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66 votes

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
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4 Comments

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... :)

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Detailed Video Solution here – GATE 2017 Question

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9 Answers

83 votes
83 votes
Best answer
$\sum_i c_i a_i = 0 \text{ with } \exists i: c_i \ne 0$ 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 $(C)$.
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@jatin khachane 1 @Cristine

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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138 votes
138 votes

   Answer is (C)

4 Comments

@amey46 , Is it valid to say that |A|=0 ,

because determinant for non square matrix does not exist.

whearas the logic that if A is linearly dependent then it is |A|=0 seems right !

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Its given $a_{i}$ is column vector so A will be like $[a_{1} a_{2}… a_{n}]$ and not like$\begin{bmatrix} a_{1}\\ a_{2}\\ .\\ .\\a_{3} \end{bmatrix}$
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They said that it is R^n . So every column  vector is n dimensional
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18 votes
18 votes

First read the question and then we'll answer gradually.

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.

Answer-(C)

1 comment

this is the correct way to answer........
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4 votes
4 votes

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/

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