Use o método de Gauss-Jordan para encontrar a matriz inversa, se existir, das seguintes matrizes.
A=\left(\begin{array}{ccc} 2 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 1 & 0 & 1 \end{array}\right), B=\left(\begin{array}{ccc} 2 & 1 & -1 \\\\ 1 & 0 & 2 \\\\ 4 & -1 & 3 \end{array}\right) C=\left(\begin{array}{ccc} -2 & 1 & -1 \\\\ 3 & 1 & 2 \\\\ 1 & -2 & 0 \end{array}\right), D=\left(\begin{array}{ccc} 2 & -1 & 1 \\\\ -3 & 2 & -2 \\\\ -2 & 1 & 0 \end{array}\right)

Passo 1

O enunciado pede: para cada uma das matrizes abaixo, usar o método de Gauss–Jordan e encontrar a matriz inversa, se existir:
A=\begin{pmatrix} 2 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 1 & 0 & 1 \end{pmatrix},\quad B=\begin{pmatrix} 2 & 1 & -1 \\\\ 1 & 0 & 2 \\\\ 4 & -1 & 3 \end{pmatrix},\quad C=\begin{pmatrix} -2 & 1 & -1 \\\\ 3 & 1 & 2 \\\\ 1 & -2 & 0 \end{pmatrix},\quad D=\begin{pmatrix} 2 & -1 & 1 \\\\ -3 & 2 & -2 \\\\ -2 & 1 & 0 \end{pmatrix}.
Informações relevantes dadas:
  • Cada matriz é 3\times3.
  • O objetivo é, para cada matriz M dentre A,B,C,D, encontrar M^{-1} usando Gauss–Jordan ou dizer que não existe caso não seja invertível.

Passo 2

Pequena explicação do método (passo-a-passo mental antes de executar): em Gauss–Jordan trabalhamos com a matriz aumentada
\bigl[\,M\mid I\,\bigr],
onde I é a identidade 3\times3. Aplicam-se operações elementares de linha até reduzir a parte esquerda a I. As operações elementares permitidas são:
  • trocar duas linhas: \,R_i\leftrightarrow R_j,
  • multiplicar uma linha por um escalar não nulo: \,R_i\leftarrow \lambda R_i,
  • somar a uma linha um múltiplo de outra: \,R_i\leftarrow R_i + \mu R_j.
    Quando a parte esquerda fica I, a parte direita será M^{-1}. Vou aplicar essas operações mostrando cada etapa e explicando as contas.

Passo 3

Começando por A. Monto a matriz aumentada:
\bigl[\,A\mid I\,\bigr]= \left[\begin{array}{ccc|ccc} 2 & 1 & 0 & 1 & 0 & 0\\\\ 1 & 1 & 1 & 0 & 1 & 0\\\\ 1 & 0 & 1 & 0 & 0 & 1 \end{array}\right].
Faço \,R_1\leftrightarrow R_2 para ter pivô 1 na primeira linha:
\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 1 & 0\\\\ 2 & 1 & 0 & 1 & 0 & 0\\\\ 1 & 0 & 1 & 0 & 0 & 1 \end{array}\right].
Elimino abaixo do pivô: \,R_2\leftarrow R_2-2R_1 e \,R_3\leftarrow R_3-R_1:
\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 1 & 0\\\\ 0 & -1 & -2 & 1 & -2 & 0\\\\ 0 & -1 & 0 & 0 & -1 & 1 \end{array}\right].
Faço \,R_2\leftarrow -R_2 (para pivô 1 em coluna 2):
\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 1 & 0\\\\ 0 & 1 & 2 & -1 & 2 & 0\\\\ 0 & -1 & 0 & 0 & -1 & 1 \end{array}\right].
Elimino os outros elementos na coluna 2: \,R_1\leftarrow R_1-R_2 e \,R_3\leftarrow R_3+R_2:
\left[\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & -1 & 0\\\\ 0 & 1 & 2 & -1 & 2 & 0\\\\ 0 & 0 & 2 & -1 & 1 & 1 \end{array}\right].
Faço \,R_3\leftarrow \tfrac{1}{2}R_3 para pivô 1 em coluna 3:
\left[\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & -1 & 0\\\\ 0 & 1 & 2 & -1 & 2 & 0\\\\ 0 & 0 & 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \end{array}\right].
Agora elimino coluna 3 acima: \,R_1\leftarrow R_1+R_3 e \,R_2\leftarrow R_2-2R_3:
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{1}{2}\\\\ 0 & 1 & 0 & 0 & 1 & -1\\\\ 0 & 0 & 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \end{array}\right].
A parte esquerda é I, então
A^{-1}= \begin{pmatrix} \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{1}{2}\\[4pt] 0 & 1 & -1\\[4pt] -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \end{pmatrix}.]

Passo 4

Agora B. Matriz aumentada inicial:
\bigl[\,B\mid I\,\bigr]= \left[\begin{array}{ccc|ccc} 2 & 1 & -1 & 1 & 0 & 0\\\\ 1 & 0 & 2 & 0 & 1 & 0\\\\ 4 & -1 & 3 & 0 & 0 & 1 \end{array}\right].
Troco \,R_1\leftrightarrow R_2 para ter pivô 1:
\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 0 & 1 & 0\\\\ 2 & 1 & -1 & 1 & 0 & 0\\\\ 4 & -1 & 3 & 0 & 0 & 1 \end{array}\right].
Elimino abaixo: \,R_2\leftarrow R_2-2R_1, \,R_3\leftarrow R_3-4R_1:
\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 0 & 1 & 0\\\\ 0 & 1 & -5 & 1 & -2 & 0\\\\ 0 & -1 & -5 & 0 & -4 & 1 \end{array}\right].
Faço \,R_3\leftarrow R_3+R_2 (elimina coluna 2 em R_3):
\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 0 & 1 & 0\\\\ 0 & 1 & -5 & 1 & -2 & 0\\\\ 0 & 0 & -10 & 1 & -6 & 1 \end{array}\right].
Faço \,R_3\leftarrow -\tfrac{1}{10}R_3 para pivô 1 em coluna 3:
\left[\begin{array}{ccc|ccc} 1 & 0 & 2 & 0 & 1 & 0\\\\ 0 & 1 & -5 & 1 & -2 & 0\\\\ 0 & 0 & 1 & -\tfrac{1}{10} & \tfrac{3}{5} & -\tfrac{1}{10} \end{array}\right].
Elimino coluna 3 nas linhas acima: \,R_1\leftarrow R_1-2R_3, \,R_2\leftarrow R_2+5R_3:
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac{1}{5} & -\tfrac{1}{5} & \tfrac{1}{5}\\\\ 0 & 1 & 0 & \tfrac{1}{2} & 1 & -\tfrac{1}{2}\\\\ 0 & 0 & 1 & -\tfrac{1}{10} & \tfrac{3}{5} & -\tfrac{1}{10} \end{array}\right].
Logo
B^{-1}= \begin{pmatrix} \tfrac{1}{5} & -\tfrac{1}{5} & \tfrac{1}{5}\\[4pt] \tfrac{1}{2} & 1 & -\tfrac{1}{2}\\[4pt] -\tfrac{1}{10} & \tfrac{3}{5} & -\tfrac{1}{10} \end{pmatrix}.]

Passo 5

Agora C. Aumentada inicial:
\bigl[\,C\mid I\,\bigr]= \left[\begin{array}{ccc|ccc} -2 & 1 & -1 & 1 & 0 & 0\\\\ 3 & 1 & 2 & 0 & 1 & 0\\\\ 1 & -2 & 0 & 0 & 0 & 1 \end{array}\right].
Troco \,R_1\leftrightarrow R_3 para ter pivô 1 na primeira linha:
\left[\begin{array}{ccc|ccc} 1 & -2 & 0 & 0 & 0 & 1\\\\ 3 & 1 & 2 & 0 & 1 & 0\\\\ -2 & 1 & -1 & 1 & 0 & 0 \end{array}\right].
Elimino abaixo: \,R_2\leftarrow R_2-3R_1, \,R_3\leftarrow R_3+2R_1:
\left[\begin{array}{ccc|ccc} 1 & -2 & 0 & 0 & 0 & 1\\\\ 0 & 7 & 2 & 0 & 1 & -3\\\\ 0 & -3 & -1 & 1 & 0 & 2 \end{array}\right].
Faço \,R_2\leftarrow \tfrac{1}{7}R_2:
\left[\begin{array}{ccc|ccc} 1 & -2 & 0 & 0 & 0 & 1\\\\ 0 & 1 & \tfrac{2}{7} & 0 & \tfrac{1}{7} & -\tfrac{3}{7}\\\\ 0 & -3 & -1 & 1 & 0 & 2 \end{array}\right].
Elimino coluna 2 nas outras linhas: \,R_1\leftarrow R_1+2R_2, \,R_3\leftarrow R_3+3R_2:
\left[\begin{array}{ccc|ccc} 1 & 0 & \tfrac{4}{7} & 0 & \tfrac{2}{7} & \tfrac{1}{7}\\\\ 0 & 1 & \tfrac{2}{7} & 0 & \tfrac{1}{7} & -\tfrac{3}{7}\\\\ 0 & 0 & -\tfrac{1}{7} & 1 & \tfrac{3}{7} & \tfrac{5}{7} \end{array}\right].
Faço \,R_3\leftarrow -7R_3 para pivô 1 na coluna 3:
\left[\begin{array}{ccc|ccc} 1 & 0 & \tfrac{4}{7} & 0 & \tfrac{2}{7} & \tfrac{1}{7}\\\\ 0 & 1 & \tfrac{2}{7} & 0 & \tfrac{1}{7} & -\tfrac{3}{7}\\\\ 0 & 0 & 1 & -7 & -3 & -5 \end{array}\right].
Elimino coluna 3 nas linhas acima: \,R_1\leftarrow R_1-\tfrac{4}{7}R_3, \,R_2\leftarrow R_2-\tfrac{2}{7}R_3. Calculando os resultados obtém-se:
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 4 & 2 & 3\\\\ 0 & 1 & 0 & 2 & 1 & 1\\\\ 0 & 0 & 1 & -7 & -3 & -5 \end{array}\right].
Portanto
C^{-1}= \begin{pmatrix} 4 & 2 & 3\\[4pt] 2 & 1 & 1\\[4pt] -7 & -3 & -5 \end{pmatrix}.]

Passo 6

Por fim D. Aumentada inicial:
\bigl[\,D\mid I\,\bigr]= \left[\begin{array}{ccc|ccc} 2 & -1 & 1 & 1 & 0 & 0\\\\ -3 & 2 & -2 & 0 & 1 & 0\\\\ -2 & 1 & 0 & 0 & 0 & 1 \end{array}\right].
Faço \,R_1\leftarrow \tfrac{1}{2}R_1 para pivô 1:
\left[\begin{array}{ccc|ccc} 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & 0 & 0\\\\ -3 & 2 & -2 & 0 & 1 & 0\\\\ -2 & 1 & 0 & 0 & 0 & 1 \end{array}\right].
Elimino abaixo: \,R_2\leftarrow R_2+3R_1, \,R_3\leftarrow R_3+2R_1:
\left[\begin{array}{ccc|ccc} 1 & -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} & 0 & 0\\\\ 0 & \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{3}{2} & 1 & 0\\\\ 0 & 0 & 1 & 1 & 0 & 1 \end{array}\right].
Uso R_3 (já tem pivô 1 em coluna 3) para zerar as entradas acima: \,R_1\leftarrow R_1-\tfrac{1}{2}R_3, \,R_2\leftarrow R_2+\tfrac{1}{2}R_3:
\left[\begin{array}{ccc|ccc} 1 & -\tfrac{1}{2} & 0 & 0 & 0 & -\tfrac{1}{2}\\\\ 0 & \tfrac{1}{2} & 0 & 2 & 1 & \tfrac{1}{2}\\\\ 0 & 0 & 1 & 1 & 0 & 1 \end{array}\right].
Faço \,R_2\leftarrow 2R_2 para pivô 1 em coluna 2:
\left[\begin{array}{ccc|ccc} 1 & -\tfrac{1}{2} & 0 & 0 & 0 & -\tfrac{1}{2}\\\\ 0 & 1 & 0 & 4 & 2 & 1\\\\ 0 & 0 & 1 & 1 & 0 & 1 \end{array}\right].
Elimino a entrada em coluna 2 na primeira linha: \,R_1\leftarrow R_1+\tfrac{1}{2}R_2:
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & 2 & 1 & 0\\\\ 0 & 1 & 0 & 4 & 2 & 1\\\\ 0 & 0 & 1 & 1 & 0 & 1 \end{array}\right].
Assim
D^{-1}= \begin{pmatrix} 2 & 1 & 0\\[4pt] 4 & 2 & 1\\[4pt] 1 & 0 & 1 \end{pmatrix}.]

Resposta

A^{-1}=\begin{pmatrix} \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{1}{2}\\[4pt] 0 & 1 & -1\\[4pt] -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \end{pmatrix}
B^{-1}=\begin{pmatrix} \tfrac{1}{5} & -\tfrac{1}{5} & \tfrac{1}{5}\\[4pt] \tfrac{1}{2} & 1 & -\tfrac{1}{2}\\[4pt] -\tfrac{1}{10} & \tfrac{3}{5} & -\tfrac{1}{10} \end{pmatrix}
C^{-1}=\begin{pmatrix} 4 & 2 & 3\\[4pt] 2 & 1 & 1\\[4pt] -7 & -3 & -5 \end{pmatrix}
D^{-1}=\begin{pmatrix} 2 & 1 & 0\\[4pt] 4 & 2 & 1\\[4pt] 1 & 0 & 1 \end{pmatrix}]

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