Solve \[ \mathbf{\dot{x}}=\begin{pmatrix} 3 & 1 & -2\\ -1 & 2 & 1\\ 4 & 1 & -3 \end{pmatrix} \mathbf{x},\qquad \mathbf{x}\left ( 0\right ) =\begin{pmatrix} 1\\ 4\\ -7 \end{pmatrix} \] Solution
The first step is to find the eigenvalues. For this we need to solve \begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} 3-\lambda & 1 & -2\\ -1 & 2-\lambda & 1\\ 4 & 1 & -3-\lambda \end{vmatrix} & =0\\ \left ( 3-\lambda \right ) \begin{vmatrix} 2-\lambda & 1\\ 1 & -3-\lambda \end{vmatrix} -\begin{vmatrix} -1 & 1\\ 4 & -3-\lambda \end{vmatrix} -2\begin{vmatrix} -1 & 2-\lambda \\ 4 & 1 \end{vmatrix} & =0\\ \left ( 3-\lambda \right ) \left ( \left ( 2-\lambda \right ) \left ( -3-\lambda \right ) -1\right ) -\left ( \left ( 3+\lambda \right ) -4\right ) -2\left ( -1-4\left ( 2-\lambda \right ) \right ) & =0\\ \lambda ^{3}-2\lambda ^{2}-\lambda +2 & =0 \end{align*}
Guessing a root at \(\lambda =1\) is verified to be correct since \(1-2-1+2=0\). Now that we know one root, we can do long division \(\frac{\left ( \lambda ^{3}-2\lambda ^{2}-\lambda +2\right ) }{\left ( \lambda -1\right ) }=\lambda ^{2}-\lambda -2\). Therefore the charaterstic polynominal factors to \begin{align*} \lambda ^{3}-2\lambda ^{2}-\lambda +2 & =\left ( \lambda -1\right ) \left ( \lambda ^{2}-\lambda -2\right ) \\ & =\left ( \lambda -1\right ) \left ( \lambda -2\right ) \left ( \lambda +1\right ) \end{align*}
Hence the eigenvalues are \(\lambda _{1}=1,\lambda _{2}=2,\lambda _{3}=-1\).
For \(\lambda _{1}=1\)\begin{align*} \left ( A-\lambda _{1}I\right ) \mathbf{v}_{1} & =0\\\begin{pmatrix} 3-\lambda & 1 & -2\\ -1 & 2-\lambda & 1\\ 4 & 1 & -3-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 3-1 & 1 & -2\\ -1 & 2-1 & 1\\ 4 & 1 & -3-1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 2 & 1 & -2\\ -1 & 1 & 1\\ 4 & 1 & -4 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Let \(v_{1}=1\). First equation gives \(2+v_{2}-2v_{3}=0\) and the second equation gives \(-1+v_{2}+v_{3}=0\). Subtracting gives \(3-3v_{3}=0\), giving \(v_{3}=1\). Therefore \(v_{2}=0\). Hence \[ \mathbf{v}^{1}=\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix} \] For \(\lambda _{2}=2\)\begin{align*} \left ( A-\lambda _{2}I\right ) \mathbf{v}^{2} & =0\\\begin{pmatrix} 3-\lambda & 1 & -2\\ -1 & 2-\lambda & 1\\ 4 & 1 & -3-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 3-2 & 1 & -2\\ -1 & 2-2 & 1\\ 4 & 1 & -3-2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 1 & 1 & -2\\ -1 & 0 & 1\\ 4 & 1 & -5 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Let \(v_{1}=1\). Hence first equation gives \(1+v_{2}-2v_{3}=0\) and second equation gives \(-1+v_{3}=0\). Therefore \(v_{3}=1\) and \(v_{2}=2v_{3}-1=1\). Hence\[ \mathbf{v}^{2}=\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} \] For \(\lambda _{3}=-1\)\begin{align*} \left ( A-\lambda _{3}I\right ) \mathbf{v}^{3} & =0\\\begin{pmatrix} 3-\lambda & 1 & -2\\ -1 & 2-\lambda & 1\\ 4 & 1 & -3-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 3+1 & 1 & -2\\ -1 & 2+1 & 1\\ 4 & 1 & -3+1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 4 & 1 & -2\\ -1 & 3 & 1\\ 4 & 1 & -2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Let \(v_{1}=1\). Hence first equation gives \(4+v_{2}-2v_{3}=0\) and second equation gives \(-1+3v_{2}+v_{3}=0\). Multiplying \(4+v_{2}-2v_{3}=0\) by \(-3\) and adding it to \(-1+3v_{2}+v_{3}=0\) gives \(\left ( -12-3v_{2}+6v_{3}+\left ( -1+3v_{2}+v_{3}\right ) \right ) =0\) or \(-13+7v_{3}=0\) Hence \(v_{3}=\frac{13}{7}\). Therefore \(v_{2}=2v_{3}-4=2\left ( \frac{13}{7}\right ) -4=-\frac{2}{7}\). Hence\[ \mathbf{v}^{3}=\begin{pmatrix} 1\\ -\frac{2}{7}\\ \frac{13}{7}\end{pmatrix} \] Therefore \begin{align*} \mathbf{x}^{1}\left ( t\right ) & =e^{\lambda _{1}t}\mathbf{v}^{1}=e^{t}\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix} \\ \mathbf{x}^{2}\left ( t\right ) & =e^{\lambda _{2}t}\mathbf{v}^{2}=e^{2t}\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} \\ \mathbf{x}^{3}\left ( t\right ) & =e^{\lambda _{3}t}\mathbf{v}^{3}=e^{-t}\begin{pmatrix} 1\\ -\frac{2}{7}\\ \frac{13}{7}\end{pmatrix} \end{align*}
Hence the general solution is\begin{align*} \mathbf{x}\left ( t\right ) & =c_{1}\mathbf{x}^{1}\left ( t\right ) +c_{2}\mathbf{x}^{2}\left ( t\right ) +c_{3}\mathbf{x}^{3}\left ( t\right ) \\ & =e^{t}\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix} +c_{2}e^{2t}\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} +c_{3}e^{-t}\begin{pmatrix} 1\\ -\frac{2}{7}\\ \frac{13}{7}\end{pmatrix} \end{align*}
Or\begin{equation} \mathbf{x}\left ( t\right ) =\begin{pmatrix} c_{1}e^{t}+c_{2}e^{2t}+c_{3}e^{-t}\\ c_{2}e^{2t}-\frac{2}{7}c_{3}e^{-t}\\ c_{1}e^{t}+c_{2}e^{2t}+\frac{13}{7}c_{3}e^{-t}\end{pmatrix} \tag{A} \end{equation} Initial conditions are now used to find \(c_{1},c_{2},c_{3}\). At \(t=0\) the above reduces to\begin{align} \mathbf{x}\left ( 0\right ) & =\begin{pmatrix} 1\\ 4\\ -7 \end{pmatrix} \nonumber \\\begin{pmatrix} c_{1}+c_{2}+c_{3}\\ c_{2}-\frac{2}{7}c_{3}\\ c_{1}+c_{2}+\frac{13}{7}c_{3}\end{pmatrix} & =\begin{pmatrix} 1\\ 4\\ -7 \end{pmatrix} \nonumber \\\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & -\frac{2}{7}\\ 1 & 1 & \frac{13}{7}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} & =\begin{pmatrix} 1\\ 4\\ -7 \end{pmatrix} \tag{1} \end{align}
Gaussian elimination on \(\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & -\frac{2}{7}\\ 1 & 1 & \frac{13}{7}\end{pmatrix}\begin{pmatrix} 1\\ 4\\ -7 \end{pmatrix} \). Replacing row 3 by row 3 - row 1 gives\[\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & -\frac{2}{7}\\ 0 & 0 & \frac{13}{7}-1 \end{pmatrix}\begin{pmatrix} 1\\ 4\\ -7-1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & -\frac{2}{7}\\ 0 & 0 & \frac{6}{7}\end{pmatrix}\begin{pmatrix} 1\\ 4\\ -8 \end{pmatrix} \] Hence (1) becomes\[\begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & -\frac{2}{7}\\ 0 & 0 & \frac{6}{7}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 1\\ 4\\ -8 \end{pmatrix} \] Back substitution gives \(\frac{6}{7}c_{3}=-8\), or \(c_{3}=-\frac{28}{3}\). From second row \begin{align*} c_{2}-\frac{2}{7}c_{3} & =4\\ c_{2} & =4+\frac{2}{7}c_{3}\\ & =4+\frac{2}{7}\left ( -\frac{28}{3}\right ) \\ & =\frac{4}{3} \end{align*}
From first row\begin{align*} c_{1}+c_{2}+c_{3} & =1\\ c_{1} & =1-c_{2}-c_{3}\\ & =1-\frac{4}{3}+\frac{28}{3}\\ & =9 \end{align*}
Using the above values of \(c_{1},c_{2},c_{3}\), Eq (A) becomes\begin{align} \mathbf{x}\left ( t\right ) & =\begin{pmatrix} c_{1}e^{t}+c_{2}e^{2t}+c_{3}e^{-t}\\ c_{2}e^{2t}-\frac{2}{7}c_{3}e^{-t}\\ c_{1}e^{t}+c_{2}e^{2t}+\frac{13}{7}c_{3}e^{-t}\end{pmatrix} \nonumber \\ & =\begin{pmatrix} 9e^{t}+\frac{4}{3}e^{2t}-\frac{28}{3}e^{-t}\\ \frac{4}{3}e^{2t}-\frac{2}{7}\left ( -\frac{28}{3}\right ) e^{-t}\\ 9e^{t}+\frac{4}{3}e^{2t}+\frac{13}{7}\left ( -\frac{28}{3}\right ) e^{-t}\end{pmatrix} \nonumber \\ & =\begin{pmatrix} 9e^{t}+\frac{4}{3}e^{2t}-\frac{28}{3}e^{-t}\\ \frac{4}{3}e^{2t}+\frac{8}{3}e^{-t}\\ 9e^{t}+\frac{4}{3}e^{2t}-\frac{52}{3}e^{-t}\end{pmatrix} \tag{2} \end{align}
Therefore\begin{align*} x_{1}\left ( t\right ) & =9e^{t}+\frac{4}{3}e^{2t}-\frac{28}{3}e^{-t}\\ x_{2}\left ( t\right ) & =\frac{4}{3}e^{2t}+\frac{8}{3}e^{-t}\\ x_{3}\left ( t\right ) & =9e^{t}+\frac{4}{3}e^{2t}-\frac{52}{3}e^{-t} \end{align*}
Find general solution of \[ \mathbf{\dot{x}}=\begin{pmatrix} 1 & -5 & 0\\ 1 & -3 & 0\\ 0 & 0 & 1 \end{pmatrix} \mathbf{x}\] Solution
The first step is to find the eigenvalues. For this we need to solve \begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} 1-\lambda & -5 & 0\\ 1 & -3-\lambda & 0\\ 0 & 0 & 1-\lambda \end{vmatrix} & =0\\ \left ( 1-\lambda \right ) \begin{vmatrix} -3-\lambda & 0\\ 0 & 1-\lambda \end{vmatrix} +5\begin{vmatrix} 1 & 0\\ 0 & 1-\lambda \end{vmatrix} & =0\\ \left ( 1-\lambda \right ) \left ( \left ( -3-\lambda \right ) \left ( 1-\lambda \right ) \right ) +5\left ( 1-\lambda \right ) & =0 \end{align*}
Factoring \(\left ( 1-\lambda \right ) \) gives\begin{align*} \left ( 1-\lambda \right ) \left ( \left ( -3-\lambda \right ) \left ( 1-\lambda \right ) +5\right ) & =0\\ \left ( 1-\lambda \right ) \left ( \lambda ^{2}+2\lambda -3+5\right ) & =0\\ \left ( 1-\lambda \right ) \left ( \lambda ^{2}+2\lambda +2\right ) & =0 \end{align*}
Hence one root is \(\lambda _{1}=1\). Now we find roots of \(\left ( \allowbreak \lambda ^{2}+2\lambda +2\right ) \). \(\lambda =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}=-1\pm \frac{1}{2}\sqrt{4-4\left ( 2\right ) }=-1\pm \frac{1}{2}\sqrt{-4}\). Hence \[ \lambda =-1\pm i \] Therefore the roots are \begin{align*} \lambda _{1} & =1\\ \lambda _{2} & =-1+i\\ \lambda _{3} & =-1-i \end{align*}
For \(\lambda _{1}=1\)\begin{align*} \left ( A-\lambda _{1}I\right ) \mathbf{v}^{1} & =0\\\begin{pmatrix} 1-\lambda _{1} & -5 & 0\\ 1 & -3-\lambda _{1} & 0\\ 0 & 0 & 1-\lambda _{1}\end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 1-1 & -5 & 0\\ 1 & -3-1 & 0\\ 0 & 0 & 1-1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 0 & -5 & 0\\ 1 & -4 & 0\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Hence \(v_{3}\) is arbitrary, say \(v_{3}=1\). And \(v_{2}=0\) from first equation. And from second equation \(v_{1}=0\). Therefore \[ \mathbf{v}^{1}=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \] Hence\begin{align*} \mathbf{x}^{1}\left ( t\right ) & =e^{\lambda _{1}t}\mathbf{v}^{1}\\ & =e^{t}\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} \end{align*}
For \(\lambda _{2}=-1+i\)\begin{align*} \left ( A-\lambda _{2}I\right ) \mathbf{v}^{2} & =0\\\begin{pmatrix} 1-\lambda _{2} & -5 & 0\\ 1 & -3-\lambda _{2} & 0\\ 0 & 0 & 1-\lambda _{2}\end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 1-\left ( -1+i\right ) & -5 & 0\\ 1 & -3-\left ( -1+i\right ) & 0\\ 0 & 0 & 1-\left ( -1+i\right ) \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} 2-i & -5 & 0\\ 1 & -2-i & 0\\ 0 & 0 & 2-i \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
From last equation \(v_{3}=0\). from second equation \(v_{1}=\left ( 2+i\right ) v_{2}\). Hence \[ \mathbf{v}^{2}=\begin{pmatrix} \left ( 2+i\right ) v_{2}\\ v_{2}\\ 0 \end{pmatrix} =v_{2}\begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \] Choosing \(v_{2}=1\) the above becomes\[ \mathbf{v}^{2}=\begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \] Hence\[ \mathbf{x}_{\lambda _{2}}^{2}\left ( t\right ) =e^{\lambda _{2}t}\mathbf{v}^{2}=e^{\left ( -1+i\right ) t}\begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \] Since this is complex root, we will now find the real and imaginary parts of the above, and use these to generate \(\mathbf{x}^{2}\left ( t\right ) ,\mathbf{x}^{3}\left ( t\right ) \) from the above. \begin{align*} e^{\left ( -1+i\right ) t}\begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} & =e^{-t}e^{it}\begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \\ & =e^{-t}\left ( \cos t+i\sin t\right ) \begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \\ & =\left ( e^{-t}\cos t+ie^{-t}\sin t\right ) \begin{pmatrix} 2+i\\ 1\\ 0 \end{pmatrix} \\ & =\begin{pmatrix} \left ( e^{-t}\cos t+ie^{-t}\sin t\right ) \left ( 2+i\right ) \\ \left ( e^{-t}\cos t+ie^{-t}\sin t\right ) \\ 0 \end{pmatrix} \\ & =\begin{pmatrix} 2e^{-t}\cos t+ie^{-t}\cos t+2ie^{-t}\sin t-e^{-t}\sin t\\ e^{-t}\cos t+ie^{-t}\sin t\\ 0 \end{pmatrix} \\ & =\begin{pmatrix} \left ( 2e^{-t}\cos t-e^{-t}\sin t\right ) +i\left ( e^{-t}\cos t+2e^{-t}\sin t\right ) \\ e^{-t}\cos t+ie^{-t}\sin t\\ 0 \end{pmatrix} \end{align*}
The real of the above is \[ \mathbf{x}^{2}\left ( t\right ) =\begin{pmatrix} 2e^{-t}\cos t-e^{-t}\sin t\\ e^{-t}\cos t\\ 0 \end{pmatrix} \] And imaginary part is \[ \mathbf{x}^{3}\left ( t\right ) =\begin{pmatrix} e^{-t}\cos t+2e^{-t}\sin t\\ e^{-t}\sin t\\ 0 \end{pmatrix} \] We have now obtain the three eigenvectors we want. Hence the general solution is\begin{align*} \mathbf{x}\left ( t\right ) & =c_{1}\mathbf{x}^{1}\left ( t\right ) +c_{2}\mathbf{x}^{2}\left ( t\right ) +c_{3}\mathbf{x}^{3}\left ( t\right ) \\ & =c_{1}e^{t}\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix} +c_{2}e^{-t}\begin{pmatrix} 2\cos t-\sin t\\ \cos t\\ 0 \end{pmatrix} +c_{3}e^{-t}\begin{pmatrix} \cos t+2\sin t\\ \sin t\\ 0 \end{pmatrix} \end{align*}
Solve\[ \mathbf{\dot{x}}=\begin{pmatrix} -4 & -4 & 0\\ 10 & 9 & 1\\ -4 & -3 & 1 \end{pmatrix} \mathbf{x},\qquad \mathbf{x}\left ( 0\right ) =\begin{pmatrix} 2\\ 1\\ -1 \end{pmatrix} \] Solution
The first step is to find the eigenvalues. For this we need to solve \begin{align*} \det \left ( A-\lambda I\right ) & =0\\\begin{vmatrix} -4-\lambda & -4 & 0\\ 10 & 9-\lambda & 1\\ -4 & -3 & 1-\lambda \end{vmatrix} & =0\\ \left ( -4-\lambda \right ) \begin{vmatrix} 9-\lambda & 1\\ -3 & 1-\lambda \end{vmatrix} +4\begin{vmatrix} 10 & 1\\ -4 & 1-\lambda \end{vmatrix} & =0\\ \left ( -4-\lambda \right ) \left ( \left ( 9-\lambda \right ) \left ( 1-\lambda \right ) +3\right ) +4\left ( \left ( 10\right ) \left ( 1-\lambda \right ) +4\right ) & =0\\ \left ( \lambda -2\right ) ^{3} & =0 \end{align*}
Hence root is \(\lambda =2\) of multiplicity \(3\).
To eigenvectors we start as before, using \(\lambda =2.\)\begin{align*} \left ( A-\lambda I\right ) \mathbf{v}^{1} & =0\\\begin{pmatrix} -4-\lambda & -4 & 0\\ 10 & 9-\lambda & 1\\ -4 & -3 & 1-\lambda \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} -4-2 & -4 & 0\\ 10 & 9-2 & 1\\ -4 & -3 & 1-2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \\\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} & =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \end{align*}
Now we check if the eigenvalue is complete or defective. Using the first 2 rows we obtain\begin{align*} -6v_{1}-4v_{2} & =0\\ 10v_{1}+7v_{2}+v_{3} & =0 \end{align*}
Solving gives \(v_{1}=2v_{3},v_{2}=-3v_{3}\). Hence\[\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} =\begin{pmatrix} 2v_{3}\\ -3v_{3}\\ v_{3}\end{pmatrix} =v_{3}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} \] Choosing \(v_{3}=1\) gives\[ \mathbf{v}^{1}=\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} \] Lets see now if we can obtain another linearly independent eigenvector. Using the first row and the third row\begin{align*} -6v_{1}-4v_{2} & =0\\ -4v_{1}-3v_{2}-v_{3} & =0 \end{align*}
Solving gives \(v_{1}=2v_{3},v_{2}=-3v_{3}\). Which is the same as the one found above. Finally using the second and third row\begin{align*} 10v_{1}+7v_{2}+v_{3} & =0\\ -4v_{1}-3v_{2}-v_{3} & =0 \end{align*}
Solving gives \(v_{1}=2v_{3},v_{2}=-3v_{3}\) which is the same as above. So the eigenvalue \(2\) is defective. \[ \mathbf{x}^{1}\left ( t\right ) =e^{2t}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} \] Since the eigenvalue is defective, to find the second and third eigenvectors we do the following. To find \(\mathbf{v}^{2}\). We need to solve\begin{equation} \left ( A-\lambda I\right ) ^{2}\mathbf{v}^{2}=\mathbf{0}\tag{1} \end{equation} But \(A-\lambda I=\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix} \) from earlier. Hence \begin{align*} \left ( A-\lambda I\right ) ^{2} & =\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix}\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix} \\ & =\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix} \end{align*}
Therefore (1) becomes\[\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix}\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \] Using the first equation \(-4v_{1}-4v_{2}-4v_{3}=0\) or equivalently \(v_{1}+v_{2}+v_{3}=0\). Therefore \(v_{1}=-v_{2}-v_{3}\). Hence\begin{align*} \mathbf{v}^{2} & =\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\end{pmatrix} \\ & =\begin{pmatrix} -v_{2}-v_{3}\\ v_{2}\\ v_{3}\end{pmatrix} \\ & =v_{2}\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} +v_{3}\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix} \end{align*}
Taking \(v_{2}=1,v_{3}=0\) gives\[ \mathbf{v}^{2}=\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} \] Let us check the above choice is valid: \(\left ( A-\lambda I\right ) \mathbf{v}^{2}=\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix}\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} =\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} \) which is not zero. Good, so we can use it. Therefore\begin{align*} \mathbf{x}^{2}\left ( t\right ) & =e^{\lambda t}\left ( \mathbf{v}^{2}+t\left ( A-\lambda I\right ) \mathbf{v}^{2}\right ) \\ & =e^{2t}\left ( \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} +t\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix}\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} \right ) \\ & =e^{2t}\left ( \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} +t\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} \right ) \\ & =e^{2t}\begin{pmatrix} -1+2t\\ 1-3t\\ t \end{pmatrix} \end{align*}
Now we find the third eigenvector \(\mathbf{v}^{3}\). We need to solve\begin{equation} \left ( A-\lambda I\right ) ^{3}\mathbf{v}^{3}=\mathbf{0}\tag{1} \end{equation} But \(\left ( A-\lambda I\right ) ^{2}=\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix} \) from earlier. Hence \begin{align*} \left ( A-\lambda I\right ) ^{3} & =\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix}\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix} \\ & =\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{align*}
Therefore \(\mathbf{v}_{3}\) is arbitrary as long as \(\left ( A-\lambda I\right ) ^{2}\mathbf{v}_{3}\neq \mathbf{0}\). Let us pick \(\mathbf{v}_{3}=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} \). Checking this choice is valid: \(\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix}\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} =\begin{pmatrix} -4\\ 6\\ -2 \end{pmatrix} \). Not zero. Good, so we can use it. Therefore\begin{align*} \mathbf{x}^{3}\left ( t\right ) & =e^{\lambda t}\left ( \mathbf{v}^{3}+t\left ( A-\lambda I\right ) \mathbf{v}^{3}+\frac{t^{2}}{2}\left ( A-\lambda I\right ) ^{2}\mathbf{v}^{3}\right ) \\ & =e^{2t}\left ( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} +t\begin{pmatrix} -6 & -4 & 0\\ 10 & 7 & 1\\ -4 & -3 & -1 \end{pmatrix}\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} +\frac{t^{2}}{2}\begin{pmatrix} -4 & -4 & -4\\ 6 & 6 & 6\\ -2 & -2 & -2 \end{pmatrix}\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} \right ) \\ & =e^{2t}\left ( \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} +t\begin{pmatrix} -6\\ 10\\ -4 \end{pmatrix} +\frac{t^{2}}{2}\begin{pmatrix} -4\\ 6\\ -2 \end{pmatrix} \right ) \\ & =e^{2t}\begin{pmatrix} 1-6t-2t^{2}\\ 10t+3t^{2}\\ -4t-t^{2}\end{pmatrix} \end{align*}
Therefore the general solution is\begin{align*} \mathbf{x}\left ( t\right ) & =c_{1}\mathbf{x}^{1}\left ( t\right ) +c_{2}\mathbf{x}^{2}\left ( t\right ) +c_{3}\mathbf{x}^{3}\left ( t\right ) \\ & =c_{1}e^{2t}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} +c_{2}e^{2t}\begin{pmatrix} -1+2t\\ 1-3t\\ t \end{pmatrix} +c_{3}e^{2t}\begin{pmatrix} 1-6t-2t^{2}\\ 10t+3t^{2}\\ -4t-t^{2}\end{pmatrix} \\ & =\begin{pmatrix} e^{2t}\left ( 2c_{1}+c_{2}\left ( -1+2t\right ) +c_{3}\left ( 1-6t-2t^{2}\right ) \right ) \\ e^{2t}\left ( -3c_{1}+c_{2}\left ( 1-3t\right ) +c_{3}\left ( 10t+3t^{2}\right ) \right ) \\ e^{2t}\left ( c_{1}+tc_{2}+c_{3}\left ( -4t-t^{2}\right ) \right ) \end{pmatrix} \end{align*}
Now we find \(c_{i}\) from initial conditions. At \(t=0\)\begin{align*} \begin{pmatrix} 2\\ 1\\ -1 \end{pmatrix} & =c_{1}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} +c_{2}\begin{pmatrix} -1\\ 1\\ 1 \end{pmatrix} +c_{3}\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} \\ & =\begin{pmatrix} 2c_{1}-c_{2}+c_{3}\\ -3c_{1}+c_{2}\\ c_{1}\end{pmatrix} \end{align*}
Or\begin{equation} \begin{pmatrix} 2 & -1 & 1\\ -3 & 1 & 0\\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\\ c_{3}\end{pmatrix} =\begin{pmatrix} 2\\ 1\\ -1 \end{pmatrix} \tag{2} \end{equation} From last row, \(c_{1}=-1\). From second row \(-3c_{1}+c_{2}=1\), hence \(c_{2}=1-3=-2\). From first row \(2c_{1}-c_{2}+c_{3}=2\), hence \(c_{3}=2-2+2=2\). Therefore the general solution becomes\begin{align*} \mathbf{x}\left ( t\right ) & =c_{1}\mathbf{x}^{1}\left ( t\right ) +c_{2}\mathbf{x}^{2}\left ( t\right ) +c_{3}\mathbf{x}^{3}\left ( t\right ) \\ & =-e^{2t}\begin{pmatrix} 2\\ -3\\ 1 \end{pmatrix} -2e^{2t}\begin{pmatrix} -1+2t\\ 1-3t\\ t \end{pmatrix} +2e^{2t}\begin{pmatrix} 1-6t-2t^{2}\\ 10t+3t^{2}\\ -4t-t^{2}\end{pmatrix} \\ & =e^{2t}\begin{pmatrix} -2-2\left ( -1+2t\right ) +2\left ( 1-6t-2t^{2}\right ) \\ 3-2\left ( 1-3t\right ) +2\left ( 10t+3t^{2}\right ) \\ -1-2t+2\left ( -4t-t^{2}\right ) \end{pmatrix} \\ & =e^{2t}\begin{pmatrix} -4t^{2}-16t+2\\ 6t^{2}+26t+1\\ -2t^{2}-10t-1 \end{pmatrix} \end{align*}
Or\begin{align*} x_{1}\left ( t\right ) & =e^{2t}\left ( -4t^{2}-16t+2\right ) \\ x_{2}\left ( t\right ) & =e^{2t}\left ( 6t^{2}+26t+1\right ) \\ x_{3}\left ( t\right ) & =e^{2t}\left ( -2t^{2}-10t-1\right ) \end{align*}
This is a plot of the solutions. The solutions all blow up in time due to positive exponential terms.