Question

In Problems 3and 4use (3) to compute e^At.

3.$A=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right)$

Short answer

The value ofe^At is$e^{At}=\left(\begin{array}{ccc}t+1& t& t\\ t& t+1& t\\ -2t& -2t& -2t+1\end{array}\right)$

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

We are given

$A=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right)$

By using

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

To compute e^At. So,

A² = A.A

$$\begin{gathered} =\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right)\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

from that equation we can easily tell that

A³= A⁴= A⁵=$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$

Using (1), we have

role="math" localid="1664262462995" $$\begin{gathered} e^{At}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right)t+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)+.... \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}t& t& t\\ t& t& t\\ -2t& -2t& -2t\end{array}\right) \\ =\left(\begin{array}{ccc}t+1& t& t\\ t& t+1& t\\ -2t& -2t& -2t+1\end{array}\right) \end{gathered}$$

Therefore, the compute ofe^At is e^{At$=\left(\begin{array}{ccc}t+1& t& t\\ t& t+1& t\\ -2t& -2t& -2t+1\end{array}\right)$}