Dada a matriz
A=\left(\begin{array}{cccc} 3 & -1 & 5 & 0 \\\\ 0 & 2 & 0 & 1 \\\\ 2 & 0 & -1 & 3 \\\\ 1 & 1 & 2 & 0 \end{array}\right),
calcule A^{-1}.

Passo 1

O enunciado fornece a matriz
A=\begin{pmatrix} 3 & -1 & 5 & 0\\\\ 0 & 2 & 0 & 1\\\\ 2 & 0 & -1 & 3\\\\ 1 & 1 & 2 & 0 \end{pmatrix}
e pede A^{-1}. Vamos usar o método de Gauss–Jordan: montamos a matriz aumentada \bigl[A\mid I_4\bigr] e aplicamos operações de linha para obter \bigl[I_4\mid A^{-1}\bigr].

Passo 2

O pivô inicial é a_{11}=3. Eliminamos as entradas abaixo de a_{11} com
R_3\to R_3 - \tfrac{2}{3}R_1,\quad R_4\to R_4 - \tfrac{1}{3}R_1.
Fica
\left[ \begin{array}{cccc|cccc} 3 & -1 & 5 & 0 & 1 & 0 & 0 & 0\\\\ 0 & 2 & 0 & 1 & 0 & 1 & 0 & 0\\\\ 0 & \tfrac{2}{3} & -\tfrac{13}{3} & 3 & -\tfrac{2}{3} & 0 & 1 & 0\\\\ 0 & \tfrac{4}{3} & \tfrac{1}{3} & 0 & -\tfrac{1}{3} & 0 & 0 & 1 \end{array} \right].

Passo 3

O pivô na coluna 2 é a_{22}=2. Eliminamos as entradas em coluna 2 usando
R_1\to R_1 + \tfrac12R_2,\quad R_3\to R_3 - \tfrac{1}{3}R_2,\quad R_4\to R_4 - \tfrac{2}{3}R_2.
O resultado é
\left[ \begin{array}{cccc|cccc} 3 & 0 & 5 & \tfrac12 & 1 & \tfrac12 & 0 & 0\\\\ 0 & 2 & 0 & 1 & 0 & 1 & 0 & 0\\\\ 0 & 0 & -\tfrac{13}{3} & \tfrac{8}{3} & -\tfrac{2}{3} & -\tfrac{1}{3} & 1 & 0\\\\ 0 & 0 & \tfrac{1}{3} & -\tfrac{2}{3} & -\tfrac{1}{3} & -\tfrac{2}{3} & 0 & 1 \end{array} \right].

Passo 4

O pivô na coluna 3 é -\tfrac{13}{3}. Eliminamos as outras entradas em coluna 3 com
R_1\to R_1 + \tfrac{15}{13}R_3,\quad R_4\to R_4 + \tfrac{1}{13}R_3.
Isso leva a
\left[ \begin{array}{cccc|cccc} 3 & 0 & 0 & \tfrac{93}{26} & \tfrac{3}{13} & \tfrac{3}{26} & \tfrac{15}{13} & 0\\\\ 0 & 2 & 0 & 1 & 0 & 1 & 0 & 0\\\\ 0 & 0 & -\tfrac{13}{3} & \tfrac{8}{3} & -\tfrac{2}{3} & -\tfrac{1}{3} & 1 & 0\\\\ 0 & 0 & 0 & -\tfrac{6}{13} & -\tfrac{5}{13} & -\tfrac{9}{13} & \tfrac{1}{13} & 1 \end{array} \right].

Passo 5

O pivô na coluna 4 é -\tfrac{6}{13}. Eliminamos as entradas acima com
R_1\to R_1 + \tfrac{31}{4}R_4,\quad R_2\to R_2 + \tfrac{13}{6}R_4,\quad R_3\to R_3 + \tfrac{52}{9}R_4.
Fica
\left[ \begin{array}{cccc|cccc} 3 & 0 & 0 & 0 & -\tfrac{143}{52} & -\tfrac{273}{52} & \tfrac{91}{52} & \tfrac{31}{4}\\\\ 0 & 2 & 0 & 0 & -\tfrac{5}{6} & -\tfrac{1}{2} & \tfrac{1}{6} & \tfrac{13}{6}\\\\ 0 & 0 & -\tfrac{13}{3} & 0 & -\tfrac{26}{9} & -\tfrac{13}{3} & \tfrac{13}{9} & \tfrac{52}{9}\\\\ 0 & 0 & 0 & -\tfrac{6}{13} & -\tfrac{5}{13} & -\tfrac{9}{13} & \tfrac{1}{13} & 1 \end{array} \right].

Passo 6

Agora basta normalizar cada linha para que o bloco esquerdo vire I_4. Dividindo
R_1\to \tfrac{1}{3}R_1,\quad R_2\to \tfrac{1}{2}R_2,\quad R_3\to -\tfrac{3}{13}R_3,\quad R_4\to -\tfrac{13}{6}R_4
obtemos
\bigl[I_4\mid A^{-1}\bigr] =\left[ \begin{array}{cccc|cccc} 1 & 0 & 0 & 0 & -\tfrac{11}{12} & -\tfrac{7}{4} & \tfrac{7}{12} & \tfrac{31}{12}\\\\ 0 & 1 & 0 & 0 & -\tfrac{5}{12} & -\tfrac{1}{4} & \tfrac{1}{12} & \tfrac{13}{12}\\\\ 0 & 0 & 1 & 0 & \tfrac{2}{3} & 1 & -\tfrac{1}{3} & -\tfrac{4}{3}\\\\ 0 & 0 & 0 & 1 & \tfrac{5}{6} & \tfrac{3}{2} & -\tfrac{1}{6} & -\tfrac{13}{6} \end{array} \right].

Resposta

A^{-1}=\begin{pmatrix} -\tfrac{11}{12} & -\tfrac{7}{4} & \tfrac{7}{12} & \tfrac{31}{12}\\[6pt] -\tfrac{5}{12} & -\tfrac{1}{4} & \tfrac{1}{12} & \tfrac{13}{12}\\[6pt] \tfrac{2}{3} & 1 & -\tfrac{1}{3} & -\tfrac{4}{3}\\[6pt] \tfrac{5}{6} & \tfrac{3}{2} & -\tfrac{1}{6} & -\tfrac{13}{6} \end{pmatrix}.]

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