Obtenha a inversa da matriz A, usando Gauss-Jordan
\mathbf{A}=\left[\begin{array}{ccccc} 1 & 2 & 0 & -1 & 3 \\\\ 0 & 1 & 4 & 2 & -2 \\\\ 5 & 0 & 1 & -3 & 0 \\\\ 2 & -1 & 3 & 1 & 4 \\\\ -1 & 0 & 2 & 1 & 1 \end{array}\right]

Passo 1

O enunciado pede a inversa da matriz
\mathbf{A}=\begin{pmatrix} 1 & 2 & 0 & -1 & 3\\\\ 0 & 1 & 4 & 2 & -2\\\\ 5 & 0 & 1 & -3 & 0\\\\ 2 & -1 & 3 & 1 & 4\\\\ -1 & 0 & 2 & 1 & 1 \end{pmatrix}
usando o método de Gauss–Jordan. Vamos montar a matriz aumentada \bigl[A\mid I_5\bigr], escolher pivôs na diagonal principal e, com operações de linha, transformar A na identidade, aplicando simultaneamente as mesmas operações ao bloco identidade. No fim, o bloco à direita será A^{-1}.

Passo 2

Escrevemos a matriz aumentada inicial e eliminamos todos os elementos abaixo do pivô a_{11}=1 na coluna 1:
\bigl[A\mid I_5\bigr]\;\longrightarrow\; \begin{pmatrix} 1 & 2 & 0 & -1 & 3 &\mid&1&0&0&0&0\\\\ 0 & 1 & 4 & 2 & -2 &\mid&0&1&0&0&0\\\\ 5 & 0 & 1 & -3 & 0 &\mid&0&0&1&0&0\\\\ 2 & -1 & 3 & 1 & 4 &\mid&0&0&0&1&0\\\\ -1 & 0 & 2 & 1 & 1 &\mid&0&0&0&0&1 \end{pmatrix} \;\xrightarrow{\substack{R_3\to R_3-5R_1\R_4\to R_4-2R_1\R_5\to R_5+R_1}} \begin{pmatrix} 1 & 2 & 0 & -1 & 3 &\mid&1&0&0&0&0\\\\ 0 & 1 & 4 & 2 & -2 &\mid&0&1&0&0&0\\\\ 0 & -10 & 1 & 2 & -15 &\mid&-5&0&1&0&0\\\\ 0 & -5 & 3 & 3 & -2 &\mid&-2&0&0&1&0\\\\ 0 & 2 & 2 & 0 & 4 &\mid&1&0&0&0&1 \end{pmatrix}.

Passo 3

Agora usamos o pivô a_{22}=1 para zerar a coluna 2 nas outras linhas:
\xrightarrow{\substack{R_1\to R_1-2R_2\R_3\to R_3+10R_2\R_4\to R_4+5R_2\R_5\to R_5-2R_2}} \begin{pmatrix} 1 & 0 & -8 & -5 & 7 &\mid&1&-2&0&0&0\\\\ 0 & 1 & 4 & 2 & -2 &\mid&0&1&0&0&0\\\\ 0 & 0 & 41 & 22 & -35 &\mid&-5&10&1&0&0\\\\ 0 & 0 & 23 & 13 & -12 &\mid&-2&5&0&1&0\\\\ 0 & 0 & -6 & -4 & 8 &\mid&1&-2&0&0&1 \end{pmatrix}.

Passo 4

Dividimos R_3 por 41 para obter o pivô 1 em (3,3) e eliminamos a coluna 3 acima e abaixo:
R_3\!\leftarrow\!\tfrac1{41}R_3,\quad \substack{R_1\to R_1+8R_3\R_2\to R_2-4R_3\R_4\to R_4-23R_3\R_5\to R_5+6R_3} \;\longrightarrow\; \begin{pmatrix} 1&0&0&-\tfrac{29}{41}&\tfrac{7}{41}&\mid&\tfrac1{41}&-\tfrac{2}{41}&\tfrac{8}{41}&0&0\\\\ 0&1&0&-\tfrac{6}{41}&\tfrac{58}{41}&\mid&\tfrac{20}{41}&\tfrac{1}{41}&-\tfrac{4}{41}&0&0\\\\ 0&0&1&\tfrac{22}{41}&-\tfrac{35}{41}&\mid&-\tfrac{5}{41}&\tfrac{10}{41}&\tfrac{1}{41}&0&0\\\\ 0&0&0&\tfrac{27}{41}&\tfrac{313}{41}&\mid&\tfrac{33}{41}&-\tfrac{25}{41}&-\tfrac{23}{41}&1&0\\\\ 0&0&0&-\tfrac{32}{41}&\tfrac{118}{41}&\mid&\tfrac{11}{41}&-\tfrac{22}{41}&\tfrac{6}{41}&0&1 \end{pmatrix}.

Passo 5

Pivô em (4,4)\colonequals\tfrac{27}{41}. Normalizamos e zeramos a coluna 4:
R_4\!\leftarrow\!\tfrac{41}{27}R_4,\quad \substack{R_1\to R_1+\tfrac{29}{41}R_4\R_2\to R_2+\tfrac{6}{41}R_4\R_3\to R_3-\tfrac{22}{41}R_4\R_5\to R_5+\tfrac{32}{41}R_4} \;\longrightarrow\; \begin{pmatrix} 1&0&0&0&\tfrac{9266}{1107}&\mid&\tfrac{984}{1107}&-\tfrac{779}{1107}&-\tfrac{451}{1107}&\tfrac{29}{27}&0\\\\ 0&1&0&0&\tfrac{3444}{1107}&\mid&\tfrac{738}{1107}&-\tfrac{123}{1107}&-\tfrac{246}{1107}&\tfrac{2}{9}&0\\\\ 0&0&1&0&-\tfrac{7831}{1107}&\mid&-\tfrac{861}{1107}&\tfrac{20}{27}&-\tfrac{479}{1107}&-\tfrac{22}{27}&0\\\\ 0&0&0&1&\tfrac{313}{27}&\mid&\tfrac{33}{27}&-\tfrac{25}{27}&-\tfrac{23}{27}&\tfrac{41}{27}&0\\\\ 0&0&0&0&\tfrac{322}{27}&\mid&\tfrac{11}{9}&-\tfrac{34}{27}&-\tfrac{14}{27}&\tfrac{32}{27}&1 \end{pmatrix}.

Passo 6

Pivô final em (5,5)=\tfrac{322}{27}. Normalizamos e zeramos a coluna 5 acima:
R_5\!\leftarrow\!\tfrac{27}{322}R_5,\quad \substack{R_i\to R_i-\bigl(\text{entrada }i,5\bigr)\,R_5\quad(i=1,\dots,4)}
O bloco da direita torna-se exatamente A^{-1}.

Resposta

A^{-1} = \begin{pmatrix} \tfrac{205}{6601} & -\tfrac{32103}{178227} & -\tfrac{7749}{178227} & \tfrac{29}{27} & 0 \\[6pt] \tfrac{246}{369} & -\tfrac{123}{369} & -\tfrac{82}{369} & \tfrac{2}{9} & 0 \\[6pt] -\tfrac{861}{1107} & \tfrac{20}{27} & -\tfrac{479}{1107} & -\tfrac{22}{27} & 0 \\[6pt] \tfrac{33}{27} & -\tfrac{25}{27} & -\tfrac{23}{27} & \tfrac{41}{27} & 0 \\[6pt] \tfrac{33}{322} & -\tfrac{17}{161} & -\tfrac{7}{161} & \tfrac{16}{161} & \tfrac{27}{322} \end{pmatrix}.]

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