Question

In Problems 35-46 find the general solution of the given system.

$X\text{'}=\left(\begin{array}{ccc}2& 4& 4\\ -1& -2& 0\\ -1& 0& -2\end{array}\right)X$

Short answer

The general solution of the given system for $X\text{'}=\left(\begin{array}{ccc}2& 4& 4\\ -1& -2& 0\\ -1& 0& -2\end{array}\right)X$is $X=c_1\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}2\sin 2t-2\cos 2t\\ \cos 2t\\ \cos 2t\end{array}\right)+c_3\left(\begin{array}{c}-2\cos 2t-2\sin 2t\\ \sin 2t\\ \sin 2t\end{array}\right)$

Step-by-step solution

Determine the complex eigen values

The polynomial of degree n of the variable $\lambda$is the characteristic polynomial of the n x n matrix A.

$p\left(\lambda \right)=det(\lambda I-A)$

Its root is the eigenvalue of A.

Determine the eigen values

Given

$X\text{'}=\left(\begin{array}{ccc}2& 4& 4\\ -1& -2& 0\\ -1& 0& -2\end{array}\right)X$

Where

$A=\left(\begin{array}{ccc}2& 4& 4\\ -1& -2& 0\\ -1& 0& -2\end{array}\right)$

Obtaining the eigenvalues,

$$\begin{gathered} det(A-\lambda I)=0 \\ \left|\begin{array}{ccc}2-\lambda & 4& 4\\ -1& -2-\lambda & 0\\ -1& 0& -2-\lambda \end{array}\right|=0 \\ (2-\lambda )\left|\begin{array}{cc}-2-\lambda & 0\\ 0& -2-\lambda \end{array}\right|-4\left|\begin{array}{cc}-1& 0\\ -1& -2-\lambda \end{array}\right|+4\left|\begin{array}{cc}-1& -2-\lambda \\ -1& 0\end{array}\right|=0 \end{gathered}$$

$$\begin{gathered} (2-\lambda )(-2-\lambda )(-2-\lambda )+4(-2-\lambda )+4(-2-\lambda )=0 \\ (2-\lambda )(-2-\lambda )(-2-\lambda )+8(-2-\lambda )=0 \\ (-2-\lambda )\left[\right(2-\lambda \left)\right(-2-\lambda )+8]=0 \\ (-2-\lambda )\left[-4-2\lambda +2\lambda +{\lambda }^2+8\right]=0 \\ (-2-\lambda )\left[{\lambda }^2+4\right]=0 \end{gathered}$$

so, our eigenvalues are ${\lambda }_1=-2,{\lambda }_2=2i$and${\lambda }_3=\overline{{\lambda }_2}=-2i$

Determine the eigenvector and corresponding vector

For ${\lambda }_{-}1=-2:$

$$\begin{gathered} (A-(2i\left)I\right|0)=\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ -1& -2-2i& 0& 0\\ -1& 0& -2-2i& 0\end{array}\right) \\ =\left(\begin{array}{cccc}4& 4& 4& 0\\ -1& 0& 0& 0\\ -1& 0& 0& 0\end{array}\right) \end{gathered}$$

from the second and third row we see that k₁=0, as for the first row we have

$$\begin{gathered} 4k_1+4k_2+4k_3=0 \\ {k}_2=-k_3 \end{gathered}$$

choosing k₃=-1 yields k₂ =1.

This gives an eigenvector and a corresponding solution vector

$K_1=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right),X_1=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)e^{-2t}$

For${\lambda }_2=2i$

$$\begin{gathered} =\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ -1& -(2+2i)& 0& 0\\ -1& 0& -(2+2i)& 0\end{array}\right) \\ (A-(2i\left)I\right|0)=\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ -1& -2-2i& 0& 0\\ -1& 0& -2-2i& 0\end{array}\right) \end{gathered}$$

Apply row operation $(2-2i)R_2+R_1\to R_2$:

$=\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ 0& -4& 4& 0\\ -1& 0& -(2+2i)& 0\end{array}\right)$

Apply row operation$(2-2i)R_3+R_1\to R_3$:

$=\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ 0& -4& 4& 0\\ 0& 4& -4& 0\end{array}\right)$

Apply row operation $R_3+R_2\to R_3$:

$=\left(\begin{array}{cccc}2-2i& 4& 4& 0\\ 0& -4& 4& 0\\ 0& 0& 0& 0\end{array}\right)$

from the second row we have

$-4k_2+4k_3=0$

$k_2=k_3$

choosing k₃= 1 yields k₂= 1 we can substitute this into the equation we obtain from the first row,

$$\begin{gathered} (2-2i)k_1+4k_2+4k_3=0 \\ (2-2i)k_1+4\left(1\right)+4\left(1\right)=0 \\ (2-2i)k_1+8=0 \\ {k}_1=-2-2i \end{gathered}$$

This gives an eigenvector

$$\begin{gathered} K_2=\left(\begin{array}{c}-2-2i\\ 1\\ 1\end{array}\right) \\ =\left(\begin{array}{c}-2\\ 1\\ 1\end{array}\right)+i\left(\begin{array}{c}-2\\ 0\\ 0\end{array}\right) \end{gathered}$$

and corresponding column vectors

$B_1=\left(\begin{array}{c}-2\\ 1\\ 1\end{array}\right),and{B}_2=\left(\begin{array}{c}-2\\ 0\\ 0\end{array}\right)$

Determine the general solution of the given system

Also,

$$\begin{gathered} \lambda =\alpha +\beta i \\ \lambda =0+2i \end{gathered}$$

where$\alpha =0and\beta =2$

Therefore,

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

And

$$\begin{gathered} X_3=\left[B_2\cos \beta t+B_1\sin \beta t\right]e^{\alpha t} \\ =\left(\begin{array}{c}-2\\ 0\\ 0\end{array}\right)\cos 2t+\left(\begin{array}{c}-2\\ 1\\ 1\end{array}\right)\sin 2t \\ =\left(\begin{array}{c}-2\cos 2t-2\sin 2t\\ \sin 2t\\ \sin 2t\end{array}\right) \end{gathered}$$

Finally, the general solution of the given system is

$$\begin{gathered} X=c_1{X}_{\mathbf{1}}+c_2{X}_2+c_3{X}_{\mathbf{3}} \\ =c_1\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}2\sin 2t-2\cos 2t\\ \cos 2t\\ \cos 2t\end{array}\right)+c_3\left(\begin{array}{c}-2\cos 2t-2\sin 2t\\ \sin 2t\\ \sin 2t\end{array}\right) \end{gathered}$$

Therefore, the general solution of the given system for $X\text{'}=\left(\begin{array}{ccc}2& 4& 4\\ -1& -2& 0\\ -1& 0& -2\end{array}\right)X$is $X=c_1\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)e^{-2t}+c_2\left(\begin{array}{c}2\sin 2t-2\cos 2t\\ \cos 2t\\ \cos 2t\end{array}\right)+c_3\left(\begin{array}{c}-2\cos 2t-2\sin 2t\\ \sin 2t\\ \sin 2t\end{array}\right)$