Resolució sistema d’equacions

Considerem el sistema d’equacions següent: $$\begin{cases}2x – 2y + z &= -5 \\3x + y + 3z &= -1 \\4x – y – 2z &= -12\end{cases}$$ Resoleu-ho pel mètode de Cramer

Resolem el sistema d’equacions amb el mètode de Cramer:

$$\begin{cases}
2x – 2y + z = -5 \\
3x + y + 3z = -1 \\
4x – y – 2z = -12
\end{cases}$$

Matriu de coeficients i vector de termes independents

$$A = \begin{pmatrix}
2 & -2 & 1 \\
3 & 1 & 3 \\
4 & -1 & -2
\end{pmatrix}, \quad
\vec{b} = \begin{pmatrix}
-5 \\
-1 \\
-12
\end{pmatrix}$$

Determinant principal $D$

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

$$= 2(1) + 2(-18) + (-7) = 2 – 36 – 7 = -41$$

Determinant $D_x$

$$D_x = \begin{vmatrix}
-5 & -2 & 1 \\
-1 & 1 & 3 \\
-12 & -1 & -2
\end{vmatrix}
= -5 \cdot \begin{vmatrix} 1 & 3 \\ -1 & -2 \end{vmatrix}(-2) \cdot \begin{vmatrix} -1 & 3 \\ -12 & -2 \end{vmatrix}1 \cdot \begin{vmatrix} -1 & 1 \\ -12 & -1 \end{vmatrix}$$

$$= -5(1) + 2(38) + 13 = -5 + 76 + 13 = 84$$

Determinant $D_y$

$$D_y = \begin{vmatrix}
2 & -5 & 1 \\
3 & -1 & 3 \\
4 & -12 & -2
\end{vmatrix}
= 2 \cdot \begin{vmatrix} -1 & 3 \\ -12 & -2 \end{vmatrix}(-5) \cdot \begin{vmatrix} 3 & 3 \\ 4 & -2 \end{vmatrix}1 \cdot \begin{vmatrix} 3 & -1 \\ 4 & -12 \end{vmatrix}$$

$$= 2(38) + 5(-18) + (-32) = 76 – 90 – 32 = -46$$

Determinant $D_z$

$$D_z = \begin{vmatrix}
2 & -2 & -5 \\
3 & 1 & -1 \\
4 & -1 & -12
\end{vmatrix}
= 2 \cdot \begin{vmatrix} 1 & -1 \\ -1 & -12 \end{vmatrix}(-2) \cdot \begin{vmatrix} 3 & -1 \\ 4 & -12 \end{vmatrix}(-5) \cdot \begin{vmatrix} 3 & 1 \\ 4 & -1 \end{vmatrix}$$

$$= 2(-13) + 2(-32) + (-5)(-7) = -26 – 64 + 35 = -55$$

Solució amb la regla de Cramer

$$x = \frac{D_x}{D} = \frac{84}{-41}, \quad
y = \frac{D_y}{D} = \frac{-46}{-41} = \frac{46}{41}, \quad
z = \frac{D_z}{D} = \frac{-55}{-41} = \frac{55}{41}$$

$$\boxed{x = -\frac{84}{41}, \quad y = \frac{46}{41}, \quad z = \frac{55}{41}}$$