## Compute the fundamental matrix of a given linear system of first-order differential equations

Let $Y$ and $A$ be matrices as defined in this previous exercise, with $\frac{d}{dt} Y = AY$. Compute the fundamental matrix of this differential equation where $A$ is one of the following matrices:

1. $A_1 = \begin{bmatrix} 2 & -2 & 14 \\ 0 & 3 & -7 \\ 0 & 0 & 2 \end{bmatrix}$
2. $A_2 = \begin{bmatrix} 9 & 4 & 5 \\ -4 & 0 & -3 \\ -6 & -4 & -2 \end{bmatrix}$
3. $A_3 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

Let $A$ be a matrix, and let $P^{-1}AP = B$ be in Jordan canonical form. Using this previous exercise, we have that $\mathsf{exp}(At) = P\mathsf{exp}(Bt)P^{-1}$ is a fundamental matrix, and by this previous exercise, $P\mathsf{exp}(Bt)$ is a fundamental matrix of our differential equation. We can use this previous result (and part 2 of this previous exercise) to compute $\mathsf{exp}(Bt)$.

We computed the Jordan canonical forms (and conjugators) for these three matrices in this previous exercise.

For the matrix $A_1$, we have $P_1 = \begin{bmatrix} -7 & -1 & -2 \\ 7 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$ and $P_1^{-1}A_1P_1 = B_1 = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$. Now $\mathsf{exp}(Bt) = \mathsf{exp}([2]t \oplus [2]t \oplus [3]t)$ $= \begin{bmatrix} e^{2t} & 0 & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{3t} \end{bmatrix}$, and $P_1 \mathsf{exp}(B_1t) = \begin{bmatrix} -7e^{2t} & -e^{2t} & -2e^{3t} \\ 7e^{2t} & 0 & e^{3t} \\ e^{2t} & 0 & 0 \end{bmatrix}$. We can easily verify that this is indeed a fundamental matrix.

For the matrix $A_2$, as computed in this previous exercise, if we let $P_2 = \begin{bmatrix} -2 & 5 & 3 \\ 1 & -3 & -2 \\ 2 & -5 & -2 \end{bmatrix}$ then $P_2^{-1}A_2P_2 = B_2 = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ is in Jordan canonical form. Now $\mathsf{exp}(B_2t) = \mathsf{exp}(\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} t) \oplus \mathsf{exp}([3]t)$ $= \begin{bmatrix} e^{2t} & te^{2t} & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{3t} \end{bmatrix}$. So $P_2\mathsf{exp}(B_2t) = \begin{bmatrix} -2e^{2t} & -2te^{2t} + 5e^{2t} & 3e^{3t} \\ e^{2t} & te^{2t} - 3e^{2t} & -2e^{3t} \\ 2e^{2t} & 2te^{2t} - 5e^{2t} & -2e^{3t} \end{bmatrix}$ is a fundamental matrix.

Finally, consider $A_3$. Letting $P_3 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \end{bmatrix}$ (Seethis previous exercise), we have $P_3^{-1}A_3P_3 = B_3 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \oplus \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ in Jordan canonical form. Then $\mathsf{exp}(B_3t) = \begin{bmatrix} e^t & te^t \\ 0 & e^t \end{bmatrix} \oplus \begin{bmatrix} e^t & te^t \\ 0 & e^t \end{bmatrix}$, and hence $P_3\mathsf{exp}(B_3t) = \begin{bmatrix} 0 & 0 & e^t & te^t \\ 0 & 0 & 0 & e^t \\ e^t & te^t & 0 & 0 \\ 0 & e^t & e^t & te^t + e^t \end{bmatrix}$ is a fundamental matrix.