Resolvendo o sistema de equações lineares
3x - y + 2z = 7
2x - 3y + z = -1
x + 2y - z = 2
encontramos y igual a
A) 1 B) 3 C) 5 D) 2 E) 4
Gabarito: d) 2
Soluções para a tarefa
Respondido por
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usando o metodo de cramer:
![D = \left[\begin{array}{ccc}3&-1&2\\2&-3&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}3&-1&\\2&-3&\\1&2&\end{array}\right] D = 3*-3*-1-1*1*1+2*2*2-2*-3*1-3*1*2+1*2*-1 = 14](https://tex.z-dn.net/?f=D+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B-1%26amp%3B2%5C%5C2%26amp%3B-3%26amp%3B1%5C%5C1%26amp%3B2%26amp%3B-1%5Cend%7Barray%7D%5Cright%5D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B-1%26amp%3B%5C%5C2%26amp%3B-3%26amp%3B%5C%5C1%26amp%3B2%26amp%3B%5Cend%7Barray%7D%5Cright%5D+D+%3D+3%2A-3%2A-1-1%2A1%2A1%2B2%2A2%2A2-2%2A-3%2A1-3%2A1%2A2%2B1%2A2%2A-1+%3D+14 "D = \left[\begin{array}{ccc}3&-1&2\\2&-3&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}3&-1&\\2&-3&\\1&2&\end{array}\right] D = 3*-3*-1-1*1*1+2*2*2-2*-3*1-3*1*2+1*2*-1 = 14")
![Dy = \left[\begin{array}{ccc}3&7&2\\2&-1&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}3&7&\\2&-1&\\1&2&\end{array}\right] Dy = 3*-1*-1+7*1*1+2*2*2-2*-1*1-3*1*2-7*2*-1 = 28 y = Dy/D = 28/14 = 2](https://tex.z-dn.net/?f=Dy+%3D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B7%26amp%3B2%5C%5C2%26amp%3B-1%26amp%3B1%5C%5C1%26amp%3B2%26amp%3B-1%5Cend%7Barray%7D%5Cright%5D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B7%26amp%3B%5C%5C2%26amp%3B-1%26amp%3B%5C%5C1%26amp%3B2%26amp%3B%5Cend%7Barray%7D%5Cright%5D+Dy+%3D+3%2A-1%2A-1%2B7%2A1%2A1%2B2%2A2%2A2-2%2A-1%2A1-3%2A1%2A2-7%2A2%2A-1+%3D+28+y+%3D+Dy%2FD+%3D+28%2F14+%3D+2 "Dy = \left[\begin{array}{ccc}3&7&2\\2&-1&1\\1&2&-1\end{array}\right] \left[\begin{array}{ccc}3&7&\\2&-1&\\1&2&\end{array}\right] Dy = 3*-1*-1+7*1*1+2*2*2-2*-1*1-3*1*2-7*2*-1 = 28 y = Dy/D = 28/14 = 2")
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