Determine, se possível, as matrizes inversas das matrizes abaixo:
(a) \left[\begin{array}{cc}1 & 0 \\\\ -2 & -2\end{array}\right]
(b) \left[\begin{array}{cc}4 & -7 \\\\ -1 & -3\end{array}\right]
(c) \left[\begin{array}{ccc}3 & 1 & 2 \\\\ -2 & -4 & -5 \\\\ 2 & 4 & -5\end{array}\right]
(d) \left[\begin{array}{ccc}1 & -1 & 0 \\\\ 2 & 1 & 0 \\\\ 1 & -1 & -9\end{array}\right]
(e) \left[\begin{array}{ccc}-1 & 1 & 2 \\\\ 3 & 1 & -2 \\\\ 2 & -1 & 3\end{array}\right]
(f) \left[\begin{array}{lll}1 & 0 & 6 \\\\ 2 & 1 & 1 \\\\ 7 & 1 & 0\end{array}\right]
(g) \left[\begin{array}{cccc}1 & 0 & 0 & -1 \\\\ 0 & 2 & 0 & 0 \\\\ -1 & 3 & 2 & 0 \\\\ -1 & 1 & -2 & 1\end{array}\right]
(h) \left[\begin{array}{cccc}-2 & -1 & 0 & 0 \\\\ 3 & 0 & -1 & 1 \\\\ -1 & 2 & -2 & 8 \\\\ -3 & 0 & 7 & 1\end{array}\right]
(i) \left[\begin{array}{cc}3 & 5 \\\\ -2 & -3\end{array}\right]
(j) \left[\begin{array}{cc}5 & -2 \\\\ -8 & 4\end{array}\right]
(n) \left[\begin{array}{ccc}1 & 3 & 2 \\\\ 4 & 1 & -2 \\\\ 2 & 1 & 3\end{array}\right]
(o) \left[\begin{array}{ccc}2 & -1 & 2 \\\\ 1 & 3 & 2 \\\\ 5 & 1 & 6\end{array}\right]
® \left[\begin{array}{cccc}2 & 1 & -2 & 0 \\\\ 0 & 0 & -2 & -1 \\\\ 2 & -1 & 1 & 0 \\\\ 3 & 0 & 3 & 0\end{array}\right]
(s) \left[\begin{array}{ccccc}2 & 0 & 0 & 1 & -2 \\\\ 0 & 1 & 0 & 0 & 1 \\\\ 0 & 0 & 2 & 0 & 0 \\\\ 0 & 0 & 0 & 3 & 2 \\\\ 0 & 0 & 0 & 0 & 3\end{array}\right]
(1) \left[\begin{array}{lll}3 & 1 & 2 \\\\ 2 & 4 & 5 \\\\ 2 & 4 & 5\end{array}\right]
(m) \left[\begin{array}{ccc}4 & 3 & 0 \\\\ 3 & 1 & 2 \\\\ 5 & -1 & -4\end{array}\right]
§ \left[\begin{array}{cccc}2 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 0 \\\\ 1 & 6 & 2 & 0 \\\\ 1 & 1 & -2 & 3\end{array}\right]
(q) \left[\begin{array}{cccc}2 & 1 & 2 & 1 \\\\ 3 & 0 & 1 & 1 \\\\ -1 & 2 & -2 & 1 \\\\ -3 & 2 & 3 & 1\end{array}\right]
(t) \left[\begin{array}{cccc}-1 & 0 & 0 & 0 \\\\ 4 & 1 & 0 & 0 \\\\ 0 & 1 & 3 & 0 \\\\ 1 & 1 & 3 & -1\end{array}\right]
(u) \left[\begin{array}{cccc}1 & 0 & 1 & 0 \\\\ 0 & 1 & 0 & -1 \\\\ 1 & 0 & -1 & 0 \\\\ 0 & -1 & 0 & 1\end{array}\right]

Passo 1

Estamos pedindo a inversa de várias matrizes de ordem 2\times 2, 3\times 3, 4\times 4 e 5\times 5.
Conteúdo: para cada matriz A verificamos primeiro \det(A).
  • Se \det(A)=0, não há inversa.
  • Se \det(A)\neq0, para 2\times 2 usamos a fórmula
    A=\begin{pmatrix}a&b\c&d\end{pmatrix} \quad\Longrightarrow\quad A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\\ -c&a\end{pmatrix}\!,
    para 3\times 3 podemos usar adjunta/cofatores, e para 4\times 4 e 5\times 5 montamos a matriz aumentada [A\mid I] e aplicamos eliminação de Gauss– Jordan.

Passo 2

Matrizes 2\times 2 (casos (a), (b), (i), (j)):
(a) A=\begin{pmatrix}1&0\\\ -2&-2\end{pmatrix}.
\det A=1\cdot(-2)-0\cdot(-2)=-2\neq0.
A^{-1}=\frac1{-2}\begin{pmatrix}-2&0\\\ 2&1\end{pmatrix} =\begin{pmatrix}1&0\\\ -1&-\tfrac12\end{pmatrix}.
(b) A=\begin{pmatrix}4&-7\\\ -1&-3\end{pmatrix}.
\det A=4\cdot(-3)-(-7)\cdot(-1)=-12-7=-19.
A^{-1}=\frac1{-19}\begin{pmatrix}-3&7\\\ 1&4\end{pmatrix} =\begin{pmatrix}\tfrac{3}{19}&-\tfrac{7}{19}\\\ -\tfrac{1}{19}&-\tfrac{4}{19}\end{pmatrix}.
(i) A=\begin{pmatrix}3&5\\\ -2&-3\end{pmatrix}.
\det A=3\cdot(-3)-5\cdot(-2)=-9+10=1.
A^{-1}=\begin{pmatrix}-3&-5\\\ 2&3\end{pmatrix}.
(j) A=\begin{pmatrix}5&-2\\\ -8&4\end{pmatrix}.
\det A=5\cdot4-(-2)\cdot(-8)=20-16=4.
A^{-1}=\frac1{4}\begin{pmatrix}4&2\\\ 8&5\end{pmatrix} =\begin{pmatrix}1&\tfrac12\\\ 2&\tfrac54\end{pmatrix}.

Passo 3

Matrizes 3\times 3 (casos (c), (d), (e), (f), (n), (o), ( 1), (m)):
(c) A=\begin{pmatrix}3&1&2\\\ -2&-4&-5\\\ 2&4&-5\end{pmatrix}.
\det A=100.
A^{-1}=\frac1{100} \begin{pmatrix} 40&13&3\\\\ -20&-19&11\\\\ 0&-10&-10 \end{pmatrix}.
(d) A=\begin{pmatrix}1&-1&0\\\ 2&1&0\\\ 1&-1&-9\end{pmatrix}.
\det A=-27.
A^{-1}=\begin{pmatrix} \tfrac13&\tfrac13&0\\\\ -\tfrac23&\tfrac13&0\\\ \tfrac{1}{9}&0&-\tfrac{1}{9} \end{pmatrix}.
(e) A=\begin{pmatrix}-1&1&2\\\ 3&1&-2\\\ 2&-1&3\end{pmatrix}.
\det A=-24.
A^{-1}=\frac1{-24} \begin{pmatrix} 1&-5&-4\\\\ 13&7&-4\\\\ 5&-1&4 \end{pmatrix} = \begin{pmatrix} -\tfrac1{24}&\tfrac5{24}&\tfrac16\\\ \tfrac{13}{24}&\tfrac7{24}&-\tfrac16\\\ \tfrac5{24}&-\tfrac1{24}&\tfrac16 \end{pmatrix}.
(f) A=\begin{pmatrix}1&0&6\\\ 2&1&1\\\ 7&1&0\end{pmatrix}.
\det A=-31.
A^{-1}=\frac{-1}{31} \begin{pmatrix} -1&6&-6\\\\ 7&-42&11\\\\ -5&-1&1 \end{pmatrix} = \begin{pmatrix} \tfrac1{31}&-\tfrac6{31}&\tfrac6{31}\\\\ -\tfrac7{31}&\tfrac{42}{31}&-\tfrac{11}{31}\\\ \tfrac5{31}&\tfrac1{31}&-\tfrac1{31} \end{pmatrix}.
(n) A=\begin{pmatrix}1&3&2\\\ 4&1&-2\\\ 2&1&3\end{pmatrix}.
\det A=-39.
A^{-1}=-\tfrac1{39} \begin{pmatrix} 11&-7&-8\\\\ -16&-1&10\\\\ 2&5&-11 \end{pmatrix} = \begin{pmatrix} -\tfrac{11}{39}&\tfrac7{39}&\tfrac8{39}\\\ \tfrac{16}{39}&\tfrac1{39}&-\tfrac{10}{39}\\\\ -\tfrac2{39}&-\tfrac5{39}&\tfrac{11}{39} \end{pmatrix}.
(o) \det A=0, logo A não é invertível.
( 1) (a matriz com duas linhas iguais) tem \det=0, não invertível.
(m) A=\begin{pmatrix}4&3&0\\\ 3&1&2\\\ 5&-1&-4\end{pmatrix}, \det A=58.
A^{-1}=\frac1{58} \begin{pmatrix} -2&12&6\\\\ 22&-16&-8\\\\ -8&19&-5 \end{pmatrix}.

Passo 4

Matriz 5\times 5 triangular (caso (s)):
A=\begin{pmatrix} 2&0&0&1&-2\\\\ 0&1&0&0&1\\\\ 0&0&2&0&0\\\\ 0&0&0&3&2\\\\ 0&0&0&0&3 \end{pmatrix},
como todos os pivôs na diagonal são não-zeros, A é invertível. Aplicando substituição retrógrada, obtém-se
A^{-1}= \begin{pmatrix} \tfrac12&0&0&-\tfrac{1}{6}&\tfrac{1}{9}\\\\ 0&1&0&0&-\tfrac13\\\\ 0&0&\tfrac12&0&0\\\\ 0&0&0&\tfrac13&-\tfrac{2}{9}\\\\ 0&0&0&0&\tfrac13 \end{pmatrix}.

Passo 5

Matrizes 4\times 4 [(g), (h), ®, §, (q), (t), (u)]: todas têm determinante \neq 0 (confira cálculo) e suas inversas se obtêm com Gauss–Jordan. O resultado final é:
(g) A^{-1}=\begin{pmatrix} 1&0&0&1\\\\ 0&\tfrac12&0&0\\\ \tfrac12&-\tfrac34&\tfrac12&0\\\\ 1&-\tfrac12&1&1 \end{pmatrix},
(h) A^{-1}=\begin{pmatrix} -1&-\tfrac12&\tfrac{1}{16}&-\tfrac{1}{16}\\\\ 3&1&-\tfrac{7}{16}&\tfrac{7}{16}\\\\ -\tfrac12&-\tfrac14&\tfrac{3}{16}&-\tfrac{1}{16}\\\\ -\tfrac32&-\tfrac34&\tfrac{7}{16}&\tfrac{9}{16} \end{pmatrix},
® A^{-1}=\begin{pmatrix} 0&1&1&-2\\\\ -1&0&0&1\\\ \tfrac12&-\tfrac12&0&\tfrac12\\\\ -\tfrac32&\tfrac12&1&-\tfrac12 \end{pmatrix},
§ A^{-1}=\begin{pmatrix} 1&0&0&-1\\\\ 0&1&0&0\\\\ -\tfrac12&3&1&-6\\\\ -\tfrac16&-\tfrac16&0&\tfrac13 \end{pmatrix},
(q) A^{-1}=\begin{pmatrix} \tfrac{1}{2}&\tfrac12&0&-1\\\ \tfrac32&-\tfrac12&1&-1\\\\ 0&1&-1&0\\\\ -\tfrac52&\tfrac52&1&-1 \end{pmatrix},
(t) A^{-1}=\begin{pmatrix} -1&0&0&0\\\\ 4&1&0&0\\\\ -\tfrac43&-\tfrac13&\tfrac13&0\\\ \tfrac13&-\tfrac13&-1&-1 \end{pmatrix},
(u) A^{-1}=\begin{pmatrix} \tfrac12&0&\tfrac12&0\\\\ 0&1&0&-1\\\ \tfrac12&0&-\tfrac12&0\\\\ 0&-1&0&1 \end{pmatrix}.

Resposta

(a) \displaystyle\begin{pmatrix}1&0\\\ -1&-\tfrac12\end{pmatrix}
(b) \displaystyle\begin{pmatrix}\tfrac{3}{19}&-\tfrac{7}{19}\\\ -\tfrac{1}{19}&-\tfrac{4}{19}\end{pmatrix}
(c) \displaystyle\frac1{100}\begin{pmatrix}40&13&3\\\ -20&-19&11\\\ 0&-10&-10\end{pmatrix}
(d) \displaystyle\begin{pmatrix}\tfrac13&\tfrac13&0\\\ -\tfrac23&\tfrac13&0\\tfrac{1}{9}&0&-\tfrac{1}{9}\end{pmatrix}
(e) \displaystyle\begin{pmatrix}-\tfrac1{24}&\tfrac5{24}&\tfrac16\\tfrac{13}{24}&\tfrac7{24}&-\tfrac16\\tfrac5{24}&-\tfrac1{24}&\tfrac16\end{pmatrix}
(f) \displaystyle\begin{pmatrix}\tfrac1{31}&-\tfrac6{31}&\tfrac6{31}\\\ -\tfrac7{31}&\tfrac{42}{31}&-\tfrac{11}{31}\\tfrac5{31}&\tfrac1{31}&-\tfrac1{31}\end{pmatrix}
(g) \displaystyle\begin{pmatrix}1&0&0&1\\\ 0&\tfrac12&0&0\\tfrac12&-\tfrac34&\tfrac12&0\\\ 1&-\tfrac12&1&1\end{pmatrix}
(h) \displaystyle\begin{pmatrix}-1&-\tfrac12&\tfrac{1}{16}&-\tfrac{1}{16}\\\ 3&1&-\tfrac{7}{16}&\tfrac{7}{16}\\\ -\tfrac12&-\tfrac14&\tfrac{3}{16}&-\tfrac{1}{16}\\\ -\tfrac32&-\tfrac34&\tfrac{7}{16}&\tfrac{9}{16}\end{pmatrix}
(i) \displaystyle\begin{pmatrix}-3&-5\\\ 2&3\end{pmatrix}
(j) \displaystyle\begin{pmatrix}1&\tfrac12\\\ 2&\tfrac54\end{pmatrix}
(n) \displaystyle\begin{pmatrix}-\tfrac{11}{39}&\tfrac7{39}&\tfrac8{39}\\tfrac{16}{39}&\tfrac1{39}&-\tfrac{10}{39}\\\ -\tfrac2{39}&-\tfrac5{39}&\tfrac{11}{39}\end{pmatrix}
(o) não invertível
§ \displaystyle\begin{pmatrix}1&0&0&-1\\\ 0&1&0&0\\\ -\tfrac12&3&1&-6\\\ -\tfrac16&-\tfrac16&0&\tfrac13\end{pmatrix}
(q) \displaystyle\begin{pmatrix}\tfrac12&\tfrac12&0&-1\\tfrac32&-\tfrac12&1&-1\\\ 0&1&-1&0\\\ -\tfrac52&\tfrac52&1&-1\end{pmatrix}
( 1) não invertível
(m) \displaystyle\frac1{58}\begin{pmatrix}-2&12&6\\\ 22&-16&-8\\\ -8&19&-5\end{pmatrix}
® \displaystyle\begin{pmatrix}0&1&1&-2\\\ -1&0&0&1\\tfrac12&-\tfrac12&0&\tfrac12\\\ -\tfrac32&\tfrac12&1&-\tfrac12\end{pmatrix}
(s) \displaystyle\begin{pmatrix}\tfrac12&0&0&-\tfrac{1}{6}&\tfrac{1}{9}\\\ 0&1&0&0&-\tfrac13\\\ 0&0&\tfrac12&0&0\\\ 0&0&0&\tfrac13&-\tfrac{2}{9}\\\ 0&0&0&0&\tfrac13\end{pmatrix}
(t) \displaystyle\begin{pmatrix}-1&0&0&0\\\ 4&1&0&0\\\ -\tfrac43&-\tfrac13&\tfrac13&0\\tfrac13&-\tfrac13&-1&-1\end{pmatrix}
(u) \displaystyle\begin{pmatrix}\tfrac12&0&\tfrac12&0\\\ 0&1&0&-1\\tfrac12&0&-\tfrac12&0\\\ 0&-1&0&1\end{pmatrix}

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