Chapter 9.RP, Problem 1RP

In Problems 1-4, find a general solution for the system $x^{\prime }\left(t\right)=Ax\left(t\right)$, where $A$ is given.

$A=\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]$

To determine

To solve: The system $x^{\prime }\left(t\right)=Ax\left(t\right)$, where $A=\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]$.

Answer to Problem 1RP

Solution:

The solution of system $x^{\prime }\left(t\right)=Ax\left(t\right)$, where $A=\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]$ is $X\left(t\right)=c_1{e}^{3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2{e}^{4t}\left[\begin{array}{c}3\\ 2\end{array}\right]$.

Explanation of Solution

Formula used:

The procedure to solve the $x^{\prime }\left(t\right)=Ax\left(t\right)$:

Step-1: Apply the condition $\left|A-\lambda I\right|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $\left|A-\lambda I\right|X=0$ in order to calculate eigenvectors $u_1$ and $u_2$ corresponding to eigenvalue ${\lambda }_1$ and ${\lambda }_2$ respectively:

Step-3:Write the general solution of system $x^{\prime }\left(t\right)=Ax\left(t\right)$ as:

$X\left(t\right)=c_1{e}^{{\lambda }_{1}t}{u}_1+c_2{e}^{{\lambda }_{2}t}{u}_2$

Calculation:

Consider the system $x^{\prime }\left(t\right)=Ax\left(t\right)$,

$A=\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]$

Apply the condition $\left|A-\lambda I\right|=0$ for matrix $A$ in order to calculate eigenvalues:

$\begin{array}{rll}\left|\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|& =& 0\\ \left|\begin{array}{cc}6-\lambda & -3\\ 2& 1-\lambda \end{array}\right|& =& 0\end{array}$

Expand the above determinant:

$\begin{array}{rll}\left(6-\lambda \right)\left(1-\lambda \right)+6& =& 0\\ 6-\lambda -6\lambda +{\lambda }^2+6& =& 0\\ 6-7\lambda +{\lambda }^2+6& =& 0\\ {\lambda }^2-7\lambda +12& =& 0\end{array}$

Factories the above quadratic equation:

$\begin{array}{rll}{\lambda }^2-4\lambda -3\lambda +12& =& 0\\ \lambda \left(\lambda -4\right)-3\left(\lambda -3\right)& =& 0\\ \left(\lambda -4\right)\left(\lambda -3\right)& =& 0\end{array}$

Equate first factor to 0:

$\begin{array}{rll}\left(\lambda -4\right)& =& 0\\ \lambda & =& 4\end{array}$

Equate second factor to 0:

$\begin{array}{rll}\left(\lambda -3\right)& =& 0\\ \lambda & =& 3\end{array}$

Apply the $\left|A-\lambda I\right|X=0$ in order to calculate eigenvector corresponding to eigenvalue $\lambda =4$:

$\begin{array}{rll}\left[\begin{array}{cc}6-4& -3\\ 2& 1-4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}2& -3\\ 2& -3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Write above matrix notation in scalar form notation:

$\begin{array}{l}2x_1-3x_2=0\\ 2x_1-3x_2=0\end{array}$

The eigenvector corresponding to eigenvalue $\lambda =4$:

$u_1=\left[\begin{array}{c}3\\ 2\end{array}\right]$

Apply the $\left|A-\lambda I\right|X=0$ in order to calculate eigenvector corresponding to eigenvalue $\lambda =3$:

$\begin{array}{rll}\left[\begin{array}{cc}6-3& -3\\ 2& 1-3\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ \left[\begin{array}{cc}3& -3\\ 2& -2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

Write above matrix notation in scalar form notation:

$\begin{array}{l}3x_1-3x_2=0\\ 2x_1-2x_2=0\end{array}$

The eigenvector corresponding to eigenvalue $\lambda =3$:

$u_2=\left[\begin{array}{c}1\\ 1\end{array}\right]$

The general solution of system corresponding to eigenvector $u_1$ and $u_2$:

$X\left(t\right)=c_1{e}^{3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2{e}^{4t}\left[\begin{array}{c}3\\ 2\end{array}\right]$

Conclusion: Therefore, the solution of system $x^{\prime }\left(t\right)=Ax\left(t\right)$, where $A=\left[\begin{array}{cc}6& -3\\ 2& 1\end{array}\right]$ is $X\left(t\right)=c_1{e}^{3t}\left[\begin{array}{c}1\\ 1\end{array}\right]+c_2{e}^{4t}\left[\begin{array}{c}3\\ 2\end{array}\right]$.