A matriz \Phi(t) cujas colunas são as soluções linearmente independentes do sistema X^{\prime}(t)=A X(t) se denomina Matriz Fundamental, isto é:
\Phi(t)=\left[X_{1}(t), X_{2}(t), \ldots, X_{n}(t)\right]=\left[\begin{array}{cccc} x_{11}(t) & x_{12}(t) & \ldots & x_{1 n}(t) \\\ x_{21}(t) & x_{22}(t) & \ldots & x_{2 n}(t) \\\ \vdots & \vdots & \vdots & \vdots \\\ x_{n 1}(t) & x_{n 2}(t) & \ldots & x_{n n}(t) \end{array}\right]
Observe que a solução geral do sistema homogêneo pode-se escrever usando esta matriz, na seguinte forma:
Z(t)=\Phi(t) C, \quad \text { onde } \quad C=\left(c_{1}, c_{2}, \ldots, c_{n}\right)^{T} .
Verifique que as matrizes \Phi_{1}(t)=\left[\begin{array}{cc}e^{2 t} & t e^{2 t} \\\\ -e^{2 t} & -t e^{2 t}+e^{2 t}\end{array}\right] e \Phi_{2}(t)=\left[\begin{array}{cc}e^{2 t}+t e^{2 t} & t e^{2 t} \\\\ -t e^{2 t} & e^{2 t}-t e^{2 t}\end{array}\right] são matrizes fundamentais do sistema X^{\prime}(t)=\left[\begin{array}{cc}3 & 1 \\\\ -1 & 1\end{array}\right] X(t).

Passo 1

O enunciado pede para mostrar que cada matriz \Phi_i(t) satisfaz a EDO matricial
\Phi_i'(t)=A\,\Phi_i(t)
com A=\begin{pmatrix}3&1\\\ -1&1\end{pmatrix} e que \det\Phi_i(t)\neq0 (isto garante que as colunas são LI). Isso faz de \Phi_i(t) uma matriz fundamental.

Passo 2

Para \Phi_{1}(t)=\begin{pmatrix}e^{2t}&t\,e^{2t}\\\ -e^{2t}&-t\,e^{2t}+e^{2t}\end{pmatrix}:\n
  1. Calculemos \Phi_{1}'(t) elemento a elemento:
\Phi_{1}'(t) =\begin{pmatrix} ( e^{2t} )'&(t\,e^{2t})'\\\ (-e^{2t})'&(-t\,e^{2t}+e^{2t})' \end{pmatrix} =\begin{pmatrix} 2e^{2t}&e^{2t}+2t\,e^{2t}\\\\ -2e^{2t}&\bigl(e^{2t}-t\,e^{2t}\bigr)' \end{pmatrix} =\begin{pmatrix} 2e^{2t}&(1+2t)e^{2t}\\\\ -2e^{2t}&(1-2t)e^{2t} \end{pmatrix}.
  1. Multipliquemos A\,\Phi_{1}(t):
A\,\Phi_{1}(t) =\begin{pmatrix}3&1\\\ -1&1\end{pmatrix} \begin{pmatrix}e^{2t}&t\,e^{2t}\\\ -e^{2t}&-t\,e^{2t}+e^{2t}\end{pmatrix} =\begin{pmatrix} 3e^{2t}-e^{2t}&3t\,e^{2t}+(-t\,e^{2t}+e^{2t})\\\\ - e^{2t}-e^{2t}&-t\,e^{2t}+(-t\,e^{2t}+e^{2t}) \end{pmatrix} =\begin{pmatrix} 2e^{2t}&(1+2t)e^{2t}\\\\ -2e^{2t}&(1-2t)e^{2t} \end{pmatrix}.
Vê-se que \Phi_{1}'(t)=A\,\Phi_{1}(t).\n
3. O determinante é
\det\Phi_{1}(t) =\begin{vmatrix}e^{2t}&t\,e^{2t}\\\ -e^{2t}&e^{2t}-t\,e^{2t}\end{vmatrix} =e^{2t}(e^{2t}-t\,e^{2t})-t\,e^{2t}(-e^{2t}) =e^{4t}\neq0.

Passo 3

Para \Phi_{2}(t)=\begin{pmatrix}e^{2t}+t\,e^{2t}&t\,e^{2t}\\\ -t\,e^{2t}&e^{2t}-t\,e^{2t}\end{pmatrix}:\n
  1. Derivando:
\Phi_{2}'(t) =\begin{pmatrix} ( e^{2t}+t\,e^{2t} )'&(t\,e^{2t})'\\\ (-t\,e^{2t})'&(e^{2t}-t\,e^{2t})' \end{pmatrix} =\begin{pmatrix} 3e^{2t}+2t\,e^{2t}&(1+2t)e^{2t}\\\\ -(1+2t)e^{2t}&(1-2t)e^{2t} \end{pmatrix}.
  1. Multiplicando A\,\Phi_{2}(t):
A\,\Phi_{2}(t) =\begin{pmatrix}3&1\\\ -1&1\end{pmatrix} \begin{pmatrix}e^{2t}+t\,e^{2t}&t\,e^{2t}\\\ -t\,e^{2t}&e^{2t}-t\,e^{2t}\end{pmatrix} =\begin{pmatrix} 3(e^{2t}+t\,e^{2t})-t\,e^{2t}&3t\,e^{2t}+(e^{2t}-t\,e^{2t})\\\\ -(e^{2t}+t\,e^{2t})-t\,e^{2t}&-t\,e^{2t}+(e^{2t}-t\,e^{2t}) \end{pmatrix}
=\begin{pmatrix} 3e^{2t}+2t\,e^{2t}&(1+2t)e^{2t}\\\\ -(1+2t)e^{2t}&(1-2t)e^{2t} \end{pmatrix}
logo \Phi_{2}'(t)=A\,\Phi_{2}(t).\n
3. O determinante:
\det\Phi_{2}(t) =\begin{vmatrix}e^{2t}(1+t)&t\,e^{2t}\\\ -t\,e^{2t}&e^{2t}(1-t)\end{vmatrix} =e^{4t}\bigl((1+t)(1-t)+t^2\bigr)=e^{4t}\neq0.

Resposta

Ambas \Phi_{1}(t) e \Phi_{2}(t) satisfazem \Phi'(t)=A\,\Phi(t) e são inversíveis em todo t, logo são matrizes fundamentais do sistema.

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