Finding the linear combinations of two vectors I am studying for my finals and I'm trying to answer the following question:

Consider the following two vectors in $\mathbb{R}^3$: $a=(1,2,3)$ and $b=(2,3,1)$. Decide
  whether it is possible to  express the vector $c=(2,4,5)$ as a linear
  combination of $a$ and $b$.

I have used the following row operations to find what $C_1$,$C_2$ and $C_3$ are equal to:
\begin{align*}
\left[ \begin{array}{ccc|c} 1 & 2 &2 & 2\\
2 & 3 & 4 & 4 \\
3 & 1 & 5 & 5 \end{array}\right]
&\overset{R_2 = 2R_1 -R_1}{\Longrightarrow}
\left[ \begin{array}{ccc|c} 
1 & 2 &2 &2 \\
0 & 1 & 0 & 0 \\
3 & 1 & 5 & 5 
\end{array}\right]
\overset{R_3=3R_1-R_3}{\Longrightarrow} 
\left[\begin{array}{ccc|c}
1 &2 & 2 & 2 \\
0 &1 & 0 & 0 \\
0 & 5 &1 & 1 
\end{array} \right]
\\
&\overset{R_3=5R_2+R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c} 
1 & 2 & 2 & 2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 
\end{array}\right]
\end{align*}
Is the following argument correct?

By looking at the last row of the matrix, we see that it says $0 = 1$, which is impossible, and the system of equations therefore has no solutions. This means that we can not find the values for $C_1$ and $C_2$, and so $c$ can not be written as a linear combination of $a$ and $b$. 

Thanks in advance!
 A: What you did is actually incorrect, because you don't want to row-reduce
$$\left(\begin{array}{ccc|c}1 & 2 & 2 & 2\\ 2 & 3 & 4 & 4 \\ 3 & 1 & 5 & 5
\end{array}\right)$$
The problem with this is you're asking if there exist constants $C_1$, $C_2$, $C_3$ such that
$C_1 a + C_2 b + C_3 c = c$, which is obvious (pick $C_3 = 1$, $C_1 = C_2 = 0$).
Instead, you want to row reduce the matrix
$$\left(\begin{array}{cc|c}1 & 2 & 2\\ 2 & 3 & 4 \\ 3 & 1 & 5
\end{array}\right)$$
So in other words, you know that the vectors $a$ and $b$ are not linear multiples of each other, so you just want to check to see if you can write $c$ as a linear combination of $a$ and $b$, or you want to know if there exist $\alpha, \beta$ such that
$\alpha a + \beta b = c$.  (This would be the case of the first matrix)
Alternatively you could row reduce the matrix (this is most similar to what you have already done)
$$\left(\begin{array}{ccc|c}1 & 2 & 2 & 0\\ 2 & 3 & 4 & 0 \\ 3 & 1 & 5 & 0
\end{array}\right)$$
This would be asking if there exist $C_1, C_2, C_3$ such that
$C_1 a + C_2 b + C_3 c = 0$
It should be clear that these are equivalent methods.  Once you have done the row-reduction correctly (just chop off the last column and replace it by a column of 0's and see what you find), then the answer should be fairly easy to see.  I'll leave the rest to you.
By the way, if you find that the only possibility is $C_1 = C_2 = C_3 = 0$, this would mean that $c$ is not a linear combination of $a$ and $b$.
A: It is probably quicker to do this without matrices.
Suppose $\lambda a + \mu b = c$. Then you get three equations in two unknowns.
$\lambda + 2\mu = 2$
$2\lambda + 3\mu = 4$
$3\lambda + \mu = 5$
Solving the first two gives $\mu = 0, \lambda = 2$ but this solution doesn't work in the third equation so there is no solution to the equations, hence no such linear combination.
A: Note that your original linear system appears to be completely wrong for the purposes of solving the problem. What you're after is constants $c_1$ and $c_2$ such that 
$$
c_1 \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix}, 
$$
or, equivalently,
$$
\begin{bmatrix} 
1 & 2 \\
2 & 3 \\
3 & 1 
\end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix} 
$$
A: $c=m\cdot a+n \cdot b$
Hence :
$(2,4,5)=m\cdot(1,2,3)+n\cdot(2,3,1)$
So we have following system of equations :
$\begin{cases}
 m+2n=2 \\
 2m+3n=4 \\
 3m+n=5
\end{cases}$
which has no solution ,therefore you cannot express $c$ as linear combination of $a$ and $b$ .
A: You can also do your example in following way,
Let $(2,4,5)=\alpha.(1,2,3)+\beta.(2,3,1)$
$(2,4,5)=(\alpha+2\beta,2\alpha+3\beta,3\alpha+\beta)$
$\therefore 2=\alpha+2\beta,4=2\alpha+3\beta,5=3\alpha+\beta$
Now, solving the first two equation, we get, $\alpha=2,\beta=0$ but it does not satisfied last equation therefore $c$ can not be expressed as a linear combination of $a$ and $b$.
