Question

If Dis designed as in (9), then find e^Dt.

Short answer

The value of e^Dtis $\left(\begin{array}{cccc}{e}^{{\lambda }_{1}t}& 0& 0.....& 0\\ 0& e^{{\lambda }_{2}t}& 0.....& 0\\ .& .& .& .\\ 0& 0& 0.....& e^{{\lambda }_{n}t}\end{array}\right)$.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

where I is the unit matrix of order n.

Therefore the infinite matrix power series is

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula,

Find the value of eDt.

To find the value of e^Dt, we will use the following formula.

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

where,

$$\begin{gathered} e^{Dt}=\left(\begin{array}{cccc}1& 0& 0...& 0\\ 0& 1& 0...& 0\\ .& .& .& .\\ 0& 0& 0& 1\end{array}\right)+\left(\begin{array}{cccc}{\lambda }_1& 0& 0...& 0\\ 0& {\lambda }_2& 0...& 0\\ .& .& .& .\\ 0& 0& 0& {\lambda }_n\end{array}\right)t+\left(\begin{array}{cccc}{\lambda }_1^2& 0& 0...& 0\\ 0& {\lambda }_2^2& 0...& 0\\ .& .& .& .\\ 0& 0& 0& {\lambda }_n^2\end{array}\right)t^2+\left(\begin{array}{cccc}{\lambda }_1^3& 0& 0...& 0\\ 0& {\lambda }_2^3& 0...& 0\\ .& .& .& .\\ 0& 0& 0& {\lambda }_n^3\end{array}\right)\frac{t^3}{3!} \\ +.... \end{gathered}$$

$=\left(\begin{array}{ccc}1+{\lambda }_{1}t+\frac{1}{2!}{\left({\lambda }_{1}t\right)}^2+...& 0& 0\\ 0& 1+{\lambda }_{2}t+\frac{1}{2!}{\left({\lambda }_{2}t\right)}^2+...& \\ 0& & 1+{\lambda }_{n}t+\frac{1}{2!}{\left({\lambda }_{n}t\right)}^2+...\end{array}\right)$

role="math" localid="1665143965792" $=\left(\begin{array}{ccc}{e}^{{\lambda }_{1}t}& 0& 0\\ 0& e^{{\lambda }_{2}t}& 0\\ 0& 0& e^{{\lambda }_{n}t}\end{array}\right)$

Hence e^Dtis " width="9" height="19" role="math">