2x+3y-4z=-9
x+y+6z012
x-y+z=4
regla de kramer 3x3
anyimendozaparr:
con procedimientos
Respuestas a la pregunta
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1
Expresarlo en matrices:
![x = \frac{ \left[\begin{array}{ccc}-9&1&1\\12&1&-1\\4&6&1\end{array}\right] }{\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-9\\12\\4\end{array}\right]](https://tex.z-dn.net/?f=x+%3D++%5Cfrac%7B++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-9%26amp%3B1%26amp%3B1%5C%5C12%26amp%3B1%26amp%3B-1%5C%5C4%26amp%3B6%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%7D%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B-4%5C%5C1%26amp%3B1%26amp%3B6%5C%5C1%26amp%3B-1%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D+%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-9%5C%5C12%5C%5C4%5Cend%7Barray%7D%5Cright%5D+ "x = \frac{ \left[\begin{array}{ccc}-9&1&1\\12&1&-1\\4&6&1\end{array}\right] }{\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }\left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-9\\12\\4\end{array}\right]")
Entonces:
![y = \frac{\left[\begin{array}{ccc}2&-9&1\\1&12&-1\\1&4&1\end{array}\right] {\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }](https://tex.z-dn.net/?f=y+%3D++%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B-9%26amp%3B1%5C%5C1%26amp%3B12%26amp%3B-1%5C%5C1%26amp%3B4%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B-4%5C%5C1%26amp%3B1%26amp%3B6%5C%5C1%26amp%3B-1%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%7D "y = \frac{\left[\begin{array}{ccc}2&-9&1\\1&12&-1\\1&4&1\end{array}\right] {\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }")
![z =\frac{ \left[\begin{array}{ccc}2&3&-9\\1&1&12\\1&-1&4\end{array}\right] }{\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }](https://tex.z-dn.net/?f=z+%3D%5Cfrac%7B+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B-9%5C%5C1%26amp%3B1%26amp%3B12%5C%5C1%26amp%3B-1%26amp%3B4%5Cend%7Barray%7D%5Cright%5D+%7D%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B3%26amp%3B-4%5C%5C1%26amp%3B1%26amp%3B6%5C%5C1%26amp%3B-1%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%7D%0A "z =\frac{ \left[\begin{array}{ccc}2&3&-9\\1&1&12\\1&-1&4\end{array}\right] }{\left[\begin{array}{ccc}2&3&-4\\1&1&6\\1&-1&1\end{array}\right] }")
Entonces:
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