Following vector in R3 are linear combination

[hkus10]

Do you think the following vector in R3 are linear combination of

v1 = [4                                v2= [2                                v3 = [-2
      2                                     1                                      -1
     -3]                                   -2]                                      0]

[-2
 -1
  1]

The answer that I get is

[ 1 1/2 -1/2 -1/2
 -3  -2   0    1
  0   0   0   0   ]

Since I get a row of zeros, this means this is a inconsistent system which has no solution.

However, the book says that this vector has a combination of v_1, v_2, and v_3.

Did I make any error on my calculation?

 

[Mark44]

Do you think the following vector in R3 are linear combination of

v1 = [4                                v2= [2                                v3 = [-2
      2                                     1                                      -1
     -3]                                   -2]                                      0]

[-2
 -1
  1]

The answer that I get is

[ 1 1/2 -1/2 -1/2
 -3  -2   0    1
  0   0   0   0   ]

Since I get a row of zeros, this means this is a inconsistent system which has no solution.

No, that's not what it means at all. A row of zeros means that you have an infinite number of solutions. If you ended up with a row with a nonzero number on the right, then you would have an inconsistent system (you're working with an augmented matrix - you don't show it, but you are).

One solution I found was c1 = -1, c2 = 1, and c3 = 0.

However, the book says that this vector has a combination of v_1, v_2, and v_3.

Did I make any error on my calculation?

 

[hkus10]

No, that's not what it means at all. A row of zeros means that you have an infinite number of solutions. If you ended up with a row with a nonzero number on the right, then you would have an inconsistent system (you're working with an augmented matrix - you don't show it, but you are).

One solution I found was c1 = -1, c2 = 1, and c3 = 0.

Can you show the way that your get the answer?

 

[Mark44]

Here's your augmented matrix:
\begin{bmatrix} 4 & 2 & -2 & | & -2\\ 2 & 1 & -1 & | & -1 \\ -3 & -2 & 0 & | 1 \end{bmatrix}

This matrix is shorthand for c₁v₁ + c₂v₂ + c₃v₃ = <-2, -1, 1>, where v₁, v₂₁, and v₃ are as you had them in post #1.

The matrix you ended with in post 1 was correct as far as you went, but it wasn't in RREF (reduced row-echelon form), so continue reducing it as far as it will go. I ended up with two nonzero rows, the first of which was 1 0 -2 | -1. This row says that c₁ - 2c₃ = -1, or equivalently, c₁ = 2c₃ - 1. You should get a solution for c₂ and c₃ is a free variable.