(a) Encontre a inversa da matriz
A=\left(\begin{array}{rrrr} 1 & -1 & 0 & 1 \\\\ -1 & 2 & 1 & -1 \\\\ 0 & 1 & 2 & 0 \\\\ 1 & -1 & 0 & 2 \end{array}\right) .
(b) Resolva o sistema linear
\left\{\begin{array}{l} x-y+w=1 \\\\ -x+2 y+z-w=-1 \\\ y+2 z=0 \\\ x-y+2 w=1 \end{array}\right.

Passo 1

O enunciado pede (a) a matriz inversa de \r
A=\begin{pmatrix} 1 & -1 & 0 & 1\\\\ -1 & 2 & 1 & -1\\\\ 0 & 1 & 2 & 0\\\\ 1 & -1 & 0 & 2 \end{pmatrix}\r
e (b) a solução do sistema linear \r
\begin{cases} x - y + w = 1,\\\\ -\,x + 2y + z - w = -1,\\\ y + 2z = 0,\\\ x - y + 2w = 1. \end{cases}\r
Note que o sistema pode ser escrito como A\mathbf{x}=\mathbf{b} com \mathbf{x}=(x,y,z,w)^T e \mathbf{b}=(1,-1,0,1)^T. Para resolver (a) faremos eliminação de Gauss em [A\mid I] e em (b) usaremos \mathbf{x}=A^{-1}\mathbf{b}.

Passo 2

Montamos a matriz aumentada [A\mid I] e aplicamos \r \mathrm{R}_2\leftarrow \mathrm{R}_2+\mathrm{R}_1, \r \mathrm{R}_4\leftarrow \mathrm{R}_4-\mathrm{R}_1: \r
\left[\begin{array}{cccc|cccc} 1 & -1 & 0 & 1 & 1 & 0 & 0 & 0\\\\ -1 & 2 & 1 & -1 & 0 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 0 & 0 & 1 & 0\\\\ 1 & -1 & 0 & 2 & 0 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{\substack{\mathrm{R}_2+\mathrm{R}_1\\mathrm{R}_4-\mathrm{R}_1}} \left[\begin{array}{cccc|cccc} 1 & -1 & 0 & 1 & 1 & 0 & 0 & 0\\\\ 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right]

Passo 3

Agora usamos \mathrm{R}_3\leftarrow \mathrm{R}_3-\mathrm{R}_2 e \mathrm{R}_1\leftarrow \mathrm{R}_1+\mathrm{R}_2: \r
\left[\begin{array}{cccc|cccc} 1 & -1 & 0 & 1 & 1 & 0 & 0 & 0\\\\ 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0\\\\ 0 & 1 & 2 & 0 & 0 & 0 & 1 & 0\\\\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{\substack{\mathrm{R}_3-\mathrm{R}_2\\mathrm{R}_1+\mathrm{R}_2}} \left[\begin{array}{cccc|cccc} 1 & 0 & 1 & 1 & 2 & 1 & 0 & 0\\\\ 0 & 1 & 1 & 0 & 1 & 1 & 0 & 0\\\\ 0 & 0 & 1 & 0 & -1 & -1 & 1 & 0\\\\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 1 \end{array}\right]

Passo 4

Eliminamos as colunas 3 e 4 acima dos pivôs: \r \mathrm{R}_1\leftarrow \mathrm{R}_1-\mathrm{R}_3, \r \mathrm{R}_2\leftarrow \mathrm{R}_2-\mathrm{R}_3, \r \mathrm{R}_1\leftarrow \mathrm{R}_1-\mathrm{R}_4. Obtemos [I\mid A^{-1}]: \r
A^{-1} = \begin{pmatrix} 4 & 2 & -1 & -1\\\\ 2 & 2 & -1 & 0\\\\ -1 & -1 & 1 & 0\\\\ -1 & 0 & 0 & 1 \end{pmatrix}.

Passo 5

Para resolver A\mathbf{x}=\mathbf{b} calculamos \mathbf{x}=A^{-1}\mathbf{b} com \mathbf{b}=(1,-1,0,1)^T: \r
\begin{pmatrix} x\y\z\w \end{pmatrix} = \begin{pmatrix} 4 & 2 & -1 & -1\\\\ 2 & 2 & -1 & 0\\\\ -1 & -1 & 1 & 0\\\\ -1 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1\\\ -1\\\ 0\\\ 1 \end{pmatrix} = \begin{pmatrix} 1\\\ 0\\\ 0\\\ 0 \end{pmatrix}.

Resposta

(a)
A^{-1} = \begin{pmatrix} 4 & 2 & -1 & -1\\\\ 2 & 2 & -1 & 0\\\\ -1 & -1 & 1 & 0\\\\ -1 & 0 & 0 & 1 \end{pmatrix}.\r
(b)
(x,y,z,w) = (1,\,0,\,0,\,0).

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