Resolver los sgtes ejercicios de sistemas de ecuaciones lineales con tres incógnitas :
1) x - 3y + 2z = - 1
2) 4x +5y+2z=6
3) -x + 2y + z = -2
1) -x +2y +4z = 1
2) 4x +6y -2z = 2
3) x - y + 6z = 2
Respuestas
Respuesta dada por:
3
Luisa,
Vamos a usar el método de Cramer
x = D/Dx y = D/Dy z = D/Dz
D = determinante del sistema
Dx, Dy, Dz = determinante de cada variable
1)
![D= \left[\begin{array}{ccc}1&-3&2\\4&5&2&-1&2&1\end{array}\right] =45 \\ \\ \\ Dx= \left[\begin{array}{ccc}-11&-3&2\\6&5&2\\-2&2&1\end{array}\right] =73 \\ \\ \\ Dy= \left[\begin{array}{ccc}1&-1&2\\4&6&2\\-1&-2&1\end{array}\right] =12 \\ \\ \\ Dz= \left[\begin{array}{ccc}1&4&-1\\-3&5&6\\-1&2&-2\end{array}\right] =-41 D= \left[\begin{array}{ccc}1&-3&2\\4&5&2&-1&2&1\end{array}\right] =45 \\ \\ \\ Dx= \left[\begin{array}{ccc}-11&-3&2\\6&5&2\\-2&2&1\end{array}\right] =73 \\ \\ \\ Dy= \left[\begin{array}{ccc}1&-1&2\\4&6&2\\-1&-2&1\end{array}\right] =12 \\ \\ \\ Dz= \left[\begin{array}{ccc}1&4&-1\\-3&5&6\\-1&2&-2\end{array}\right] =-41](https://tex.z-dn.net/?f=+D%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B-3%26amp%3B2%5C%5C4%26amp%3B5%26amp%3B2%26amp%3B-1%26amp%3B2%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D45+%5C%5C++%5C%5C++%5C%5C++++++++++++++++++++++++++Dx%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-11%26amp%3B-3%26amp%3B2%5C%5C6%26amp%3B5%26amp%3B2%5C%5C-2%26amp%3B2%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D73+%5C%5C++%5C%5C++%5C%5C++Dy%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B-1%26amp%3B2%5C%5C4%26amp%3B6%26amp%3B2%5C%5C-1%26amp%3B-2%26amp%3B1%5Cend%7Barray%7D%5Cright%5D+%3D12+%5C%5C++%5C%5C++%5C%5C+Dz%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B4%26amp%3B-1%5C%5C-3%26amp%3B5%26amp%3B6%5C%5C-1%26amp%3B2%26amp%3B-2%5Cend%7Barray%7D%5Cright%5D+%3D-41)
x = 45/73 y = 15/4 z = - 45/41
2)
![D=\left[\begin{array}{ccc}-1&2&4\\4&6&-2\\1&-1&6\end{array}\right] =-126 \\ \\ \\ Dx= \left[\begin{array}{ccc}1&2&4\\4&6&-2\\1&-1&6\end{array}\right] -54 \\ \\ \\ Dy= \left[\begin{array}{ccc}-1&1&4\\4&2&-2\\1&2&6\end{array}\right] =-18 \\ \\ \\ Dz= \left[\begin{array}{ccc}-1&4&1\\4&6&2\\1&-1&2\end{array}\right] =-36 D=\left[\begin{array}{ccc}-1&2&4\\4&6&-2\\1&-1&6\end{array}\right] =-126 \\ \\ \\ Dx= \left[\begin{array}{ccc}1&2&4\\4&6&-2\\1&-1&6\end{array}\right] -54 \\ \\ \\ Dy= \left[\begin{array}{ccc}-1&1&4\\4&2&-2\\1&2&6\end{array}\right] =-18 \\ \\ \\ Dz= \left[\begin{array}{ccc}-1&4&1\\4&6&2\\1&-1&2\end{array}\right] =-36](https://tex.z-dn.net/?f=++D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B2%26amp%3B4%5C%5C4%26amp%3B6%26amp%3B-2%5C%5C1%26amp%3B-1%26amp%3B6%5Cend%7Barray%7D%5Cright%5D+%3D-126+%5C%5C++%5C%5C++%5C%5C++Dx%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B4%5C%5C4%26amp%3B6%26amp%3B-2%5C%5C1%26amp%3B-1%26amp%3B6%5Cend%7Barray%7D%5Cright%5D+-54+%5C%5C++%5C%5C++%5C%5C++Dy%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B1%26amp%3B4%5C%5C4%26amp%3B2%26amp%3B-2%5C%5C1%26amp%3B2%26amp%3B6%5Cend%7Barray%7D%5Cright%5D+%3D-18+%5C%5C++%5C%5C++%5C%5C+Dz%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B4%26amp%3B1%5C%5C4%26amp%3B6%26amp%3B2%5C%5C1%26amp%3B-1%26amp%3B2%5Cend%7Barray%7D%5Cright%5D+%3D-36)
x = -126/-54 = 7/3
y = -126/-18 = 7
z = -126/-36 = 7/2
Vamos a usar el método de Cramer
x = D/Dx y = D/Dy z = D/Dz
D = determinante del sistema
Dx, Dy, Dz = determinante de cada variable
1)
x = 45/73 y = 15/4 z = - 45/41
2)
x = -126/-54 = 7/3
y = -126/-18 = 7
z = -126/-36 = 7/2
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