Sistema d’equacions. Sistema compatible determinat

Per resoldre el sistema següent: {x+2y−3z=−2,3x+z=0,2x−y+2z=3,\begin{cases} x + 2y – 3z = -2, \\ 3x + z = 0, \\ 2x – y + 2z = 3, \end{cases}

Per resoldre el sistema següent:

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

1. Representació matricial

Escrivim el sistema en forma de matriu:
$$A \cdot \mathbf{X} = \mathbf{b},$$
on:
$$A = \begin{pmatrix}
1 & 2 & -3 \\
3 & 0 & 1 \\
2 & -1 & 2
\end{pmatrix}, \quad
\mathbf{X} = \begin{pmatrix} x \ y \ z \end{pmatrix}, \quad
\mathbf{b} = \begin{pmatrix} -2 \ 0 \ 3 \end{pmatrix}.$$

2. Regla de Cramer

Per trobar $x$, $y$, i $z$, utilitzem la fórmula:
$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}.$$

On:

• $\Delta = \det(A)$: determinant de la matriu de coeficients.
• $\Delta_x, \Delta_y, \Delta_z$: determinants de les matrius resultants de substituir la columna corresponent per (\mathbf{b}).

3. Càlculs dels determinants

Per a cada cas:
$$\Delta = \begin{vmatrix}
1 & 2 & -3 \\
3 & 0 & 1 \\
2 & -1 & 2
\end{vmatrix}, \quad
\Delta_x = \begin{vmatrix}
-2 & 2 & -3 \\
0 & 0 & 1 \\
3 & -1 & 2
\end{vmatrix}, \quad
\Delta_y = \begin{vmatrix}
1 & -2 & -3 \\
3 & 0 & 1 \\
2 & 3 & 2
\end{vmatrix}, \quad
\Delta_z = \begin{vmatrix}
1 & 2 & -2 \\
3 & 0 & 0 \\
2 & -1 & 3
\end{vmatrix}.$$

Determinant principal \(\Delta\):
\[
\Delta = \begin{vmatrix}
1 & 2 & -3 \\
3 & 0 & 1 \\
2 & -1 & 2
\end{vmatrix}.
\]

Expansió per la primera fila:
\[
\Delta = 1 \cdot \begin{vmatrix}
0 & 1 \\
-1 & 2
\end{vmatrix}
– 2 \cdot \begin{vmatrix}
3 & 1 \\
2 & 2
\end{vmatrix}
– 3 \cdot \begin{vmatrix}
3 & 0 \\
2 & -1
\end{vmatrix}.
\]

Calculem els menors:
\[
\begin{vmatrix}
0 & 1 \\
-1 & 2
\end{vmatrix} = (0)(2) – (1)(-1) = 1,
\quad
\begin{vmatrix}
3 & 1 \\
2 & 2
\end{vmatrix} = (3)(2) – (1)(2) = 6 – 2 = 4,
\quad
\begin{vmatrix}
3 & 0 \\
2 & -1
\end{vmatrix} = (3)(-1) – (0)(2) = -3.
\]

Substituint:
\[
\Delta = 1(1) – 2(4) – 3(-3) = 1 – 8 + 9 = 2.
\]

Determinant auxiliar \(\Delta_x\):
\[
\Delta_x = \begin{vmatrix}
-2 & 2 & -3 \\
0 & 0 & 1 \\
3 & -1 & 2
\end{vmatrix}.
\]

Expansió per la primera fila:
\[
\Delta_x = (-2) \cdot \begin{vmatrix}
0 & 1 \\
-1 & 2
\end{vmatrix}
– 2 \cdot \begin{vmatrix}
0 & 1 \\
3 & 2
\end{vmatrix}
– 3 \cdot \begin{vmatrix}
0 & 0 \\
3 & -1
\end{vmatrix}.
\]

Els menors:
\[
\begin{vmatrix}
0 & 1 \\
-1 & 2
\end{vmatrix} = 1, \quad
\begin{vmatrix}
0 & 1 \\
3 & 2
\end{vmatrix} = (0)(2) – (1)(3) = -3, \quad
\begin{vmatrix}
0 & 0 \\
3 & -1
\end{vmatrix} = 0.
\]

Substituint:
\[
\Delta_x = (-2)(1) – 2(-3) – 3(0) = -2 + 6 + 0 = 4.
\]

Determinant auxiliar \(\Delta_y\):
\[
\Delta_y = \begin{vmatrix}
1 & -2 & -3 \\
3 & 0 & 1 \\
2 & 3 & 2
\end{vmatrix}.
\]

Expansió per la primera fila:
\[
\Delta_y = 1 \cdot \begin{vmatrix}
0 & 1 \\
3 & 2
\end{vmatrix}
+ 2 \cdot \begin{vmatrix}
3 & 1 \\
2 & 2
\end{vmatrix}
– 3 \cdot \begin{vmatrix}
3 & 0 \\
2 & 3
\end{vmatrix}.
\]

Els menors:
\[
\begin{vmatrix}
0 & 1 \\
3 & 2
\end{vmatrix} = -3, \quad
\begin{vmatrix}
3 & 1 \\
2 & 2
\end{vmatrix} = 4, \quad
\begin{vmatrix}
3 & 0 \\
2 & 3
\end{vmatrix} = 9.
\]

Substituint:
\[
\Delta_y = 1(-3) + 2(4) – 3(9) = -3 + 8 – 27 = -22.
\]

Determinant auxiliar \(\Delta_z\):
\[
\Delta_z = \begin{vmatrix}
1 & 2 & -2 \\
3 & 0 & 0 \\
2 & -1 & 3
\end{vmatrix}.
\]

Expansió per la primera fila:
\[
\Delta_z = 1 \cdot \begin{vmatrix}
0 & 0 \\
-1 & 3
\end{vmatrix}
– 2 \cdot \begin{vmatrix}
3 & 0 \\
2 & 3
\end{vmatrix}
+ (-2) \cdot \begin{vmatrix}
3 & 0 \\
2 & -1
\end{vmatrix}.
\]

Els menors:
\[
\begin{vmatrix}
0 & 0 \\
-1 & 3
\end{vmatrix} = 0, \quad
\begin{vmatrix}
3 & 0 \\
2 & 3
\end{vmatrix} = 9, \quad
\begin{vmatrix}
3 & 0 \\
2 & -1
\end{vmatrix} = -3.
\]

Substituint:
\[
\Delta_z = 1(0) – 2(9) + (-2)(-3) = 0 – 18 + 6 = -12.
\]

Solucions:
\[
x = \frac{\Delta_x}{\Delta} = \frac{4}{2} = 2, \quad
y = \frac{\Delta_y}{\Delta} = \frac{-22}{2} = -11, \quad
z = \frac{\Delta_z}{\Delta} = \frac{-12}{2} = -6.
\]

### Resultat final:
\[
x = 2, \quad y = -11, \quad z = -6.
\]