Question

Resolva simultaneamente o SL Ax = b nos seguintes casos :


a) $A = \left[\begin{array}{ccc}1&2&3\\2&5&3\\1&0&8\end{array}\right], b = \left[\begin{array}{ccc}1\\0\\0\end{array}\right] , b = \left[\begin{array}{ccc}0\\1\\0\end{array}\right] e \ b = \left[\begin{array}{ccc}0\\0\\1\end{array}\right]$




b) $A = \left[\begin{array}{ccc}1&6&4\\2&4&1\\-1&2&5\end{array}\right] , b = \left[\begin{array}{ccc}1\\0\\0\end{array}\right] , b = \left[\begin{array}{ccc}0\\1\\0\end{array}\right] e \ b = \left[\begin{array}{ccc}0\\0\\1\end{array}\right]$

Answer 1

Temos diversas maneiras de resolver sistemas lineares, uma delas é o Método da Matriz Inversa, isso é, dado um sistema AX = B, invertemos a matriz A e multiplicamos dos dois lados, ou seja

                                      $\Large\displaystyle\begin{gathered}AX = B\Rightarrow X = A^{-1}B\\ \\\end{gathered}$

Logo vamos achar a matriz inversa, por eliminação de Gauss temos:

                                          $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\2 & 5 & 3 & 0 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\\\end{array}\right]\end{gathered}$

Agora, através de operações básicas

          $\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\2 & 5 & 3 & 0 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\\\end{array}\right]\xrightarrow{L_2 - 2 L_1 \rightarrow L_2}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\\\end{array}\right]\end{gathered}$

    $\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\1 & 0 & 8 & 0 & 0 & 1\\\end{array}\right]\xrightarrow{L_3 - L_1 \rightarrow L_3}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\0 & -2 & 5 & -1 & 0 & 1\\\end{array}\right]\end{gathered}$  

$\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\0 & -2 & 5 & -1 & 0 & 1\\\end{array}\right]\xrightarrow{2L_2 - L_3 \rightarrow L_3}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\end{gathered}$

$\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & -3 & -2 & 1 & 0\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\xrightarrow{L_2 + 3L_3 \rightarrow L_2}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\end{gathered}$

 $\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 3 & 1 & 0 & 0\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\xrightarrow{L_1 - 3L_3 \rightarrow L_1}\left[\begin{array}{c c c | c c c}1 & 2 & 0 & -14 & 6 & 3\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\end{gathered}$

$\large\displaystyle\begin{gathered}\left[\begin{array}{c c c | c c c}1 & 2 & 0 & -14 & 6 & 3\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\xrightarrow{L_1 - 2L_2 \rightarrow L_1}\left[\begin{array}{c c c | c c c}1 & 0 & 0 & -40 & 16 & 9\\0 & 1 & 0 & 13 & -5 & -3\\0 & 0 & 1 & 5 & -2 & -1\\\end{array}\right]\end{gathered}$

Portanto a matriz inversa é:

                         $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c}1 & 2 & 3\\2 & 5 & 3\\1 & 0 & 8\\\end{array}\right]^{-1} = \left[\begin{array}{c c c}-40 & 16 & 9\\13 & -5 & -3\\5 & -2 & -1\\\end{array}\right]\end{gathered}$

Faça o mesmo processo para a matriz do item b e você deve encontrar

                      $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c}1 & 6 & 4\\2 & 4 & 1\\-1 & 2 & 5\\\end{array}\right]^{-1} = \left[\begin{array}{c c c}-\frac{9}{8} & \frac{11}{8} & \frac{5}{8} \\ \\\frac{11}{16} & -\frac{9}{16} & -\frac{7}{16} \\ \\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{array}\right]\end{gathered}$

Agora que temos todas as matrizes inversas, podemos aplicar simplesmente

                                  $\Large\displaystyle\begin{gathered}AX = B\Rightarrow X = A^{-1}B\\ \\\end{gathered}$

E achar as soluções.

Os resultados do sistema para os 3b's diferentes já estão na matriz inversa, pois

                       $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\\\end{array}\right] \left[\begin{array}{c}1 \\0 \\0 \\\end{array}\right] = \left[\begin{array}{c}a_{11} \\a_{21} \\a_{31} \\\end{array}\right]\end{gathered}$

De maneira análoga

                       $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\\\end{array}\right] \left[\begin{array}{c}0 \\1 \\0 \\\end{array}\right] = \left[\begin{array}{c}a_{12} \\a_{22} \\a_{32} \\\end{array}\right]\end{gathered}$

Por fim

                       $\Large\displaystyle\begin{gathered}\left[\begin{array}{c c c}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\\\end{array}\right] \left[\begin{array}{c}0 \\0 \\1 \\\end{array}\right] = \left[\begin{array}{c}a_{13} \\a_{23} \\a_{33} \\\end{array}\right]\end{gathered}$

Para achar as soluções basta olhar as colunas da matriz inversa.

                                                             

Espero ter ajudado

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