Encontre as inversas das matrizes dadas, se possível.
\left[\begin{array}{rrrr}1 & 1 & 1 & 1 \\\\ 1 & 2 & -1 & 2 \\\\ 1 & -1 & 2 & 1 \\\\ 1 & 3 & 3 & 2\end{array}\right]

Passo 1

O enunciado pede para encontrar a inversa da matriz
M = \begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & 2 & -1 & 2\\\\ 1 & -1 & 2 & 1\\\\ 1 & 3 & 3 & 2 \end{pmatrix}.
O conteúdo usado será eliminação de Gauss–Jordan, montando a matriz aumentada (M\,|\,I) e transformando M em I.

Passo 2

Formamos a matriz aumentada e zeramos abaixo do pivô em M na coluna 1, subtraindo a primeira linha das outras três:
\left[\begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0\\\\ 1 & 2 & -1 & 2 & 0 & 1 & 0 & 0\\\\ 1 & -1 & 2 & 1 & 0 & 0 & 1 & 0\\\\ 1 & 3 & 3 & 2 & 0 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{R_i\gets R_i - R_1 (i=2,3,4)}\; \left[\begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0\\\\ 0 & 1 & -2 & 1 & -1 & 1 & 0 & 0\\\\ 0 & -2 & 1 & 0 & -1 & 0 & 1 & 0\\\\ 0 & 2 & 2 & 1 & -1 & 0 & 0 & 1 \end{array}\right].
Em seguida, usamos a linha 2 como pivô da coluna 2 para zerar as entradas embaixo:
R_3\gets R_3 + 2\,R_2,\quad R_4\gets R_4 - 2\,R_2 \;\longrightarrow\; \left[\begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0\\\\ 0 & 1 & -2 & 1 & -1 & 1 & 0 & 0\\\\ 0 & 0 & -3 & 2 & -3 & 2 & 1 & 0\\\\ 0 & 0 & 6 & -1 & 1 & -2 & 0 & 1 \end{array}\right].

Passo 3

Agora pivotamos na coluna 3 usando a linha 3. Primeiro dividimos R_3 por -3:
R_3\gets -\tfrac13 R_3 \;\longrightarrow\; \left[\begin{array}{cccc|cccc} 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0\\\\ 0 & 1 & -2 & 1 & -1 & 1 & 0 & 0\\\\ 0 & 0 & 1 & -\tfrac23 & 1 & -\tfrac23 & -\tfrac13 & 0\\\\ 0 & 0 & 6 & -1 & 1 & -2 & 0 & 1 \end{array}\right].
Em seguida, zeramos as colunas 3 em R_1,R_2,R_4:
R_1\gets R_1 - R_3,\quad R_2\gets R_2 + 2\,R_3,\quad R_4\gets R_4 - 6\,R_3,
resultando em
\left[\begin{array}{cccc|cccc} 1 & 1 & 0 & \tfrac{5}{3} & 0 & \tfrac{2}{3} & \tfrac{1}{3} & 0\\\\ 0 & 1 & 0 & -\tfrac{1}{3} & 1 & -\tfrac{1}{3} & -\tfrac{2}{3} & 0\\\\ 0 & 0 & 1 & -\tfrac{2}{3} & 1 & -\tfrac{2}{3} & -\tfrac{1}{3} & 0\\\\ 0 & 0 & 0 & 3 & -5 & 2 & 2 & 1 \end{array}\right].

Passo 4

Pivotamos na coluna 4 com R_4, dividindo-a por 3, e então zeramos acima:
R_4\gets \tfrac{1}{3}R_4,\quad R_1\gets R_1 - \tfrac{5}{3}R_4,\quad R_2\gets R_2 + \tfrac{1}{3}R_4,\quad R_3\gets R_3 + \tfrac{2}{3}R_4.
Isso leva a
\left[\begin{array}{cccc|cccc} 1 & 1 & 0 & 0 & \tfrac{25}{9} & -\tfrac{4}{9} & -\tfrac{7}{9} & -\tfrac{5}{9}\\\\ 0 & 1 & 0 & 0 & \tfrac{4}{9} & -\tfrac{1}{9} & -\tfrac{4}{9} & \tfrac{1}{9}\\\\ 0 & 0 & 1 & 0 & -\tfrac{1}{9} & -\tfrac{2}{9} & \tfrac{1}{9} & \tfrac{2}{9}\\\\ 0 & 0 & 0 & 1 & -\tfrac{5}{3} & \tfrac{2}{3} & \tfrac{2}{3} & \tfrac{1}{3} \end{array}\right].

Passo 5

Por fim, zeramos a entrada em (1,2) subtraindo R_2 de R_1:
R_1\gets R_1 - R_2 \;\longrightarrow\; \left[\begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & \tfrac{7}{3} & -\tfrac{1}{3} & -\tfrac{1}{3} & -\tfrac{2}{3}\\\\ 0 & 1 & 0 & 0 & \tfrac{4}{9} & -\tfrac{1}{9} & -\tfrac{4}{9} & \tfrac{1}{9}\\\\ 0 & 0 & 1 & 0 & -\tfrac{1}{9} & -\tfrac{2}{9} & \tfrac{1}{9} & \tfrac{2}{9}\\\\ 0 & 0 & 0 & 1 & -\tfrac{5}{3} & \tfrac{2}{3} & \tfrac{2}{3} & \tfrac{1}{3} \end{array}\right].
Agora M virou I e o bloco à direita é M^{-1}.

Resposta

M^{-1} = \begin{pmatrix} \frac{7}{3} & -\frac{1}{3} & -\frac{1}{3} & -\frac{2}{3}\\[6pt] \frac{4}{9} & -\frac{1}{9} & -\frac{4}{9} & \frac{1}{9}\\[6pt] -\frac{1}{9} & -\frac{2}{9} & \frac{1}{9} & \frac{2}{9}\\[6pt] -\frac{5}{3} & \frac{2}{3} & \frac{2}{3} & \frac{1}{3} \end{pmatrix}.]

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