Question

In Problems 5-14solve the given linear system.

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

Short answer

The solution for the linear system$X\text{'}=\left(\begin{array}{cc}1& -2\\ -2& 1\end{array}\right)X$is $X=c_1\left(\begin{array}{c}\cos 2t\\ -\sin 2t\end{array}\right)e^t+c_2\left(\begin{array}{c}\sin 2t\\ \cos 2t\end{array}\right)e^t$.

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. 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, $X\text{'}=\left(\begin{array}{cc}1& 2\\ -2& 1\end{array}\right)X$where the matrix is $A=\left(\begin{array}{cc}1& 2\\ -2& 1\end{array}\right)$.

Now, the eigen values are find as,

$\det (\mathbf{A}-\lambda \mathbf{I})=0$

$$\begin{gathered} \left(\begin{array}{cc}1-\lambda & 2\\ -2& 1-\lambda \end{array}\right)=0 \\ (1-\lambda )(1-\lambda )+4=0 \\ 1-\lambda -\lambda +{\lambda }^2+4=0 \\ {\lambda }^2-2\lambda +5=0 \end{gathered}$$

Further solve the above expression,

$$\begin{gathered} \lambda =\frac{2\pm \sqrt{4-4\left(1\right)\left(5\right)}}{2} \\ =\frac{2\pm 4i}{2} \\ =1\pm 2i \end{gathered}$$

So, it has eigenvalues of ${\lambda }_1=1+2i,{\lambda }_2=\overline{{\lambda }_1}=1-2i$.

Find the eigenvector of the given linear system.

Solve, this for eigenvectors;

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

Apply row operation, it gives

$R_2-R_1\to R_2:$

$$\begin{gathered} =\left(\begin{array}{ccc}-2i& 2& 0\\ 0& 0& 0\end{array}\right) \\ -2i{k}_1+2k_2=0 \\ {k}_2=ik \end{gathered}$$Choosing it K1 = 1 has K2 =i.

This gives an eigenvector $K=\left(\begin{array}{c}1\\ i\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)+i\left(\begin{array}{c}0\\ 1\end{array}\right)$.

And column vectors are written as;

$$\begin{gathered} B_1=\left(\begin{array}{c}1\\ 0\end{array}\right),B_2=\left(\begin{array}{c}0\\ 1\end{array}\right) \\ \lambda =\alpha +\beta i \\ \lambda =1+2i \end{gathered}$$

where, it gives $\alpha =1,\beta =2$.

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

Find the general solution of the given linear system.

Therefore, the eigenvector is,

$$\begin{gathered} {\mathbf{X}}_{\mathbf{1}}=\left[{\mathbf{B}}_1\cos \beta t-{\mathbf{B}}_2\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}1\\ 0\end{array}\right)\cos 2t-\left(\begin{array}{c}0\\ 1\end{array}\right)\sin 2t\right]e^t \\ =\left[\begin{array}{c}\cos 2t\\ -\sin 2t\end{array}\right]e^t \end{gathered}$$

And the second eigen vector is

$$\begin{gathered} {\mathbf{X}}_2=\left[{\mathbf{B}}_2\cos \beta t+{\mathbf{B}}_1\sin \beta t\right]e^{\alpha t} \\ =\left[\left(\begin{array}{c}0\\ 1\end{array}\right)\cos 2t+\left(\begin{array}{c}1\\ 0\end{array}\right)\sin 2t\right]e^t \\ =\left[\begin{array}{c}\sin 2t\\ \cos 2t\end{array}\right]e^t \end{gathered}$$

Hence the general solution of the system is $X=c_1\left(\begin{array}{c}\cos 2t\\ -\sin 2t\end{array}\right)e^t+c_2\left(\begin{array}{c}\sin 2t\\ \cos 2t\end{array}\right)e^t$.