obtenha caso exista a inversa de cada matriz
c=-12 6
18 -9
d=4 4
8 0
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Para que a inversa de uma matriz exista, o seu determinante deve ser diferente de zero.
Para encontrar a matriz inversa, usaremos o fato de que se multiplicarmos uma matriz, pela sua inversa, devemos obter uma madriz identidade.
A . A^-1 = I
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![c = \left[\begin{array}{cc}-12&6\\18&-9\\\end{array}\right] c = \left[\begin{array}{cc}-12&6\\18&-9\\\end{array}\right]](https://tex.z-dn.net/?f=++c+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-12%26amp%3B6%5C%5C18%26amp%3B-9%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
det(c) = -12.(-9) - (18.6) = 0
Como o determinante é zero, a inversa não existe.
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![d = \left[\begin{array}{cc}4&4\\8&0\\\end{array}\right] d = \left[\begin{array}{cc}4&4\\8&0\\\end{array}\right]](https://tex.z-dn.net/?f=++d+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26amp%3B4%5C%5C8%26amp%3B0%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
det(d) = 4 . 0 - 4 . 8 = -32
Logo a matriz inversa existe. Então vamos calcular.
Pela definição lá de cima
![\left[\begin{array}{cc}4&4\\8&0\\\end{array}\right] . \left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right] \left[\begin{array}{cc}4&4\\8&0\\\end{array}\right] . \left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26amp%3B4%5C%5C8%26amp%3B0%5C%5C%5Cend%7Barray%7D%5Cright%5D++.++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26amp%3By%5C%5Cz%26amp%3Bk%5C%5C%5Cend%7Barray%7D%5Cright%5D++%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26amp%3B0%5C%5C0%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
Sabendo que,
será a nossa matriz inversa de d.
![\left[\begin{array}{cc}4&4\\8&0\\\end{array}\right] . \left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right] \\ \\ \left[\begin{array}{cc}4x + 4z&4y+4k\\8x&8k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right] \\ \\ 8x = 0 \\ x = 0 \\ \\ 4x + 4z = 1 \\ 4.0 + 4z = 1 \\ 4z = 1 \\ z = 1/4 \left[\begin{array}{cc}4&4\\8&0\\\end{array}\right] . \left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right] \\ \\ \left[\begin{array}{cc}4x + 4z&4y+4k\\8x&8k\\\end{array}\right] = \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right] \\ \\ 8x = 0 \\ x = 0 \\ \\ 4x + 4z = 1 \\ 4.0 + 4z = 1 \\ 4z = 1 \\ z = 1/4](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26amp%3B4%5C%5C8%26amp%3B0%5C%5C%5Cend%7Barray%7D%5Cright%5D+.+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26amp%3By%5C%5Cz%26amp%3Bk%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26amp%3B0%5C%5C0%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D++%5C%5C++%5C%5C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4x+%2B+4z%26amp%3B4y%2B4k%5C%5C8x%26amp%3B8k%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26amp%3B0%5C%5C0%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C+8x+%3D+0+%5C%5C+x+%3D+0+%5C%5C++%5C%5C+4x+%2B+4z+%3D+1+%5C%5C+4.0+%2B+4z+%3D+1+%5C%5C+4z+%3D+1+%5C%5C+z+%3D+1%2F4+)

Logo a inversa de
será:
![\left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}0& -\frac{1}{8} \\\frac{1}{4} &\frac{1}{8} \\\end{array}\right] \left[\begin{array}{cc}x&y\\z&k\\\end{array}\right] = \left[\begin{array}{cc}0& -\frac{1}{8} \\\frac{1}{4} &\frac{1}{8} \\\end{array}\right]](https://tex.z-dn.net/?f=+%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Dx%26amp%3By%5C%5Cz%26amp%3Bk%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D0%26amp%3B+-%5Cfrac%7B1%7D%7B8%7D+%5C%5C%5Cfrac%7B1%7D%7B4%7D+%26amp%3B%5Cfrac%7B1%7D%7B8%7D+%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Para encontrar a matriz inversa, usaremos o fato de que se multiplicarmos uma matriz, pela sua inversa, devemos obter uma madriz identidade.
A . A^-1 = I
----------------
det(c) = -12.(-9) - (18.6) = 0
Como o determinante é zero, a inversa não existe.
---------------------------
det(d) = 4 . 0 - 4 . 8 = -32
Logo a matriz inversa existe. Então vamos calcular.
Pela definição lá de cima
Sabendo que,
Logo a inversa de
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