Question

To find the general solution of the given system.

8.$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)\mathit{X}$

Short answer

The general solution for the given matrix is$X=C_1\left(\begin{array}{c}1\\ 3t\\ \frac{3}{2}{t}^2+5t\end{array}\right)+C_2\left(\begin{array}{c}0\\ 1\\ t\end{array}\right)+C_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$

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,
• $X=e^{At}C$
  

Calculate the Matrix exponential eAt

Given that,

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

By using

$X=e^{At}C$ (1)

To find the general solution of the given system. Now, let

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

By using

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

Compute e^At . So,

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 (2), we have

localid="1664290749477" $$\begin{gathered} e^{At}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 3& 0& 0\\ 5& 1& 0\end{array}\right)t+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 3& 0& 0\end{array}\right)\frac{t^2}{2!}+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)+.... \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 3t& 0& 0\\ 5t& t& 0\end{array}\right)+\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \frac{3}{2}{t}^2& 0& 0\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 3t& 1& 0\\ \frac{3}{2}{t}^2+5t& t& 1\end{array}\right) \end{gathered}$$

Find the General solution of given matrix.

Finally, using (1) to find the general solution of the system,

$$\begin{gathered} X=\left(\begin{array}{ccc}1& 0& 0\\ 3t& 1& 0\\ \frac{3}{2}{t}^2+5t& t& 1\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\\ C_3\end{array}\right) \\ =C_1\left(\begin{array}{c}1\\ 3t\\ \frac{3}{2}{t}^2+5t\end{array}\right)+C_2\left(\begin{array}{c}0\\ 1\\ t\end{array}\right)+C_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right) \end{gathered}$$

Therefore, the general solution for the given matrix is,

$X=C_1\left(\begin{array}{c}1\\ 3t\\ \frac{3}{2}{t}^2+5t\end{array}\right)+C_2\left(\begin{array}{c}0\\ 1\\ t\end{array}\right)+C_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)$