Respuestas
Respuesta dada por:
1
A* X + B = C
A *X = C - B
conseguimos la matriz identidad multiplicando


C - B =![\left[\begin{array}{ccc}1&-1&2\\0&1&3\\\end{array}\right] - \left[\begin{array}{ccc}-1&0&1\\2&1&0\\\end{array}\right] = \left[\begin{array}{ccc}2&-1&1\\-2&0&3\\\end{array}\right] \left[\begin{array}{ccc}1&-1&2\\0&1&3\\\end{array}\right] - \left[\begin{array}{ccc}-1&0&1\\2&1&0\\\end{array}\right] = \left[\begin{array}{ccc}2&-1&1\\-2&0&3\\\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B-1%26amp%3B2%5C%5C0%26amp%3B1%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+-++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26amp%3B0%26amp%3B1%5C%5C2%26amp%3B1%26amp%3B0%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B-1%26amp%3B1%5C%5C-2%26amp%3B0%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+)

det A =![\left[\begin{array}{ccc}3&5\\-1&-2\\\end{array}\right] =(3*-2) - (5*-1) = -1 \left[\begin{array}{ccc}3&5\\-1&-2\\\end{array}\right] =(3*-2) - (5*-1) = -1](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B5%5C%5C-1%26amp%3B-2%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D%283%2A-2%29+-+%285%2A-1%29+%3D+-1)
Para hallar el adjunto de la traspuesta.
![\left[\begin{array}{ccc}3&5\\-1&-2\\\end{array}\right] \left[\begin{array}{ccc}3&5\\-1&-2\\\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B5%5C%5C-1%26amp%3B-2%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
Se elimina fila 1 columna 1 y queda -2
Se elimina fila 1 columna 2 y queda -1
Se elimina fila columna 1 fila 2 y queda 5
Se elimina fila columna 2 fila 1 y queda 3
![\left[\begin{array}{ccc}-2&1\\-5&3\\\end{array}\right] \left[\begin{array}{ccc}-2&1\\-5&3\\\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B1%5C%5C-5%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
trasponemos.
![\left[\begin{array}{ccc}-2&-5\\1&3\\\end{array}\right] \left[\begin{array}{ccc}-2&-5\\1&3\\\end{array}\right]](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B-5%5C%5C1%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
![= \frac{ \left[\begin{array}{ccc}-2&-5\\1&3\\\end{array}\right] }{-1} = \left[\begin{array}{ccc}2&5\\-1&-3\\\end{array}\right] = \frac{ \left[\begin{array}{ccc}-2&-5\\1&3\\\end{array}\right] }{-1} = \left[\begin{array}{ccc}2&5\\-1&-3\\\end{array}\right]](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26amp%3B-5%5C%5C1%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+%7D%7B-1%7D+%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B5%5C%5C-1%26amp%3B-3%5C%5C%5Cend%7Barray%7D%5Cright%5D+)

![X = \left[\begin{array}{ccc}2&5\\-1&-3\\\end{array}\right] * \left[\begin{array}{ccc}2&-1&1\\-2&0&3\\\end{array}\right] = \left[\begin{array}{ccc}-6&-2&17\\4&1&-10\\\end{array}\right] X = \left[\begin{array}{ccc}2&5\\-1&-3\\\end{array}\right] * \left[\begin{array}{ccc}2&-1&1\\-2&0&3\\\end{array}\right] = \left[\begin{array}{ccc}-6&-2&17\\4&1&-10\\\end{array}\right]](https://tex.z-dn.net/?f=++X+%3D+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B5%5C%5C-1%26amp%3B-3%5C%5C%5Cend%7Barray%7D%5Cright%5D+%2A+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%26amp%3B-1%26amp%3B1%5C%5C-2%26amp%3B0%26amp%3B3%5C%5C%5Cend%7Barray%7D%5Cright%5D+%3D+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%26amp%3B-2%26amp%3B17%5C%5C4%26amp%3B1%26amp%3B-10%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
2* 2 + 5 * -2 = -6 2*-1 + 5*0 = -2 2*1 + 5*3 = 17
-1*2 + (-3*-2) = 4 -1*-1+(-3*0)= 1 -1*1+(-3*3)= -10
A *X = C - B
conseguimos la matriz identidad multiplicando
C - B =
det A =
Para hallar el adjunto de la traspuesta.
Se elimina fila 1 columna 1 y queda -2
Se elimina fila 1 columna 2 y queda -1
Se elimina fila columna 1 fila 2 y queda 5
Se elimina fila columna 2 fila 1 y queda 3
trasponemos.
2* 2 + 5 * -2 = -6 2*-1 + 5*0 = -2 2*1 + 5*3 = 17
-1*2 + (-3*-2) = 4 -1*-1+(-3*0)= 1 -1*1+(-3*3)= -10
star78:
Perfecto, muchas gracias
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