Question

In Problems 5-14solve the given linear system.

$\mathit{X}\mathbf{\text{'}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{-}\mathbf{2}& \mathbf{5}\\ \mathbf{-}\mathbf{2}& \mathbf{4}\end{array}\mathbf{\right)}\mathit{X}$

Short answer

The solution for the linear system $X\text{'}=\left(\begin{array}{cc}-2& 5\\ -2& 4\end{array}\right)X$is $X=c_1\left(\begin{array}{c}5\cos t\\ 3\cos t-\sin t\end{array}\right)e^t+c_2\left(\begin{array}{c}5\sin t\\ \cos t+3\sin t\end{array}\right)e^t$.

Step-by-step solution

Define matrix exponential.

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

$A^0=I,A^1=A,A^2=A.A,A^3=A^2.A,.....,A^k=A.\underset{ktimes}{A}.....$where is the unit matrix of order.

Therefore, the infinite matrix power series is $I+\frac{t}{1!}A+\frac{t^2}{2!}{A}^2+\frac{t^3}{3!}{A}^3+\cdots +\frac{t^k}{k!}{A}^k+\cdots$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,$e^{tA}=\sum _{k=0}^{\infty }\frac{t^k}{k!}{A}^k$.

Find the eigenvalue of the given linear system.

It has given that, $X\text{'}=\left(\begin{array}{cc}-2& 5\\ -2& 4\end{array}\right)X$where, $A=\left(\begin{array}{cc}-2& 5\\ -2& 4\end{array}\right)$.

Now, for eigen values the calculated as:

$$\begin{gathered} det(A-\lambda I)=0 \\ \left|\begin{array}{cc}-2-\lambda & 5\\ -2& 4-\lambda \end{array}\right|=0 \\ (-2-\lambda )(4-\lambda )+10=0 \\ -8+2\lambda -4\lambda +{\lambda }^2+10=0 \\ {\lambda }^2-2\lambda +2=0 \end{gathered}$$

Further, solve the above expression,

$$\begin{gathered} \lambda =\frac{2\pm \sqrt{4-4\left(1\right)\left(2\right)}}{2} \\ =\frac{2\pm 2i}{2} \\ =1\pm i \end{gathered}$$

So, the eigenvalues are${\lambda }_1=1+i,{\lambda }_2=\overline{{\lambda }_1}=1-i$

Find the eigenvector of the given linear system.

Solve, for eigenvectors

$$\begin{gathered} (\mathbf{A}-(1+i)\mathbf{I}\mid 0)=\left|\begin{array}{cc}2-\left(1+i\right)& 50\\ -2& 4-\left(1+i\right)0\end{array}\right| \\ =\left(\begin{array}{ccc}-3-i& 5& 0\\ -2& 3-i& 0\end{array}\right) \end{gathered}$$

Apply row operation,

$$\begin{gathered} \left(\frac{3}{2}+\frac{1}{2}i\right)R_2-R_1\to R_2: \\ =\left(\begin{array}{ccc}-3-i& 5& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

$$\begin{gathered} -(3+i)k_1+5k_2=0 \\ {k}_1=\frac{-5}{-(3+i)}{k}_2 \end{gathered}$$

Choosing $k_2=3+i$ it has k₁ =5; This gives an eigenvector $K=\left(\begin{array}{c}5\\ 3+i\end{array}\right)=\left(\begin{array}{c}5\\ 3\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right)$. And column vectors are,

$$\begin{gathered} B_1=\left(\begin{array}{c}5\\ 3\end{array}\right),B_2=\left(\begin{array}{c}0\\ 1\end{array}\right) \\ \lambda =\alpha +\beta i \\ \lambda =1+i \end{gathered}$$

where,$\alpha =1,\beta =1$.

Hence the eigenvector is $K=\left(\begin{array}{c}5\\ 3+i\end{array}\right)=\left(\begin{array}{c}5\\ 3\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right)$.

Find the general solution of the given linear system.

Therefore, the eigenvector is found as:

$$\begin{gathered} {\mathbf{X}}_1=\left[{\mathbf{B}}_1\cos \beta t-{\mathbf{B}}_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}5\\ 3\end{array}\right)\cos t-\left(\begin{array}{c}0\\ 1\end{array}\right)\sin t\right]e^t \\ =\left[\begin{array}{c}5\cos t\\ 3\cos t-\sin t\end{array}\right]e^t \end{gathered}$$

And, the second eigenvector is found as:

$$\begin{gathered} {\mathbf{X}}_2=\left[{\mathbf{B}}_2\cos \beta t-{\mathbf{B}}_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ 1\end{array}\right)\cos t+\left(\begin{array}{c}5\\ 3\end{array}\right)\sin t\right]e^t \\ =\left[\begin{array}{c}5\sin t\\ \cos t+3\sin t\end{array}\right]e^t \end{gathered}$$

Hence the general solution of the system is $X=c_1\left(\begin{array}{c}5\cos t\\ 3\cos t-\sin t\end{array}\right)e^t+c_2\left(\begin{array}{c}5\sin t\\ \cos t+3\sin t\end{array}\right)e^t$.