### 1.32 Find $$e^{A t}$$ where A is a matrix

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 Mathematicaâ€™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);

`

$\left [\begin {array}{cc} -{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{-t} & {\mathrm e}^{-t}-{\mathrm e}^{-2 t} \\ -2 \,{\mathrm e}^{-t}+2 \,{\mathrm e}^{-2 t} & 2 \,{\mathrm e}^{-2 t}-{\mathrm e}^{-t} \end {array}\right ]$