Linear Algebra - Vector Equations

[GreenPrint]

Homework Statement

See attachment for the question

Homework Equations

The Attempt at a Solution

I don't understand...

the back of the book says that the answer is no

when I solve reduce row echelon form the augmented matrix Ab I get that

x1 = 2 - 5x3
x2 = 3 - 4x3
x3 is free
i thought thought that the answer was yes because the system is consistent and x3 can just be anything i don't understand why the answer is no

thanks for any help

 

[fauboca]

Homework Statement

See attachment for the question

Homework Equations

The Attempt at a Solution

I don't understand...

the back of the book says that the answer is no

when I solve reduce row echelon form the augmented matrix Ab I get that

x1 = 2 - 5x3
x2 = 3 - 4x3
x3 is free
i thought thought that the answer was yes because the system is consistent and x3 can just be anything i don't understand why the answer is no

thanks for any help

If the determinant is 0, they are linearly dependent. If the determinant is non-zero, then the vectors are independent.

If they are dependent, the reduce row echelon form will tell you av_1+cv_2+dv_3 = b

 

[DaleSwanson]

I came here to ask the exact same question. Since x3 is free there are an infinite number of solutions. This has spurred a debate over on http://www.cramster.com/solution/solution/1204287, which is why I came here. I think that any of the solutions will work. For example, the values x1 = -3, x2 = -1, x3 = 1 cause the equation x1(a1) + x2(a2) + x3(a3) = b to be true.

 

[GreenPrint]

I'm confused... is the answer yes or no and I'm not exactly sure why
although i know about linear independence and dependence and determinants and crap we haven't gotten to that part in the course yet (i prestudied the course over the break) so i think I'm suppose to come up with a sollution that doesn't involve that stuff

 

[Deveno]

Homework Statement

See attachment for the question

Homework Equations

The Attempt at a Solution

I don't understand...

the back of the book says that the answer is no

when I solve reduce row echelon form the augmented matrix Ab I get that

x1 = 2 - 5x3
x2 = 3 - 4x3
x3 is free
i thought that the answer was yes because the system is consistent and x3 can just be anything i don't understand why the answer is no

thanks for any help

books can be wrong (it might be helpful to know what else the "back of the book" says besides just "no").

the situation here is that a₃ is a linear combination of a₁ and a₂: (5,-6,8) = 5(1,-2,0) + 4(0,1,2). so we may as well ask if b is in the span of a₁ and a₂. and it is:

b = 2a₁ + 3a₂

= 2a₁ + 3a₂ + 0a₃

(if x₃ is free, we should be able to ignore one of the column vectors, since row rank = column rank).

(note to moderators: the link provided earlier in this thread provides a step-by-step solution to the problem posted here, and i am assuming the OP is already familiar with this solution).