Encontre, se existir, a inversa das matrizes abaixo.
(f) A=\left[\begin{array}{llll}4 & 0 & 1 & 1 \\\\ 3 & 1 & 3 & 1 \\\\ 0 & 1 & 2 & 0 \\\\ 3 & 2 & 4 & 1\end{array}\right].

Passo 1

O enunciado pede a inversa da matriz
A = \begin{pmatrix} 4 & 0 & 1 & 1\\\\ 3 & 1 & 3 & 1\\\\ 0 & 1 & 2 & 0\\\\ 3 & 2 & 4 & 1 \end{pmatrix}.
O conteúdo necessário é a inversão de matriz 4\times4. Para simplificar a conta, vamos usar partição em blocos 2\times2 e a fórmula do inverso via complemento de Schur.

Passo 2

Particionamos
A = \begin{pmatrix} M & N \\[6pt] P & Q \end{pmatrix},
onde
M = \begin{pmatrix}4&0\\\ 3&1\end{pmatrix},\quad N = \begin{pmatrix}1&1\\\ 3&1\end{pmatrix},\quad P = \begin{pmatrix}0&1\\\ 3&2\end{pmatrix},\quad Q = \begin{pmatrix}2&0\\\ 4&1\end{pmatrix}.
Se M e o complemento de Schur S = Q - P\,M^{-1}N forem invertíveis,
A^{-1} = \begin{pmatrix} M^{-1} + M^{-1}N\,S^{-1}P\,M^{-1} & -\,M^{-1}N\,S^{-1}\\[4pt] -\,S^{-1}P\,M^{-1} & S^{-1} \end{pmatrix}.]

Passo 3

Calculamos
\det(M)=4\cdot1-0\cdot3=4, \quad M^{-1}=\frac1{4}\begin{pmatrix}1&0\\\ -3&4\end{pmatrix}.

Passo 4

Computamos o complemento de Schur
S = Q - P\,M^{-1}N.
Primeiro
P\,M^{-1} =\frac1{4}\begin{pmatrix}-3&4\\\ -3&8\end{pmatrix},
P\,M^{-1}N =\frac1{4}\begin{pmatrix}9&1\\\ 21&5\end{pmatrix} =\begin{pmatrix}9/4&1/4\\\ 21/4&5/4\end{pmatrix}.
Logo
S =\begin{pmatrix}2&0\\\ 4&1\end{pmatrix} -\begin{pmatrix}9/4&1/4\\\ 21/4&5/4\end{pmatrix} =\frac1{4}\begin{pmatrix}-1&-1\\\ -5&-1\end{pmatrix}.

Passo 5

Determinante de S:
\det(S) =\frac1{16}\bigl((-1)(-1)-(-1)(-5)\bigr) =-\tfrac14,
então
S^{-1} =\begin{pmatrix}1&-1\\\ -5&1\end{pmatrix}.

Passo 6

Agora calculamos os demais blocos:
M^{-1}N =\frac1{4}\begin{pmatrix}1&1\\\ 9&1\end{pmatrix}, \quad P\,M^{-1} =\begin{pmatrix}-\tfrac34&1\\\ -\tfrac34&2\end{pmatrix}.
-\,M^{-1}N\,S^{-1} = -\frac1{4}\begin{pmatrix}1&1\\\ 9&1\end{pmatrix} \begin{pmatrix}1&-1\\\ -5&1\end{pmatrix} =\begin{pmatrix}1&0\\\ -1&2\end{pmatrix}.
-\,S^{-1}P\,M^{-1} = -\begin{pmatrix}1&-1\\\ -5&1\end{pmatrix} \begin{pmatrix}-\tfrac34&1\\\ -\tfrac34&2\end{pmatrix} =\begin{pmatrix}0&1\\\ -3&3\end{pmatrix}.
M^{-1} + M^{-1}N\,S^{-1}P\,M^{-1} = \frac1{4}\begin{pmatrix}1&0\\\ -3&4\end{pmatrix} + \frac1{4}\begin{pmatrix}1&1\\\ 9&1\end{pmatrix} \begin{pmatrix}1&-1\\\ -5&1\end{pmatrix} \begin{pmatrix}-\tfrac34&1\\\ -\tfrac34&2\end{pmatrix} = \begin{pmatrix}1&-1\\\ 0&-2\end{pmatrix}.

Passo 7

Reunindo tudo,
A^{-1} = \begin{pmatrix} 1 & -1 & 1 & 0\\\\ 0 & -2 & -1 & 2\\\\ 0 & 1 & 1 & -1\\\\ -3 & 3 & -5 & 1 \end{pmatrix}.

Resposta

A^{-1} = \begin{pmatrix} 1 & -1 & 1 & 0\\\\ 0 & -2 & -1 & 2\\\\ 0 & 1 & 1 & -1\\\\ -3 & 3 & -5 & 1 \end{pmatrix}.

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