1. (Simon & Blume - Exercício 8.19) Use o método de Gauss-Jordan para inverter cada matriz abaixo ou para provar que ela é singular:
    \left[\begin{array}{cccc}2 & 6 & 0 & 5 \\\\ 6 & 21 & 8 & 17 \\\\ 4 & 12 & -4 & 13 \\\\ 0 & -3 & -12 & 2\end{array}\right]

Passo 1

O enunciado pede para inverter a matriz
A = \begin{pmatrix} 2 & 6 & 0 & 5 \\[6pt] 6 & 21 & 8 & 17 \\[6pt] 4 & 12 & -4 & 13 \\[6pt] 0 & -3 & -12 & 2 \end{pmatrix}
usando o método de Gauss–Jordan. Vamos montar a matriz aumentada \bigl[A\mid I_4\bigr], aplicar pivoteamento por colunas e linhas, normalizar os pivôs e zerar, etapa por etapa, até obter \bigl[I_4\mid A^{-1}\bigr].]

Passo 2

Fazemos pivô em \,(1,1). Dividimos a linha 1 por 2 e em seguida zeramos abaixo:
\begin{array}{c|cccc|cccc} & 1 & 2 & 3 & 4 & 1 & 2 & 3 & 4\\hline R_1&2 &6 &0 &5 &1 &0 &0 &0\\\ R_2&6 &21&8 &17&0 &1 &0 &0\\\ R_3&4 &12&-4&13&0 &0 &1 &0\\\ R_4&0 &-3&-12&2 &0 &0 &0 &1 \end{array} \quad\longrightarrow\quad \begin{array}{c|cccc|cccc} & 1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &3 &0 &\tfrac52&\tfrac12&0 &0 &0\\\ R_2&0 &3 &8 &2 &-\!3&1 &0 &0\\\ R_3&0 &0 &-4&3 &-\!2&0 &1 &0\\\ R_4&0 &-3&-12&2 &0 &0 &0 &1 \end{array}

Passo 3

Fazemos pivô em \,(2,2). Dividimos R_2 por 3 e zeramos acima e abaixo:
\begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &3 &0 &\tfrac52&\tfrac12&0 &0 &0\\\ R_2&0 &3 &8 &2 &-\!3&1 &0 &0\\\ R_4&0 &-3&-12&2 &0 &0 &0 &1 \end{array} \quad\longrightarrow\quad \begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &0 &-8&\tfrac12&\tfrac72&-1 &0 &0\\\ R_2&0 &1 &\tfrac83&\tfrac23&-1&\tfrac13&0 &0\\\ R_4&0 &0 &-4&4 &-3&1 &0 &1 \end{array}

Passo 4

Fazemos pivô em \,(3,3). Dividimos R_3 por -4 e zeramos nas outras linhas:
\begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &0 &-8&\tfrac12&\tfrac72&-1 &0 &0\\\ R_2&0 &1 &\tfrac83&\tfrac23&-1&\tfrac13&0 &0\\\ R_3&0 &0 &-4&3 &-2&0 &1 &0\\\ R_4&0 &0 &-4&4 &-3&1 &0 &1 \end{array} \quad\longrightarrow\quad \begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &0 &0 &-\tfrac{11}2&\tfrac{15}2&-1&-2&0\\\ R_2&0 &1 &0 &\tfrac{8}3&-\tfrac73&\tfrac13&\tfrac23&0\\\ R_3&0 &0 &1 &-\tfrac34&\tfrac12&0 &-\tfrac14&0\\\ R_4&0 &0 &0 &1 &-1&1 &-1&1 \end{array}

Passo 5

Fazemos pivô em \,(4,4) (já está 1) e zeramos acima com combinações de R_4:
\begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline R_1&1 &0 &0 &-\tfrac{11}2&\tfrac{15}2&-1 &-2&0\\\ R_2&0 &1 &0 &\tfrac{8}3&-\tfrac73&\tfrac13&\tfrac23&0\\\ R_3&0 &0 &1 &-\tfrac34&\tfrac12&0 &-\tfrac14&0\\\ R_4&0 &0 &0 &1 &-1&1 &-1&1 \end{array} \quad\longrightarrow\quad \begin{array}{c|cccc|cccc} &1 &2 &3 &4 &1 &2 &3 &4\\hline I_4&1 &0 &0 &0 &2 &\tfrac92&-\tfrac{15}2&\tfrac{11}2\\\ &0 &1 &0 &0 &\tfrac13&-\tfrac73&\tfrac{10}3&-\tfrac{8}3\\\ &0 &0 &1 &0 &-\tfrac14&\tfrac34&-1&\tfrac34\\\ &0 &0 &0 &1 &-1&1&-1&1 \end{array}

Resposta

A inversa de A é
A^{-1} = \begin{pmatrix} 2 & \tfrac92 & -\tfrac{15}{2} & \tfrac{11}{2}\\[6pt] \tfrac13 & -\tfrac73 & \tfrac{10}{3} & -\tfrac{8}{3}\\[6pt] -\tfrac14 & \tfrac34 & -1 & \tfrac34\\[6pt] -1 & 1 & -1 & 1 \end{pmatrix}.]

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