Sistema d’equacions. Resolució pel mètode de Cramer

Siga el següent sistema de tres equacions amb tres incògnites: $$\begin{cases} \frac{x}{2} + \frac{y}{2} – \frac{z}{3} = 3 \\ \frac{x}{3} + \frac{y}{6} – \frac{z}{2} = -5 \\ \frac{x}{6} – \frac{y}{3} + \frac{z}{6} = 0 \end{cases}$$

Resolem el sistema d’equacions amb el mètode de Cramer:

$$\begin{cases}
\frac{x}{2} + \frac{y}{2} – \frac{z}{3} = 3 \\
\frac{x}{3} + \frac{y}{6} – \frac{z}{2} = -5 \\
\frac{x}{6} – \frac{y}{3} + \frac{z}{6} = 0
\end{cases}$$

Escrivim la matriu de coeficients $A$, el vector d’incògnites $\vec{x}$, i el vector de termes independents $\vec{b}$:

$$A = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & -\frac{1}{3} \\
\frac{1}{3} & \frac{1}{6} & -\frac{1}{2} \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{6}
\end{pmatrix}, \quad
\vec{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad
\vec{b} = \begin{pmatrix} 3 \\ -5 \\ 0 \end{pmatrix}$$

\section*{Determinant principal $D$}

$$D = \begin{vmatrix}
\frac{1}{2} & \frac{1}{2} & -\frac{1}{3} \\
\frac{1}{3} & \frac{1}{6} & -\frac{1}{2} \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{6}
\end{vmatrix}
= -\frac{5}{72} – \frac{5}{72} + \frac{5}{108} = -\frac{10}{72} + \frac{5}{108} = -\frac{5}{54}$$

Determinant $D_x$

$$D_x = \begin{vmatrix}
3 & \frac{1}{2} & -\frac{1}{3} \\
-5 & \frac{1}{6} & -\frac{1}{2} \\
0 & -\frac{1}{3} & \frac{1}{6}
\end{vmatrix}
= -\frac{5}{12} + \frac{5}{12} – \frac{5}{9} = -\frac{5}{9}$$

Determinant $D_y$

$$D_y = \begin{vmatrix}
\frac{1}{2} & 3 & -\frac{1}{3} \\
\frac{1}{3} & -5 & -\frac{1}{2} \\
\frac{1}{6} & 0 & \frac{1}{6}
\end{vmatrix}
= -\frac{5}{12} – \frac{5}{12} – \frac{5}{18} = -\frac{10}{12} – \frac{5}{18} = -\frac{10}{9}$$

Determinant $D_z$

$$D_z = \begin{vmatrix}
\frac{1}{2} & \frac{1}{2} & 3 \\
\frac{1}{3} & \frac{1}{6} & -5 \\
\frac{1}{6} & -\frac{1}{3} & 0
\end{vmatrix}
= \frac{1}{2} \cdot \left( \frac{1}{6} \cdot 0 – (-\frac{1}{3}) \cdot (-5) \right)\frac{1}{2} \cdot \left( \frac{1}{3} \cdot 0 – \frac{1}{6} \cdot (-5) \right)3 \cdot \left( \frac{1}{3} \cdot (-\frac{1}{3}) – \frac{1}{6} \cdot \frac{1}{6} \right)$$

$$= \frac{1}{2} \cdot (-\frac{5}{3}) – \frac{1}{2} \cdot \frac{5}{6} + 3 \cdot (-\frac{5}{36})
= -\frac{5}{6} – \frac{5}{12} – \frac{15}{36} = -\frac{5}{6} – \frac{5}{12} – \frac{5}{12} = -\frac{5}{3}$$

Solució amb la regla de Cramer

$$x = \frac{D_x}{D} = \frac{-\frac{5}{9}}{-\frac{5}{54}} = 6, \quad
y = \frac{D_y}{D} = \frac{-\frac{10}{9}}{-\frac{5}{54}} = 12, \quad
z = \frac{D_z}{D} = \frac{-\frac{5}{3}}{-\frac{5}{54}} = 18$$

$$\boxed{x = 6, \quad y = 12, \quad z = 18}$$