-1/6)  = -1               8 * (-1/8) = -1                       2+ (-1) + (-1)  =0

3                          9* (-1/6)    = -3/2           12 * (-1/8) = -3/2                    3 +(-3/2) + (-3/2) = 0

in short this are system of non homogeneous linear equation

x + 3y +4z = 8

2x +6y + 8z = 16

3x + 9y + 12z = 24

Please solve it yourself by bringing it to echelon form

(rank ) = (rank1 ) =1 < n=3

rank means rank of 3*3 matrix

rank1 means rank of 3*4 matrix

as rank=rank1 this system has at least one solution.

but rank=rank1 <n means this system has infinite solution.

by Loyal (5.3k points)
edited