Consider the system
$$
x^{\prime}=A x=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1}
\\\ {-3} & {2} & {4}\end{array}\right) x
$$
(a) Show that \(r=2\) is an eigenvalue of multiplicity 3 of the coefficient
matrix \(\mathbf{A}\) and that there is only one corresponding cigenvector,
namely,
$$
\xi^{(1)}=\left(\begin{array}{r}{0} \\ {1} \\ {-1}\end{array}\right)
$$
(b) Using the information in part (a), write down one solution
\(\mathbf{x}^{(1)}(t)\) of the system (i). There is no other solution of the
purely exponential form \(\mathbf{x}=\xi e^{y t}\).
(c) To find a second solution assume that \(\mathbf{x}=\xi t e^{2
t}+\mathbf{\eta} e^{2 t} .\) Show that \(\xi\) and \(\mathbf{\eta}\) satisfy the
equations
$$
(\mathbf{A}-2 \mathbf{I}) \xi=\mathbf{0}, \quad(\mathbf{A}-2 \mathbf{I})
\mathbf{n}=\mathbf{\xi}
$$
since \(\xi\) has already been found in part (a), solve the second equation for
\(\eta\). Neglect the multiple of \(\xi^{(1)}\) that appears in \(\eta\), since it
leads only to a multiple of the first solution \(\mathbf{x}^{(1)}\). Then write
down a second solution \(\mathbf{x}^{(2)}(t)\) of the system (i).
(d) To find a third solution assume that \(\mathbf{x}=\xi\left(t^{2} / 2\right)
e^{2 t}+\mathbf{\eta} t e^{2 t}+\zeta e^{2 t} .\) Show that \(\xi, \eta,\) and
\(\zeta\) satisfy the equations
$$
(\mathbf{A}-2 \mathbf{l}) \xi=\mathbf{0}, \quad(\mathbf{\Lambda}-2 \mathbf{I})
\mathbf{\eta}=\mathbf{\xi}, \quad(\mathbf{A}-2 \mathbf{l}) \zeta=\mathbf{\eta}
$$
The first two equations are the same as in part (c), so solve the third
equation for \(\zeta,\) again neglecting the multiple of \(\xi^{(1)}\) that
appears. Then write down a third solution \(\mathbf{x}^{(3)}(t)\) of the system
(i).
(e) Write down a fundamental matrix \(\boldsymbol{\Psi}(t)\) for the system (i).
(f) Form a matrix \(\mathbf{T}\) with the cigenvector \(\xi^{(1)}\) in the first
column, and the generalized eigenvectors \(\eta\) and \(\zeta\) in the second and
third columns. Then find \(T^{-1}\) and form the product
\(\mathbf{J}=\mathbf{T}^{-1} \mathbf{A} \mathbf{T}\). The matrix \(\mathbf{J}\) is
the Jordan form of \(\mathbf{A}\).
