Question

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

4.$A=\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)$

Short answer

The value ofe^At is$\left(\begin{array}{ccc}1& 0& 0\\ 3t& 1& 0\\ \frac{3}{2}{t}^2+5& t& 1\end{array}\right)$

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

$A=\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)$

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,

A² = A.A

$$\begin{gathered} =\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 3& 0& 0\end{array}\right) \end{gathered}$$

A³ = A².A

$$\begin{gathered} =\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 3& 0& 0\end{array}\right)\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right) \\ =\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

Since,

$$\begin{gathered} A^3=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \\ {A}^4=A^5=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

Using (1), we have

$e^{At}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)t+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 3& 0& 0\end{array}\right)\frac{t^2}{2!}+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)+......$

$$\begin{gathered} =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 3t& 0& 0\\ 5t& t& 0\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \frac{3}{2}{t}^2& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 3t& 1& 0\\ \frac{3}{2}{t}^2+5t& t& 1\end{array}\right) \end{gathered}$$

Therefore, the value ofe^At is$=\left(\begin{array}{ccc}1& 0& 0\\ 3t& 1& 0\\ \frac{3}{2}{t}^2+5t& t& 1\end{array}\right)$