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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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9 Answers

1 votes
1 votes

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.

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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).

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intuitive answer :

given scalars :c1,c2,c3,c4,…………,cn  and there exist a ci which is not zero .

now,

c1a1+c2a2+c3a3+………...+cnan=0 .if one scalar in solution vector is non zero ,then i can write the column vector as a linear combination of other; hence ai’s are linear dependent.

lets say c2=5

then i can write a2= -1/5(c1a1+c2a2+c3a3+……...+cnan)

 

now come to Ax=b;

as given in question b is the linear combination of columns of A .(b=a1+a2+a3+……..an) , from this i can infer that there exist a solution for sure .and we also proved above that ai’s are linear dependent . then there will always be infinite solution possible.

**** If b is the linear combination of A and columns of A are linearly independent then there will always be unique solution otherwise infinite solution**** .

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