I'm having a hard time finding the $c_1$ and $c_2$ coordinates of a system after finding the Eigenvalues and Eigenvectors. My task is to solve a system given the initial value.
Solve System : $\frac{{d\vec{x}}}{dt} = \left | \matrix{-4 & 3 \\ 2 & -3 } \right |$ with init value $\vec{x_0} = \left | \matrix{1 \\ 0}\right | $
What I have done so far:
So I have a matrix :
$\frac{{d\vec{x}}}{dt} = $$\left | \matrix{-4 & 3 \\ 2 & -3 } \right | $
I found the Eigenvalues, $\lambda_1 = -6$ , $\lambda_2 = -1$
I found the Eigenvectors by solving the system of the Eigenvalues
Which are $\vec{v1} = span(\left | \matrix{3 \\ -2} \right |)$
and $\vec{v2} = span(\left | \matrix{1 \\ 1} \right |)$
So now, I have to find a c(t) = $S^{-1}\vec{x(t)}$
What I did, then was applied my eigenvectors to a new matrix
$S =\left | \matrix{3 && 1 \\-2 && 1} \right |$
and get the $S^{-1}=\det(\left | \matrix{3 && 1 \\-2 && 1} \right | )$
which is $\frac{1}{5}$
The point where I am confused:
So, according to the theorem in my book, to find the scalars $c_1 , c_2,...,c_n$ according to the basis
$\vec{v_1}, \space \vec{v_2}, \space ,...\vec{v_n}$ I have to write the matrix form as.
$S \lambda^{t}_t S^{-1}\vec{x_0} = S =\left | \matrix{3 && 1 \\-2 && 1} \right |\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right| \frac{1}{5}\left | \matrix{3 && 1 \\-2 && 1} \right |\left|\matrix{1 \\ 0}\right| $
with $\lambda^{t}_t$ as a stand in for
$\left | \matrix{e^{\lambda_1t}_t & 0 \\ 0 & e^{\lambda_2t}_t} \right|$
So now, I would do the math and end up with :
$\left| \matrix{\frac{9}{5}e^{-6t} & \frac{-2}{5}e^{-t} \\ \frac{-6}{5}e^{-6t} & \frac{-2}{5}e^{-t}}\right|$
But I look in the back of the book, aaaand the coordinates $c_1$ && $c_2$ with respect to $\vec{v1}$ &&  $ \vec{v2}$ are actually $c_1 = \frac{-1}{5}$ and $c_2 = \frac{2}{5}$
I'm clearly doing something wrong here, but I do not know what exactly.
 A: To find this answer, all we need to do is abide by the initial value point.
$\vec{x_0} = \left |\matrix{1 \\ 0} \right |$
and therefore,
$x(t)=c_1e^{-6t}\left | \matrix{3 \\ -2} \right | + c_2e^{-1t} \left| \matrix{1\\1} \right|$
so,
$x(0)=c_1e^{-6(0)}\left | \matrix{3 \\ -2} \right | + c_2e^{-1(0)} \left| \matrix{1\\1} \right|$
Which reduces our equation to :
$x(0) = c_1\left | \matrix{3 \\ -2} \right | + c_2\left | \matrix{1\\1} \right|$
Now recall since we have an initial value $x(0) = \left | \matrix{1\\1} \right|$
$ \left | \matrix{1\\0} \right| = c_1\left | \matrix{3 \\ -2} \right | + c_2\left | \matrix{1\\1} \right|$
So now apply c1 and c2 to the matrices and we have a system that can be represented by an augmented matrix.
$\left | \matrix{3c1 & 1c2 & | 1 \\ -2c1 & 1c2 & | 0} \right|$
Now solve using row reduction and we get $\left| \matrix{1 & 0 & | \frac{1}{5}\\ 0 & 1 & |\frac{2}{5}}\right|$
Plug these values into the equation $3c_1 + c_2 = 1$
and $-2c_1 + c_2 =0$ and you get
$\frac{3}{5} + \frac{2}{5} = 1$
and
$\frac{-2}{5} + \frac{2}{5} = 0$
So our finished general equation is
$\frac{1}{5}e^{-6t}\left | \matrix{3 \\ -2} \right | + \frac{2}{5}e^{-t} \left |\matrix{1 \\ 1} \right|$
and our coordinates $c_1 = \frac{1}{5} \space  c_2 = \frac{2}{5}$
A: $\frac {dx}{dt} = Ax\\
\frac {dx}{dt} = e^{At}x_0\\
A = PDP^{-1}\\
e^{At} = P e^{Dt}P^{-1}\\
\frac{dx}{dt} = P e^{Dt}P^{-1}x_0\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&-2\\1&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\&e^{-t}\end{bmatrix}\begin{bmatrix} 1& -1\\2&3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&1\\-2&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\&e^{-t}\end{bmatrix}\begin{bmatrix} 1\\2\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3&1\\-2&1\end{bmatrix}\begin{bmatrix} e^{-6t}\\2e^{-t}\end{bmatrix}\\
\frac{dx}{dt} = \frac 15 \begin{bmatrix} 3e^{-6t}+2e^{-t}\\-2e^{-6t}+2e^{-t}\end{bmatrix}\\
\frac{dx}{dt} = \begin{bmatrix} \frac {3}{5}e^{-6t}+\frac {2}{5}e^{-t}\\-\frac{2}{5}e^{-6t}+\frac {2}{5}e^{-t}\end{bmatrix}$
$
