Question

To find the general solution of the given system.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ -2& -2& -2\end{array}\right)\mathit{X}$

Short answer

The general solution of the given matrix is$X=C_1\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)+C_2\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+C_3\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$

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,
• $X=e^{At}C$
  

Calculate the Matrix exponential eAt.

Given that,

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

By using

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

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

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

By using this equation,

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

Compute e^At. So,

A² = A.A

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

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

Using (2), we have

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

Find the General solution of given matrix.

Finally, using (1) to find the general solution of the system,$$\begin{gathered} X=\left(\begin{array}{ccc}t+1& t& t\\ t& t+1& t\\ -2t& -2t& -2t+1\end{array}\right)\left(\begin{array}{c}{C}_1\\ C_2\\ C_3\end{array}\right) \\ =C_1\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)+C_2\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+C_3\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right) \end{gathered}$$

Therefore, the general solution of the given matrix is

$X=C_1\left(\begin{array}{c}t+1\\ t\\ -2t\end{array}\right)+C_2\left(\begin{array}{c}t\\ t+1\\ -2t\end{array}\right)+C_3\left(\begin{array}{c}t\\ t\\ -2t+1\end{array}\right)$