Question

Solve each of the following linear systems.

(a)$\mathit{X}\mathbf{\text{'}}\mathbf{=}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\mathit{X}$

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

Short answer

(a) The general solution of the given system for $X\text{'}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)XisX=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t}$.

(b) The general solution of the given system for$X\text{'}=\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)XisX=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left[\left(\begin{array}{c}-1\\ 1\end{array}\right)t+\left(\begin{array}{c}-1\\ 0\end{array}\right)\right]$

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 .

localid="1668174538060" $\mathrm{p}\left(\mathrm{\lambda }\right)=\det (\mathrm{\lambda I}-\mathrm{A})$

Its root is the eigenvalue of A.

Determine the eigen values

(a) Given

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

where

$A=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$

Obtaining the eigenvalues of the coefficient matrix,

$det(A-\lambda I)=0\to \left|\begin{array}{cc}1-\lambda & 1\\ 1& 1-\lambda \end{array}\right|=0$

$(1-\lambda )(1-\lambda )-1=0$

$$\begin{gathered} 1-\lambda -\lambda +{\lambda }^2-1=0 \\ {\lambda }^2-2\lambda =0 \\ \lambda (\lambda -2)=0 \end{gathered}$$

$Sooureigenvaluesare{\lambda }_1=0and{\lambda }_2=2.$

Determine the eigen vector and a corresponding vector

For${\lambda }_1=0$:

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

Apply row operation $R_2-R_1\to R_2$:

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

here we have

$k_1+k_2=0\to k_1=-k_2$

choosing k₂=1 yields k1= -1.

This gives an eigenvector and a corresponding solution vector

$K_1=\left(\begin{array}{c}-1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-1\\ 1\end{array}\right)$

For $\lambda =2$

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

Apply row operation localid="1668083160516" $R_2+R_1\to R_2$:

localid="1668083169820" $=\left(\begin{array}{ccc}-1& 1& 0\\ 0& 0& 0\end{array}\right)$

Here we have

localid="1668083176894" $-k_1+k_2=0\to k_1=k_2$

choosing K₂=1 yields k₁=1.

This gives an eigenvector and a corresponding solution vector

localid="1668083184768" $K_1=\left(\begin{array}{c}1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t}$

Therefore, the general solution of the system is

localid="1668083210324" $$\begin{gathered} X=c_1{X}_{\mathbf{1}}+c_2{X}_{\mathbf{2}} \\ =c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t} \end{gathered}$$

Phase portrait of the system is shown below.

Trajectories of the system are localid="1668083222073" $x=-c_1+c_2{e}^{2t}andy=c_1+c_2{e}^{2t}ory=c_1+x+c_1$localid="1668083233114" $=2c_1+x$

However, y=-x is not a trajectory of the system.

The family of lines are all perpendicular to y=-x but all the constant solutions lie on the line y=-x .

Determine the eigen values

Given

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

where

$A=\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)$

Obtaining the eigenvalues of the coefficient matrix,

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

we have $\lambda =0$, which is an eigenvalue of multiplicity two.

Determine the eigen vector and a corresponding vector

Solving,

$$\begin{gathered} (A-\lambda I)K=0\to \left(\begin{array}{cc}1-0& 1\\ -1& -1-0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \end{gathered}$$

Apply row operation $R_2+R_1\to R_2$ :

localid="1668174547596" $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{k}_1\\ k_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

so here we have

$k_1+k_2=0\to k_1=-k_2$choosing k2=1 yields k1=-1.

This gives an eigenvector and a corresponding solution vector

localid="1668174551201" $K_1=\left(\begin{array}{c}-1\\ 1\end{array}\right),X_1=\left(\begin{array}{c}-1\\ 1\end{array}\right)$

localid="1668174560200" $\begin{array}{rcl}(A-\lambda I)P& =& K=\left(\begin{array}{cc}1-0& 1\\ -1& -1-0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}-1\\ 1\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}-1\\ 1\end{array}\right)\end{array}$

Apply row operation localid="1668174563808" $R_2+R_1\to R_2$:

localid="1668174568215" $=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{p}_1\\ p_2\end{array}\right)=\left(\begin{array}{c}-1\\ 1\end{array}\right)$

Here we have

localid="1668174573040" $k_1+k_2=-1$

choosing k2=0

This gives an eigenvector and a corresponding solution vector

localid="1668174576679" $P_1=\left(\begin{array}{c}-1\\ 0\end{array}\right),X_2=\left(\begin{array}{c}-1\\ 1\end{array}\right)t+\left(\begin{array}{c}-1\\ 0\end{array}\right)$

Finally, the general solution of the system is

localid="1668174586252" $$\begin{gathered} X=c_1{X}_1+c_2{X}_2 \\ =c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left[\left(\begin{array}{c}-1\\ 1\end{array}\right)t+\left(\begin{array}{c}-1\\ 0\end{array}\right)\right] \end{gathered}$$

Phase portrait of the system is shown below. y=-x is not a trajectory, however, all trajectories are parallel to y=-x.

And there are constant solutions that lie on y=-x.

a) Therefore, the general solution of the given system for $X\text{'}=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)XisX=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left(\begin{array}{c}1\\ 1\end{array}\right)e^{2t}$.

(b) Therefore, the general solution of the given system for$X\text{'}=\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right)XisX=c_1\left(\begin{array}{c}-1\\ 1\end{array}\right)+c_2\left[\left(\begin{array}{c}-1\\ 1\end{array}\right)t+\left(\begin{array}{c}-1\\ 0\end{array}\right)\right]$