30.2. FIRST ORDER SYSTEMS, CONSTANT COEFFICIENTS 587

Note how the entries are all rational functions. This will ALWAYS happen no matter whatmatrix you use and this follows from that method for finding the inverse in terms of thetranspose of the cofactor method. Since theoretically, by the fundamental theorem of alge-bra, one can always factor a polynomial into a product of linear and irreducible quadraticsin the denominator of those rational functions, this process will ALWAYS work with thecaveat that one might not be able to actually carry out the factorization of the polynomialsin the denominator. However this shows that the fundamental matrix does exist and thatyour ability to explicitly compute it is exactly as good as your ability to factor a polyno-mial. In this case, I can take the inverse Laplace transform of that matrix on the right andget

F (s) =

(3

s−1 −2

s−22

s−2 −2

s−13

s−1 −3

s−23

s−2 −2

s−1

)

Φ(t) =

(3et −2e2t 2e2t −2et

3et −3e2t 3e2t −2et

)Does it work?

Dt

(3et −2e2t 2e2t −2et

3et −3e2t 3e2t −2et

)=

(3et −4e2t 4e2t −2et

3et −6e2t 6e2t −2et

)(−1 2−3 4

)(3et −2e2t 2e2t −2et

3et −3e2t 3e2t −2et

)=

(3et −4e2t 4e2t −2et

3et −6e2t 6e2t −2et

)so yes, it solves the equation. Also Φ(0) = I. Thus this is indeed the fundamental matrix.

30.2.1 Some Technical Considerations∗

Now if F (s) = (sI−A)−1 = L (Φ(t)) , is Φ′ (t) = AΦ(t)? is Φ(0) = I? Here we are as-suming that the entries of Φ(t) have exponential growth. Then multiplying through by(sI−A) ,

I = (sI−A)∫

∞

0e−st

Φ(t)dt = (I−A/s)∫

∞

0se−st

Φ(t)dt (30.5)

(I−A/s)∫

∞

0se−st (Φ(t)−Φ(0))dt +(I−A/s)Φ(0) (30.6)

because∫

∞

0 se−stdt = 1, this being true for all large enough s. Letting s→ ∞, the first termconverges to 0. Here is roughly why this is so. Letting δ > 0, be so small that all entries ofΦ(t) are closer than ε to the entries of Φ(0) whenever t < δ ,∥∥∥∥∫ ∞

δ

se−st (Φ(t)−Φ(0))dt∥∥∥∥≤ ∫ ∞

δ

se−st(

Ceλ t +C)

dt

where ∥A∥ will denote the maximum of the absolute values of all entries of A and it isassumed that each of these is no more than Ceλ t . Now that integral on the right can becomputed and it equals the following for large s(

1−s+λ

e−st+λ ts− e−st)|∞δ=

(1

s−λe−sδ+λδ s+ e−sδ

)