Question

In Problems 1and 2use (3) to compute e^Atande^-At.

$A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

Short answer

The value of e^At and e^-At is,

$e^{At}=\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)$and$e^{-At}=\left(\begin{array}{cc}\cos ht& -\sin ht\\ -\sin ht& \cos ht\end{array}\right)$

Step-by-step solution

Definition of Matrix Exponential.

• Exponential matrix is a square matrix function similar to a regular exponential function. Used to solve a system of linear differential equations.
• In Lie group theory, the matrix exponential function provides the connection between the matrix Lie algebra and the corresponding Lie group.
• For any $n\times n$matrix A
• $e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$,

Calculate the Matrix exponential eAt  and e-At .

Given that,

$A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

By using

$e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$

To compute e^At and e^-At.

So,

A² = A.A

role="math" localid="1664260384774" $$\begin{gathered} =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =l \end{gathered}$$

A³= A². A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =A \end{gathered}$$

A⁴ = A².A²

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =I \end{gathered}$$

A⁵ = A⁴. A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ =A \end{gathered}$$

Thus,

$$\begin{gathered} A^k=A,fork=1,3,5,...... \\ I,fork=2,4,6,....... \end{gathered}$$

Substitute the values in Matrix Exponential equation.

By using equation (1), we have

$e^{At}=I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$

$=1\left(\begin{array}{c}1+\frac{t^2}{2!}+\frac{t^3}{3!}+.....\end{array}\right)+A\left(\begin{array}{c}\begin{array}{c}t+\frac{t^3}{3!}+\frac{t^5}{5!}.....\end{array}\end{array}\right)$

Notice that these are Taylor series expansions of the hyperbolic function cosht and sinht

$$\begin{gathered} =\mathbf{I}\left(\cosh t\right)+\mathbf{A}\left(\sinh t\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\cos ht+\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\sin ht \\ =\left(\begin{array}{cc}\cos ht& 0\\ 0& \cos ht\end{array}\right)\left(\begin{array}{cc}0& \sin ht\\ \sin ht& 0\end{array}\right) \\ =\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right) \end{gathered}$$

$$\begin{gathered} e^{-At}=\left(\begin{array}{cc}\cos h(-t)& \sin h(-t)\\ \sin h(-t)& \cos h(-t)\end{array}\right) \\ =\left(\begin{array}{cc}\cos ht& -\sin ht\\ -\sin ht& \cos ht\end{array}\right) \end{gathered}$$

Therefore, the e^At and e^-At is,

$e^{At}=\left(\begin{array}{cc}\cos ht& \sin ht\\ \sin ht& \cos ht\end{array}\right)and{e}^{-At}==\left(\begin{array}{cc}\cos ht& -\sin ht\\ -\sin ht& \cos ht\end{array}\right)$