Linear independence in systems of ordinary differential equations. I have a question about how to show that these are linearly independent. This is in the chapter of my Elementary Differential Equations book titled, "Linear Systems of ODE's" 
I first approached this by using an augmented matrix of the following form:
\begin{bmatrix}
1 & 1 & 1 & | & 0\\
1 & -1 &-1 &| & 0\\
0 & 1 & 1 & | & 0\end{bmatrix}
I then row reduced it to:
\begin{bmatrix}
1 & 0 & 0 & | & 0\\
0 & 1 & 1 &| & 0\\
0 & 0 & 0 & | & 0\end{bmatrix}
So I'm originally had:
$c_1 e^t + c_2 e^t + c_3 e^2t = 0$
$c_1 e^t - c_2 e^t - c_3 e^2t = 0$
$c_2 e^t + c_3 e^2t = 0$
And then with the augmented matrix, got:
$c_1 e^t=0$
$c_2 e^t + c_3 e^2t = 0$
I don't know where to go from here.
Also, I am unsure if I am allowed to do all of the things that I did. Any assistance would be appreciated.
 A: You're almost there. 
\begin{equation}
c_1e^t = 0
\end{equation}
Note that $e^t$ will never equal zero. It follows that $c_1$ = 0.
You're then left with \begin{equation} c_2e^t + c_3e^{2t} = 0
\end{equation}
This implies that $e^t$ and $e^{2t}$ are solutions to the differential equation. Then what you can do is calculate the Wronskian, given by
\begin{equation}
W(y_1, y_2) = det( \begin{matrix} y_1 & y_2 \\ y'_1 & y'_2
\end{matrix} )
\end{equation} 
If the Wronskian is non-zero, then that tells you your solutions are linearly independent. 
Otherwise, your solutions are linearly dependent.
A: $e^{2t}=e^te^t\ne ce^t$ because there is no constant $c=e^t$, similarly 
for all vector ${\bf v}$, ${\bf v}e^{2t}={\bf v}e^te^t\ne c{\bf v}e^t$.
Now all you need to show is that the two vectors multiplied by $e^t$ are independent. It will be enough to show that the following matrix has rank 2.
$\begin{bmatrix}
1 & 1 \\
1 & -1\\
0 & 1 \end{bmatrix}\rightarrow\begin{bmatrix}
1 & 1 \\
0 & -2\\
0 & 1 \end{bmatrix}\rightarrow\begin{bmatrix}
1 & 1 \\
0 & 1\\
0 & 0 \end{bmatrix}$
