Resolució sistema d’equacions pel mètode de Cramer

Resolem el sistema d’equacions amb el mètode de Cramer: $$\begin{cases}-2x + 4y + z = 6 \\x – 4y – 3z = -2 \\-3x + 4y – 2z = 4\end{cases}$$

Matriu de coeficients i vector de termes independents

$$A = \begin{pmatrix}
-2 & 4 & 1 \\
1 & -4 & -3 \\
-3 & 4 & -2
\end{pmatrix}, \quad
\vec{b} = \begin{pmatrix}
6 \\
-2 \\
4
\end{pmatrix}$$

Determinant principal $D$

$$D = \begin{vmatrix}
-2 & 4 & 1 \\
1 & -4 & -3 \\
-3 & 4 & -2
\end{vmatrix}= -2 \cdot \begin{vmatrix} -4 & -3 \\ 4 & -2 \end{vmatrix}4 \cdot \begin{vmatrix} 1 & -3 \\ -3 & -2 \end{vmatrix}1 \cdot \begin{vmatrix} 1 & -4 \\ -3 & 4 \end{vmatrix}$$

$$= -2(8 + 12) – 4(-2 – 9) + (-8)
= -2(20) + 4(11) – 8 = -40 + 44 – 8 = \boxed{-4}$$

Determinant $D_x$

$$D_x = \begin{vmatrix}
6 & 4 & 1 \\
-2 & -4 & -3 \\
4 & 4 & -2
\end{vmatrix}
= 6 \cdot \begin{vmatrix} -4 & -3 \\ 4 & -2 \end{vmatrix}4 \cdot \begin{vmatrix} -2 & -3 \\ 4 & -2 \end{vmatrix}1 \cdot \begin{vmatrix} -2 & -4 \\ 4 & 4 \end{vmatrix}$$

$$= 6(8 + 12) – 4(4 + 12) + 8 = 6(20) – 4(16) + 8 = 120 – 64 + 8 = \boxed{64}$$

Determinant $D_y$

$$D_y = \begin{vmatrix}
-2 & 6 & 1 \\
1 & -2 & -3 \\
-3 & 4 & -2
\end{vmatrix}
= -2 \cdot \begin{vmatrix} -2 & -3 \\ 4 & -2 \end{vmatrix}6 \cdot \begin{vmatrix} 1 & -3 \\ -3 & -2 \end{vmatrix}1 \cdot \begin{vmatrix} 1 & -2 \\ -3 & 4 \end{vmatrix}$$

$$= -2(4 + 12) – 6(-2 – 9) + (-2)
= -2(16) + 6(11) – 2 = -32 + 66 – 2 = \boxed{32}$$

Determinant $D_z$

$$D_z = \begin{vmatrix}
-2 & 4 & 6 \\
1 & -4 & -2 \\
-3 & 4 & 4
\end{vmatrix}
= -2 \cdot \begin{vmatrix} -4 & -2 \\4 & 4 \end{vmatrix}4 \cdot \begin{vmatrix} 1 & -2 \\ -3 & 4 \end{vmatrix}6 \cdot \begin{vmatrix} 1 & -4 \\ -3 & 4 \end{vmatrix}$$

$$= -2(-16 + 8) – 4(4 – 6) + 6(4 – 12)
= -2(-8) – 4(-2) + 6(-8) = 16 + 8 – 48 = \boxed{-24}$$

Solució amb la regla de Cramer

$$x = \frac{D_x}{D} = \frac{64}{-4} = -16, \quad
y = \frac{D_y}{D} = \frac{32}{-4} = -8, \quad
z = \frac{D_z}{D} = \frac{-24}{-4} = 6$$

$$\boxed{x = -16, \quad y = -8, \quad z = 6}$$