Determining if a vector b is a linear combination of vectors a1,a2,a3

[nglatz]

given a1=[1,-2.0] , a2=[0,1,2] , a3=[5,-6,8] , b=[2,-1,6] determine if b is a linear combination of a1, a2, a3.

I put these vectors into an augmented matrix and row reduced. My result was columns 1 and 2 having pivots and the last row being all zeros. Please help me understand how this results in b NOT being a linear combination of a1, a2, a3.

The only thing i can think of is that vectors a1, a2, a3 do not span all of R3 and b is not part of the plane that a1, a2, a3 span. If this is the case, please explain how i can see this from the augmented matrix in RREF.

 

[Alchemista]

You've misinterpreted the results of the row reduction.

First let's start with the result:
$\left[ \begin {array}{cccc} 1&0&5&2\\ -2&1&-6&-1<br /> \\ 0&2&8&6\end {array} \right]<br /> \to \left[ \begin {array}{cccc} 1&0&5&2\\ 0&1&4&3<br /> \\ 0&0&0&0\end {array} \right]$

So the coefficients of a linear combination $x_1a_1 + x_2a_2 + x_3a_3 = b$ take the form:
$\left[\begin{array}{c}<br /> x_1 \\ x_2 \\ x_3<br /> \end{array}\right] \in<br /> \left\{<br /> \left[\begin{array}{c}<br /> 2 \\ 3 \\ 0<br /> \end{array}\right] +<br /> t\left[\begin{array}{c}<br /> -5 \\ -4 \\ 1<br /> \end{array}\right] \|<br /> t \in \mathbb{F}<br /> \right\}$

 

[nglatz]

ok. that clears things up. thanks so much