1. Plantegem el problema: Resoldre el sistema lineal per el mètode de Gauss: $$\begin{cases} x + 2y - z = 0 \\ 2x - y + z = -1 \\ -x + y + z = 6 \end{cases}$$ 2. Escrivim la matriu augmentada del sistema: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 2 & -1 & 1 & -1 \\ -1 & 1 & 1 & 6 \end{array}\right]$$ 3. Fem operacions per portar la matriu a forma triangular superior. - Fem $R_2 \leftarrow R_2 - 2R_1$: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ \cancel{2} - 2\times 1 & -1 - 2\times 2 & 1 - 2\times (-1) & -1 - 2\times 0 \\ -1 & 1 & 1 & 6 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & -1 \\ -1 & 1 & 1 & 6 \end{array}\right]$$ - Fem $R_3 \leftarrow R_3 + R_1$: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & -1 \\ \cancel{-1} + 1 & 1 + 2 & 1 + (-1) & 6 + 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & -1 \\ 0 & 3 & 0 & 6 \end{array}\right]$$ - Fem $R_3 \leftarrow R_3 - \frac{3}{-5} R_2 = R_3 + \frac{3}{5} R_2$: $$\left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & -1 \\ 0 & 3 + \cancel{3} & 0 + \frac{3}{5} \times 3 & 6 + \frac{3}{5} \times (-1) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & -5 & 3 & -1 \\ 0 & 0 & \frac{9}{5} & \frac{27}{5} \end{array}\right]$$ 4. Resolem el sistema triangular: - De la tercera fila: $$\frac{9}{5} z = \frac{27}{5} \implies z = \frac{27}{5} \div \frac{9}{5} = 3$$ - De la segona fila: $$-5 y + 3 z = -1 \implies -5 y + 3 \times 3 = -1 \implies -5 y + 9 = -1 \implies -5 y = -10 \implies y = 2$$ - De la primera fila: $$x + 2 y - z = 0 \implies x + 2 \times 2 - 3 = 0 \implies x + 4 - 3 = 0 \implies x + 1 = 0 \implies x = -1$$ 5. Solució final: $$\boxed{(x,y,z) = (-1, 2, 3)}$$