Use o método de Gauss-Jordan para encontrar a matriz inversa, se existir, das seguintes matrizes:
A=\left[\begin{array}{lll} 2 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 1 & 0 & 1 \end{array}\right], \quad B=\left[\begin{array}{ccc} 2 & 1 & -1 \\\\ 1 & 0 & 2 \\\\ 4 & -1 & 3 \end{array}\right]

Passo 1

O problema pede para encontrar A^{-1} e B^{-1} usando o método de Gauss–Jordan. A ideia é montar a matriz aumentada
\bigl[M \mid I\bigr]
e aplicar operações elementares de linha até transformá-la em \bigl[I\mid M^{-1}\bigr]. Vamos fazer isso primeiro para A e depois para B.

Passo 2

Para A, montamos
\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].
Agora, dividimos R_{1} por 2 (para ter pivô 1) e zeramos abaixo:
R_{1}\leftarrow \tfrac12R_{1},\quad R_{2}\leftarrow R_{2}-R_{1},\quad R_{3}\leftarrow R_{3}-R_{1}
resultando em
\left[\begin{array}{ccc|ccc} 1 & \tfrac12 & 0 & \tfrac12 & 0 & 0\\\\ 0 & \tfrac12 & 1 & -\tfrac12 & 1 & 0\\\\ 0 & -\tfrac12 & 1 & -\tfrac12 & 0 & 1 \end{array}\right].

Passo 3

Agora pivotamos na segunda coluna. Multiplicamos R_{2} por 2 e eliminamos nas outras linhas:
R_{2}\leftarrow 2R_{2},\quad R_{1}\leftarrow R_{1}-\tfrac12R_{2},\quad R_{3}\leftarrow R_{3}+\tfrac12R_{2}
o que dá
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac12 & -\tfrac12 & \tfrac12\\\\ 0 & 1 & 0 & 0 & 1 & -1\\\\ 0 & 0 & 2 & -1 & 1 & 1 \end{array}\right].

Passo 4

Finalmente, pivotamos na terceira coluna dividindo R_{3} por 2 e zerando acima:
R_{3}\leftarrow \tfrac12R_{3},\quad R_{1}\leftarrow R_{1}-0\cdot R_{3},\quad R_{2}\leftarrow R_{2}-0\cdot R_{3}
(neste caso R_{1} e R_{2} já têm zeros na coluna 3), obtendo
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac12 & -\tfrac12 & \tfrac12\\\\ 0 & 1 & 0 & 0 & 1 & -1\\\\ 0 & 0 & 1 & -\tfrac12 & \tfrac12 & \tfrac12 \end{array}\right].
Daí lemos
A^{-1}=\begin{pmatrix} \tfrac12 & -\tfrac12 & \tfrac12\\\\ 0 & 1 & -1\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix}.

Passo 5

Para B, começamos com
\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].
Dividimos R_{1} por 2 e zeramos abaixo:
R_{1}\leftarrow \tfrac12R_{1},\quad R_{2}\leftarrow R_{2}-R_{1},\quad R_{3}\leftarrow R_{3}-4R_{1}
ficando
\left[\begin{array}{ccc|ccc} 1 & \tfrac12 & -\tfrac12 & \tfrac12 & 0 & 0\\\\ 0 & -\tfrac12 & \tfrac52 & -\tfrac12 & 1 & 0\\\\ 0 & -3 & 5 & -2 & 0 & 1 \end{array}\right].

Passo 6

Pivotamos na segunda coluna: R_{2}\leftarrow -2R_{2} e eliminamos nas outras:
R_{2}\leftarrow -2R_{2},\quad R_{1}\leftarrow R_{1}-\tfrac12R_{2},\quad R_{3}\leftarrow R_{3}+3R_{2}
resultando em
\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].

Passo 7

Por fim, dividimos R_{3} por -10 e eliminamos acima:
R_{3}\leftarrow -\tfrac1{10}R_{3},\quad R_{1}\leftarrow R_{1}-2R_{3},\quad R_{2}\leftarrow R_{2}+5R_{3}
chegando a
\left[\begin{array}{ccc|ccc} 1 & 0 & 0 & \tfrac15 & -\tfrac15 & \tfrac15\\\\ 0 & 1 & 0 & \tfrac12 & 1 & -\tfrac12\\\\ 0 & 0 & 1 & -\tfrac1{10} & \tfrac35 & -\tfrac1{10} \end{array}\right],
logo
B^{-1}=\begin{pmatrix} \tfrac15 & -\tfrac15 & \tfrac15\\\ \tfrac12 & 1 & -\tfrac12\\\\ -\tfrac1{10} & \tfrac35 & -\tfrac1{10} \end{pmatrix}.

Resposta

A^{-1}=\displaystyle\begin{pmatrix} \tfrac12 & -\tfrac12 & \tfrac12\\\\ 0 & 1 & -1\\\\ -\tfrac12 & \tfrac12 & \tfrac12 \end{pmatrix} B^{-1}=\displaystyle\begin{pmatrix} \tfrac15 & -\tfrac15 & \tfrac15\\\ \tfrac12 & 1 & -\tfrac12\\\\ -\tfrac1{10} & \tfrac35 & -\tfrac1{10} \end{pmatrix}

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