Considere a matriz
A=\left[\begin{array}{lll} 1 & 1 & 0 \\\\ 2 & 1 & 0 \\\\ 3 & 3 & 1 \end{array}\right]
(a) Encontre a inversa de A;
(b) Use a matriz inversa encontrada no item anterior para resolver o sistema A X=B, onde
X=\left[\begin{array}{l} x \\\ y \\\ z \end{array}\right] \text { e } B=\left[\begin{array}{l} 1 \\\\ 0 \\\\ 1 \end{array}\right]

Passo 1

O enunciado dá
A=\begin{pmatrix} 1 & 1 & 0\\\\ 2 & 1 & 0\\\\ 3 & 3 & 1 \end{pmatrix}
e pede (a) a inversa de A e (b) a solução de A\,X=B, com
X=\begin{pmatrix}x\y\z\end{pmatrix},\quad B=\begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix}.
Vamos usar eliminação de Gauss–Jordan para encontrar A^{-1} e depois calcular X=A^{-1}B.

Passo 2

Montamos o sistema ampliado \bigl[A\mid I\bigr] e eliminamos as entradas abaixo do pivô da primeira coluna:
\left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0\\\\ 2 & 1 & 0 & 0 & 1 & 0\\\\ 3 & 3 & 1 & 0 & 0 & 1 \end{array}\right] \;\xrightarrow{R_2\to R_2-2R_1,\;R_3\to R_3-3R_1}\; \left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0\\\\ 0 & -1 & 0 & -2 & 1 & 0\\\\ 0 & 0 & 1 & -3 & 0 & 1 \end{array}\right].

Passo 3

Normalizamos o pivô da segunda linha e zeramos acima dele, obtendo a forma [I\mid A^{-1}]:
\left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0\\\\ 0 & -1 & 0 & -2 & 1 & 0\\\\ 0 & 0 & 1 & -3 & 0 & 1 \end{array}\right] \xrightarrow{R_2\to -R_2} \left[\begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0\\\\ 0 & 1 & 0 & 2 & -1 & 0\\\\ 0 & 0 & 1 & -3 & 0 & 1 \end{array}\right] \xrightarrow{R_1\to R_1-R_2} \left[\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 1 & 0\\\\ 0 & 1 & 0 & 2 & -1 & 0\\\\ 0 & 0 & 1 & -3 & 0 & 1 \end{array}\right].
Daí
A^{-1}=\begin{pmatrix} -1 & 1 & 0\\\\ 2 & -1 & 0\\\\ -3 & 0 & 1 \end{pmatrix}.

Passo 4

Para resolver A\,X=B basta calcular
X = A^{-1}B = \begin{pmatrix} -1 & 1 & 0\\\\ 2 & -1 & 0\\\\ -3 & 0 & 1 \end{pmatrix} \begin{pmatrix}1\\\ 0\\\ 1\end{pmatrix} =\begin{pmatrix} -1\cdot1 + 1\cdot0 + 0\cdot1\\\\ 2\cdot1 + (-1)\cdot0 + 0\cdot1\\\\ -3\cdot1 + 0\cdot0 + 1\cdot1 \end{pmatrix} =\begin{pmatrix}-1\\\ 2\\\ -2\end{pmatrix}.

Resposta

(a)
A^{-1}=\begin{pmatrix} -1 & 1 & 0\\\\ 2 & -1 & 0\\\\ -3 & 0 & 1 \end{pmatrix}.
(b)
(x,y,z)=(-1,\,2,\,-2).

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