To solve the given system of linear equations, we can represent it in matrix form and use a numerical method to find the values of \(x\), \(y\), and \(z\). We will use NumPy's linear algebra solver to find the solution. The matrix form of the equations is \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the column matrix of variables, and \(B\) is the column matrix of constants.
Dado el sistema de ecuaciones:
\[
\begin{align*}
- & \quad 2x - y + kz = 1 \\
- & \quad -x + y - kz = 0 \\
- & \quad 2x - y + 2kz = -1
\end{align*}
\]
Para simplificar el sistema, podemos eliminar una variable. Vamos a eliminar \(y\) sumando las ecuaciones (1) y (2):
\[
(2x - y + kz) + (-x + y - kz) = 1 + 0
\]
Esto simplifica a:
\[
x = 1
\]
Sustituimos \(x = 1\) en las ecuaciones (1) y (3):
Para la ecuación (1):
\[
2(1) - y + kz = 1 \implies 2 - y + kz = 1 \implies -y + kz = -1 \implies y = kz + 1
\]
Para la ecuación (3):
\[
2(1) - y + 2kz = -1 \implies 2 - y + 2kz = -1 \implies -y + 2kz = -3 \implies y = 2kz + 3
\]
Igualamos las dos expresiones obtenidas para \(y\):
\[
kz + 1 = 2kz + 3
\]
Resolviendo para \(z\):
\[
kz + 1 = 2kz + 3 \implies kz - 2kz = 3 - 1 \implies -kz = 2 \implies z = -\frac{2}{k}
\]
Sustituimos \(z = -\frac{2}{k}\) en \(y = kz + 1\):
\[
y = k\left(-\frac{2}{k}\right) + 1 = -2 + 1 = -1
\]
Las soluciones del sistema son:
\[
\boxed{x = 1}, \quad \boxed{y = -1}, \quad \boxed{z = -\frac{2}{k}}
\]
Answered
