There is a differential equation:
$$\left[\begin{array}{}\dot{T}_1\\\dot{T}_2\end{array}\right]=\left[\begin{array}{}0&1\\-.00097387&-0.169476\\\end{array}\right]\left[\begin{array}{}T_1\\T_2\end{array}\right]+\left[\begin{array}{}0&0\\0.040938&0.00097387\\\end{array}\right]\left[\begin{array}{}q\\ T_3\end{array}\right]$$
with initial conditions: $$\begin{array}{}T_1(0)=T_3\\T_2(0)=5.2687\end{array}$$
The solution to the differential equations above is:
$$T_1-T_3=42^{\circ}C-10.0866^{\circ}C e^{\frac{-t}{167.9077}}-24.2457^{\circ}C e^{\frac{-t}{6.1154}}$$
I would like to transform the coordinate bases $T_1$ and $T_2$ to an alternate coordinate basis
$$\left[\begin{array}{}\dot{T}_1\\\dot{T}_{alt}\end{array}\right]=A\left[\begin{array}{}T_1\\T_{alt}\end{array}\right]+\left[\begin{array}{}k_1&0\\0&k_2\\\end{array}\right]\left[\begin{array}{}q\\ T_3\end{array}\right]$$
such that when the initial conditions are as follows, $T_1$ has the same solution as it does in the first differential equation formulation.
$$\begin{array}{}T_1(0)=T_3\\T_{alt}(0)=T_3\end{array}$$
How does one determine the A matrix and the values for $k_1$ and $k_2$?
Thanks!