LEMNISCATA
Matemàtiques, física, química…
Apliquem el mètode sobre la matriu ampliada:
$$\begin{bmatrix}
1 & 1 & 1 & | & 0 \\
2 & -5 & -2 & | & -2 \\
3 & 4 & 1 & | & 8
\end{bmatrix}$$
Fila 2 ← F2 − 2·F1
$$\begin{bmatrix}
0 & -7 & -4 & | & -2
\end{bmatrix}$$
Fila 3 ← F3 − 3·F1
$$\begin{bmatrix}
0 & 1 & -2 & | & 8
\end{bmatrix}$$
Matriu resultant:
$$\begin{bmatrix}
1 & 1 & 1 & | & 0 \\
0 & -7 & -4 & | & -2 \\
0 & 1 & -2 & | & 8
\end{bmatrix}$$
Fila 2 ← F2 ÷ (−7)
$$\begin{bmatrix}
0 & 1 & \frac{4}{7} & | & \frac{2}{7}
\end{bmatrix}$$
Matriu:
$$\begin{bmatrix}
1 & 1 & 1 & | & 0 \\
0 & 1 & \frac{4}{7} & | & \frac{2}{7} \\
0 & 1 & -2 & | & 8
\end{bmatrix}$$
Fila 3 ← F3 − F2
Càlculs:
$$-2 – \frac{4}{7} = -\frac{18}{7}$$
$$8 – \frac{2}{7} = \frac{54}{7}$$
Fila 3:
$$\begin{bmatrix}
0 & 0 & -\frac{18}{7} & | & \frac{54}{7}
\end{bmatrix}$$
Matriu:
$$\begin{bmatrix}
1 & 1 & 1 & | & 0 \\
0 & 1 & \frac{4}{7} & | & \frac{2}{7} \\
0 & 0 & -\frac{18}{7} & | & \frac{54}{7}
\end{bmatrix}$$
$$-\frac{18}{7} z = \frac{54}{7}$$
$$z = -3$$
$$\boxed{z = -3}$$
$$y + \frac{4}{7}(-3) = \frac{2}{7}$$
$$y – \frac{12}{7} = \frac{2}{7}
\quad\Rightarrow\quad
y = \frac{14}{7} = 2$$
$$\boxed{y = 2}$$
$$x + y + z = 0$$
$$x + 2 – 3 = 0 \quad\Rightarrow\quad x = 1$$
$$\boxed{x = 1}$$
$$\boxed{(x,y,z) = (1,2,-3)}$$