GATE CSE 2017 | Set 1 | Question: 3

Question

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

• A. a unique solution at $x=J_{n}$ where $J_{n}$ denotes a $n$-dimensional vector of all 1.
• B. no solution
• C. infinitely many solutions
• D. finitely many solutions

Comments

det(A) = 0?
    │
   ├── NO  → Unique Solution
    │
    └── YES → Particular solution exists?
                  │
                  ├── NO  → No Solution
                  │
                  └── YES → Infinite Solutions

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give this guy @goku4199 a raise👾👾👾

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Answer 1

$\sum_i c_i a_i = 0 \text{ with } \exists i: c_i \ne 0$ 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 $(C)$.

Comments

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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For this case how do we chech that the rank of A and A|b is equal?

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convert matrix A  and Augmented matrix(i.e A|B ) into Row Reduced Echelon Form  and check

Answer 2

   Answer is (C)

Comments

They said that it is R^n . So every column  vector is n dimensional

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Exactly,  even I have same doubt

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How b=a1+a2+a3....an given?

 

Answer 3

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)

Comments

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

Answer 4

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:

Answer 5

In the question, it is given to us that: $c_1{}, ..., c_n{}$ are scalers and not all are zero.

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

$\Rightarrow$ $c_1a_1{} + c_2a_2{} + … + c_na_n = 0$

This means the Linear Combination of above $a_i$ results in 0 and there is at least one $c_i$ which is non-zero.

We can infer that the vectors $a_i$ are thus Linearly dependent.  $...(1)$

Considering $Ax = b$, where $A = [a_1, a_2, …, a_n]$ and $b = \sum_{i=1}^{n} a_{i}$

Upon expanding the $b = \sum_{i=1}^{n} a_{i}$

$\Rightarrow$ $b = a_1{} + a_2{} + … + a_n$    $...(2)$

But since, from (1) we inferred that vectors $a_i$ are Linearly dependent, we can say that for (2) it can be infinitely many possible ways. 

So the correct answer is (C).

Answer 6

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.