Use o método de Gauss-Jordan para encontrar A^{-1}, onde A=\left(\begin{array}{rrr}2 & -1 & 1 \\\\ 1 & -2 & 1 \\\\ 3 & 1 & 0\end{array}\right).

Passo 1

O enunciado pede o inverso da matriz
A = \begin{pmatrix} 2 & -1 & 1\\\\ 1 & -2 & 1\\\\ 3 & 1 & 0 \end{pmatrix}
usando o método de Gauss–Jordan. A ideia é montar a matriz aumentada [A\mid I] e aplicar operações elementares de linha até obter [I\mid A^{-1}].

Passo 2

Formamos a matriz aumentada e usamos o pivô na posição (1,1). Primeiro dividimos a linha 1 por 2 para obter 1 na posição (1,1), depois zeramos as entradas abaixo:
\left[\begin{array}{ccc|ccc} 2 & -1 & 1 & 1 & 0 & 0\\\\ 1 & -2 & 1 & 0 & 1 & 0\\\\ 3 & 1 & 0 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{L_1\leftarrow \tfrac12L_1}\; \left[\begin{array}{ccc|ccc} 1 & -\tfrac12 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 1 & -2 & 1 & 0 & 1 & 0\\\\ 3 & 1 & 0 & 0 & 0 & 1 \end{array}\right]
\xrightarrow{\substack{L_2\leftarrow L_2 - L_1\L_3\leftarrow L_3 - 3L_1}} \left[\begin{array}{ccc|ccc} 1 & -\tfrac12 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & -\tfrac32 & \tfrac12 & -\tfrac12 & 1 & 0\\\\ 0 & \tfrac52 & -\tfrac32 & -\tfrac32 & 0 & 1 \end{array}\right]

Passo 3

Agora escolhemos o pivô na posição (2,2). Dividimos a linha 2 por -\tfrac32 para obter 1, e zeramos as entradas acima e abaixo na coluna 2:
\left[\begin{array}{ccc|ccc} 1 & -\tfrac12 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & -\tfrac32 & \tfrac12 & -\tfrac12 & 1 & 0\\\\ 0 & \tfrac52 & -\tfrac32 & -\tfrac32 & 0 & 1 \end{array}\right] \;\xrightarrow{L_2\leftarrow -\tfrac23\,L_2}\; \left[\begin{array}{ccc|ccc} 1 & -\tfrac12 & \tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & 1 & -\tfrac13 & \tfrac13 & -\tfrac23 & 0\\\\ 0 & \tfrac52 & -\tfrac32 & -\tfrac32 & 0 & 1 \end{array}\right]
\xrightarrow{\substack{L_1\leftarrow L_1 + \tfrac12L_2\L_3\leftarrow L_3 - \tfrac52L_2}} \left[\begin{array}{ccc|ccc} 1 & 0 & \tfrac13 & \tfrac23 & -\tfrac13 & 0\\\\ 0 & 1 & -\tfrac13 & \tfrac13 & -\tfrac23 & 0\\\\ 0 & 0 & -\tfrac23 & -\tfrac{7}{3} & \tfrac53 & 1 \end{array}\right]

Passo 4

Por fim, usamos o pivô na posição (3,3). Dividimos a linha 3 por -\tfrac23 para obter 1, e zeramos as entradas acima na coluna 3:
\left[\begin{array}{ccc|ccc} 1 & 0 & \tfrac13 & \tfrac23 & -\tfrac13 & 0\\\\ 0 & 1 & -\tfrac13 & \tfrac13 & -\tfrac23 & 0\\\\ 0 & 0 & -\tfrac23 & -\tfrac{7}{3} & \tfrac53 & 1 \end{array}\right] \;\xrightarrow{L_3\leftarrow -\tfrac32\,L_3}\; \left[\begin{array}{ccc|ccc} 1 & 0 & \tfrac13 & \tfrac23 & -\tfrac13 & 0\\\\ 0 & 1 & -\tfrac13 & \tfrac13 & -\tfrac23 & 0\\\\ 0 & 0 & 1 & \tfrac72 & -\tfrac52 & -\tfrac32 \end{array}\right]
\xrightarrow{\substack{L_1\leftarrow L_1 - \tfrac13L_3\L_2\leftarrow L_2 + \tfrac13L_3}} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -\tfrac12 & \tfrac12 & \tfrac12\\\\ 0 & 1 & 0 & \tfrac32 & -\tfrac32 & -\tfrac12\\\\ 0 & 0 & 1 & \tfrac72 & -\tfrac52 & -\tfrac32 \end{array}\right]

Resposta

A inversa de A é
A^{-1} = \begin{pmatrix} -\tfrac12 & \tfrac12 & \tfrac12\\\ \tfrac32 & -\tfrac32 & -\tfrac12\\\ \tfrac72 & -\tfrac52 & -\tfrac32 \end{pmatrix}.

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