Question

In problem and use a CAS OR Linear algebra software as an aid in finding the general solution of the given system

$\mathit{X}\mathbf{\text{'}}\mathbf{\left(}\begin{array}{ccc}\mathbf{0}\mathbf{.}\mathbf{9}& \mathbf{2}\mathbf{.}\mathbf{1}& \mathbf{3}\mathbf{.}\mathbf{2}\\ \mathbf{0}\mathbf{.}\mathbf{7}& \mathbf{6}\mathbf{.}\mathbf{5}& \mathbf{4}\mathbf{.}\mathbf{2}\\ \mathbf{1}\mathbf{.}\mathbf{1}& \mathbf{1}\mathbf{.}\mathbf{7}& \mathbf{3}\mathbf{.}\mathbf{4}\end{array}\mathbf{\right)}\mathit{X}$

Short answer

Finding the general solution of the given system.

$X\text{'}\left(\begin{array}{ccc}0.9& 2.1& 3.2\\ 0.7& 6.5& 4.2\\ 1.1& 1.7& 3.4\end{array}\right)X$

$$\begin{gathered} A=\left(\begin{array}{ccc}0.9& 2.1& 3.2\\ 0.7& 6.5& 4.2\\ 1.1& 1.7& 3.4\end{array}\right) \\ A=\begin{array}{ccc}0.9000& 2.1000& 3.2000\\ 0.7000& 6.5000& 4.2000\\ 1.1000& 1.7000& 3.4000\end{array} \\ x\left(t\right)=c_1\left(-0.9236 \\ -0.1322 \\ -0.3599\right)e^{8.5898t}+c_2\left(-0.9236 \\ -0.1322 \\ -0.3599\right)e^{-0.0466t}+c_3\left(0.4052 \\ -0.16760 \\ -0.6155\right)e^{2.2568t} \end{gathered}$$

Step-by-step solution

definition of Linear algebra

Linear algebra is the branch of mathematics that deals with linear equations and linear maps, as well as their representations in vector spaces and via matrices.

Given,

$X\text{'}\left(\begin{array}{ccc}0.9& 2.1& 3.2\\ 0.7& 6.5& 4.2\\ 1.1& 1.7& 3.4\end{array}\right)X$

$>>A=\left[0.92.13.2;0.76.54.2;1.11.73.4\right]$

The eigenvalues and eigenvectors of the given system using MATLAB, as shown below, and then use that information to find the general solution.

$$\begin{gathered} A=\begin{array}{ccc}0.9000& 2.1000& 3.2000\\ 0.7000& 6.5000& 4.2000\\ 1.1000& 1.7000& 3.4000\end{array} \\ eigvec=\begin{array}{ccc}-0.3822& -0.9236& 3.2000\\ -0.8512& -0.1322& 4.2000\\ -0.3598& 0.3599& 3.4000\end{array} \\ eigval=\begin{array}{ccc}8.5898& 0& 0\\ 0& -0.0466& 0\\ 0& 0& 2.2568\end{array} \end{gathered}$$

General solution

K₁and k₂ and x₁ and x₂

Now,

$$\begin{gathered} {\mathrm{K}}_1=\left(-0.3822 \\ -0.8512 \\ -0.3598\right),X_1=\left(-0.3822 \\ -0.8512 \\ -0.3598\right)e^{8.5898t} \\ {\mathrm{K}}_2=\left(-0.9236 \\ -0.1322 \\ -0.3599\right),X_2=\left(-0.9236 \\ -0.1322 \\ -0.3599\right)e^{-0.0466t} \\ {\mathrm{K}}_3=\left(0.4052 \\ -0.1676 \\ -0.6155\right),X_3=\left(0.4052 \\ -0.16760 \\ -0.6155\right)e^{2.2568t} \end{gathered}$$

Therefore, the general solution of the given system is

$$\begin{gathered} x\left(t\right)=c_1\left(-0.3822 \\ -0.8512l \\ -0.3598\right)e^{8.5898t}+c_2\left(-0.9236 \\ -0.1322 \\ -0.3599\right)e^{-0.0466t}+c_3\left(0.4052 \\ -0.1676 \\ -0.6155\right)e^{2.2568t} \end{gathered}$$

Hence,

$$\begin{gathered} x\left(t\right)=c_1\left(-0.3822 \\ -0.8512l \\ -0.3598\right)e^{8.5898t}+c_2\left(-0.9236 \\ -0.1322 \\ -0.3599\right)e^{-0.0466t}+c_3\left(0.4052 \\ -0.1676 \\ -0.6155\right)e^{2.2568t} \end{gathered}$$