Sistema d’equacions. Mètode de Cramer

Plantegem el sistema: $$\begin{cases}2x – y + 3z = 9 \\x + 4y – z = 1 \\3x + 2y + 2z = 12\end{cases}$$

1. Matriu de coeficients i vector de termes independents $$A =
\begin{pmatrix}
2 & -1 & 3 \\
1 & 4 & -1 \\
3 & 2 & 2
\end{pmatrix},
\quad
\mathbf{b} =
\begin{pmatrix}
9 \\ 1 \\ 12
\end{pmatrix}$$

2. Determinant principal
$$\Delta =
\begin{vmatrix}
2 & -1 & 3 \\
1 & 4 & -1 \\
3 & 2 & 2
\end{vmatrix}
= (2\cdot 4 \cdot 2) + (-1 \cdot -1 \cdot 3) + (3 \cdot 1 \cdot 2) – (3 \cdot 4 \cdot 3) – (2 \cdot -1 \cdot 2) – (-1 \cdot 1 \cdot 2)$$
$$\Delta = 16 + 3 + 6 – 36 + 4 + 2 = -5$$

3. Determinants per a cada variable

Per $x$: substituïm la 1a columna per $\mathbf{b}$:
$$A_x =
\begin{pmatrix}
9 & -1 & 3 \\
1 & 4 & -1 \\
12 & 2 & 2
\end{pmatrix}$$
$$\Delta_x = (9\cdot 4 \cdot 2) + (-1 \cdot -1 \cdot 12) + (3 \cdot 1 \cdot 2) – (3 \cdot 4 \cdot 12) – (9 \cdot -1 \cdot 2) – (-1 \cdot 1 \cdot 2)$$
$$\Delta_x = 72 + 12 + 6 – 144 + 18 + 2 = -34$$

Per $y$: substituïm la 2a columna per $\mathbf{b}$:
$$A_y =
\begin{pmatrix}
2 & 9 & 3 \\
1 & 1 & -1 \\
3 & 12 & 2
\end{pmatrix}$$
$$\Delta_y = (2\cdot 1 \cdot 2) + (9 \cdot -1 \cdot 3) + (3 \cdot 1 \cdot 12) – (3 \cdot 1 \cdot 3) – (2 \cdot -1 \cdot 12) – (9 \cdot 1 \cdot 2)$$
$$\Delta_y = 4 – 27 + 36 – 9 + 24 – 18 = 10$$

Per $z$: substituïm la 3a columna per $\mathbf{b}$:
$$A_z =
\begin{pmatrix}
2 & -1 & 9 \\
1 & 4 & 1 \\
3 & 2 & 12
\end{pmatrix}$$
$$\Delta_z = (2\cdot 4 \cdot 12) + (-1 \cdot 1 \cdot 3) + (9 \cdot 1 \cdot 2) – (9 \cdot 4 \cdot 3) – (2 \cdot 1 \cdot 2) – (-1 \cdot 1 \cdot 12)$$
$$\Delta_z = 96 – 3 + 18 – 108 – 4 + 12 = 11$$

4. Solució amb la fórmula de Cramer
$$x = \frac{\Delta_x}{\Delta} = \frac{-34}{-5} = 6.8, \quad
y = \frac{\Delta_y}{\Delta} = \frac{10}{-5} = -2, \quad
z = \frac{\Delta_z}{\Delta} = \frac{11}{-5} = -2.2$$

Solució final
$$\boxed{x = 6.8, \quad y = -2, \quad z = -2.2}$$