Question

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

$A=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

Short answer

The compute of e^At and e^-At is $e^{At}=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right),and{e}^{-At}=\left(\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right)$

Step-by-step solution

Determine the matrix

• A matrix is a collection of integers that are organized in rows and columns to form a rectangular array.
• The numbers are referred to as matrix elements or entries.
• Matrices are widely used in engineering, physics, economics, and statistics, as well as in many disciplines of mathematics.

Determine the compute matrix

We are given

$A=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

Where we will be using

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

$e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$

To compute e^At and e^-At .

So,

A² =A*A

=$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \end{gathered}$$

A³ = A²* A

$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 8\end{array}\right) \end{gathered}$$

A⁴= A² * A²

^{$$\begin{gathered} =\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right) \\ =\left(\begin{array}{cc}1& 0\\ 0& 16\end{array}\right) \end{gathered}$$}

We notice that

For k=1,2,3,4.....

Therefore, using (1), we have

localid="1664257862424" $e^{At}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)t+\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\frac{t^2}{2!}+\left(\begin{array}{cc}1& 0\\ 0& 4\end{array}\right)\frac{t^3}{3!}+\left(\begin{array}{cc}1& 0\\ 0& 16\end{array}\right)\frac{t^4}{4!}+...$

$=\left(\begin{array}{cc}1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}......& 0\\ 0& 1+2t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}......\end{array}\right)$

The power series in the first and second rows of the matrix represent, respectively, e^t and e^2t therefore,

$=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)$

And we can obtain e^-At directly, where

$e^{-At}==\left(\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right)$

The compute of e^At and e^-At is $e^{At}==\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)$and role="math" localid="1664258267305" $e^{-At}=\left(\begin{array}{cc}{e}^{-t}& 0\\ 0& e^{-2t}\end{array}\right)$