Nos Exercícios de 48 a 63, use o método de Gauss-Jordall para achar a inversa da matriz dada (se existir).
\left[\begin{array}{rrrr}0 & -1 & 1 & 0 \\\\ 2 & 1 & 0 & 2 \\\\ 1 & -1 & 3 & 0 \\\\ 0 & 1 & 1 & -1\end{array}\right]

Passo 1

O enunciado pede a inversa de
A = \begin{pmatrix} 0 & -1 & 1 & 0\\\\ 2 & 1 & 0 & 2\\\\ 1 & -1 & 3 & 0\\\\ 0 & 1 & 1 & -1 \end{pmatrix}
pelo método de Gauss–Jordan. Vamos montar a matriz aumentada \bigl[A\mid I_4\bigr] e aplicar operações elementares até obter \bigl[I_4\mid A^{-1}\bigr].

Passo 2

Escrevemos a matriz aumentada inicial e trocamos a linha 1 com a linha 2 para ter pivô não nulo em (1,1):
\left[\begin{array}{rrrr|rrrr} 0 & -1 & 1 & 0 & 1 & 0 & 0 & 0\\\\ 2 & 1 & 0 & 2 & 0 & 1 & 0 & 0\\\\ 1 & -1 & 3 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 1 & 1 & -1 & 0 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{R_1\leftrightarrow R_2}\; \left[\begin{array}{rrrr|rrrr} 2 & 1 & 0 & 2 & 0 & 1 & 0 & 0\\\\ 0 & -1 & 1 & 0 & 1 & 0 & 0 & 0\\\\ 1 & -1 & 3 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 1 & 1 & -1 & 0 & 0 & 0 & 1 \end{array}\right].

Passo 3

Normalizamos o pivô em (1,1) e eliminamos a coluna 1 nas outras linhas:
R_1 \gets \tfrac12\,R_1 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & \tfrac12 & 0 & 1 & \tfrac12 & 0 & 0 & 0\\\\ 0 & -1 & 1 & 0 & 1 & 0 & 0 & 0\\\\ 1 & -1 & 3 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 1 & 1 & -1 & 0 & 0 & 0 & 1 \end{array}\right]
R_3\gets R_3 - R_1, \quad R_2,R_4\text{ já têm zero em (2,1) e (4,1)} \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & \tfrac12 & 0 & 1 & \tfrac12 & 0 & 0 & 0\\\\ 0 & -1 & 1 & 0 & 1 & 0 & 0 & 0\\\\ 0 & -\tfrac32 & 3 & -1 & -\tfrac12 & 0 & 1 & 0\\\\ 0 & 1 & 1 & -1 & 0 & 0 & 0 & 1 \end{array}\right].áê

Passo 4

Normalizamos o pivô em (2,2) e eliminamos coluna 2:
R_2\gets -1\,R_2 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & \tfrac12 & 0 & 1 & \tfrac12 & 0 & 0 & 0\\\\ 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0\\\\ 0 & -\tfrac32 & 3 & -1 & -\tfrac12 & 0 & 1 & 0\\\\ 0 & 1 & 1 & -1 & 0 & 0 & 0 & 1 \end{array}\right]
R_1\gets R_1 - \tfrac12R_2,\; R_3\gets R_3 + \tfrac32R_2,\; R_4\gets R_4 - R_2 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & \tfrac12 & 1 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0\\\\ 0 & 0 & \tfrac32 & -1 & -\tfrac12 & -\tfrac32 & 1 & 0\\\\ 0 & 0 & 2 & -1 & 0 & 1 & 0 & 1 \end{array}\right].

Passo 5

Para o pivô de coluna 3 trocamos as linhas 3 e 4 (para usar 2 em vez de \tfrac32), normalizamos e eliminamos coluna 3:
R_3\leftrightarrow R_4 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & \tfrac12 & 1 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0\\\\ 0 & 0 & 2 & -1 & 0 & 1 & 0 & 1\\\\ 0 & 0 & \tfrac32 & -1 & -\tfrac12 & -\tfrac32 & 1 & 0 \end{array}\right]
R_3\gets \tfrac12R_3 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & \tfrac12 & 1 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & 1 & -1 & 0 & 0 & -1 & 0 & 0\\\\ 0 & 0 & 1 & -\tfrac12 & 0 & \tfrac12 & 0 & \tfrac12\\\\ 0 & 0 & \tfrac32 & -1 & -\tfrac12 & -\tfrac32 & 1 & 0 \end{array}\right]
R_1\gets R_1 - \tfrac12R_3,\; R_2\gets R_2 + R_3,\; R_4\gets R_4 - \tfrac32R_3 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & 0 & \tfrac54 & -2 & -11 & 5 & -4\\\\ 0 & 1 & 0 & -\tfrac12 & 1 & 4 & -2 & 2\\\\ 0 & 0 & 1 & -\tfrac12 & 0 & \tfrac12 & 0 & \tfrac12\\\\ 0 & 0 & 0 & -\tfrac14 & -\tfrac12 & -\tfrac94 & 1 & -\tfrac34 \end{array}\right].

Passo 6

Finalmente normalizamos o pivô em (4,4) e zeramos a coluna 4:
R_4\gets -4\,R_4 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & 0 & \tfrac54 & -2 & -11 & 5 & -4\\\\ 0 & 1 & 0 & -\tfrac12 & 1 & 4 & -2 & 2\\\\ 0 & 0 & 1 & -\tfrac12 & 0 & \tfrac12 & 0 & \tfrac12\\\\ 0 & 0 & 0 & 1 & 2 & 9 & -4 & 3 \end{array}\right]
R_1\gets R_1 - \tfrac54R_4,\; R_2\gets R_2 + \tfrac12R_4,\; R_3\gets R_3 + \tfrac12R_4 \;\Longrightarrow\; \left[\begin{array}{rrrr|rrrr} 1 & 0 & 0 & 0 & -2 & -11 & 5 & -4\\\\ 0 & 1 & 0 & 0 & 1 & 4 & -2 & 2\\\\ 0 & 0 & 1 & 0 & 1 & 5 & -2 & 2\\\\ 0 & 0 & 0 & 1 & 2 & 9 & -4 & 3 \end{array}\right].
Agora a parte esquerda é I_4 e a direita é A^{-1}.

Resposta

A^{-1} = \begin{pmatrix} -2 & -11 & 5 & -4\\\\ 1 & 4 & -2 & 2\\\\ 1 & 5 & -2 & 2\\\\ 2 & 9 & -4 & 3 \end{pmatrix}.

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