Mathematica
ClearAll["Global`*"] mat={{0,1}, {-2,-3}} MatrixExp[mat t]; MatrixForm[%]
\[ \left ( {\begin {array}{cc} -e^{-2 t}+2 e^{-t} & -e^{-2 t}+e^{-t} \\ 2 e^{-2 t}-2 e^{-t} & 2 e^{-2 t}-e^{-t} \\ \end {array}} \right ) \]
Now verify the result by solving for \(e^{At}\) using the method would one would do by hand, if a computer was not around. There are a number of methods to do this by hand. The eigenvalue method, based on the Cayley Hamilton theorem will be used here. Find the eigenvalues of \(|A-\lambda I|\)
m = mat-lambda IdentityMatrix[Length[mat]]
\[ \left ( {\begin {array}{cc} -\lambda & 1 \\ -2 & -\lambda -3 \\ \end {array}} \right ) \]
Det[m]
\[ \lambda ^2+3 \lambda +2 \]
sol=Solve[%==0,lambda]
Out[15]= {{lambda->-2},{lambda->-1}}
eig1=lambda/.sol[[1]] eig2=lambda/.sol[[2]]
Out[16]= -2 Out[17]= -1
(*setup the equations to find b0,b1*) eq1 = Exp[eig1 t]==b0+b1 eig1; eq2 = Exp[eig2 t]==b0+b1 eig2; sol = First@Solve[{eq1,eq2},{b0,b1}]
\[ \left \{\text {b0}\to e^{-2 t} \left (2 e^t-1\right ),\text {b1}\to e^{-2 t} \left (e^t-1\right )\right \} \]
(*Now find e^At*) b0=b0/.sol[[1]]
\[ e^{-2 t} \left (2 e^t-1\right ) \]
b1=b1/.sol[[2]]
\[ e^{-2 t} \left (e^t-1\right ) \]
b0 IdentityMatrix[Length[mat]]+b1 mat; Simplify[%] MatrixForm[%]
\[ \left ( {\begin {array}{cc} e^{-2 t} \left (-1+2 e^t\right ) & e^{-2 t} \left (-1+e^t\right ) \\ -2 e^{-2 t} \left (-1+e^t\right ) & -e^{-2 t} \left (-2+e^t\right ) \\ \end {array}} \right ) \] The answer is the same given by Mathematica’s command MatrixExp[]
Matlab
syms t A=[0 1;-2 -3]; expm(t*A)
ans = [2/exp(t)-1/exp(2*t),1/exp(t)-1/exp(2*t)] [2/exp(2*t)-2/exp(t),2/exp(2*t)-1/exp(t)]
pretty(ans)
+- -+ | 2 exp(-t)- exp(-2 t),exp(-t)-exp(-2 t) | | | | 2 exp(-2 t)-2 exp(-t),2 exp(-2 t)-exp(-t)| +- -+
Maple
restart; A:=Matrix([[0,1],[-2,-3]]); LinearAlgebra:-MatrixExponential(A,t);
\[ \left [\begin {array}{cc} -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} & {\mathrm e}^{-t}-{\mathrm e}^{-2 t} \\ -2 \,{\mathrm e}^{-t}+2 \,{\mathrm e}^{-2 t} & 2 \,{\mathrm e}^{-2 t}-{\mathrm e}^{-t} \end {array}\right ] \]