Càlcul de la matriu inversa pel mètode de Gauss-Jordan

Donada la matriu: $$C = \begin{pmatrix} 2 & 1 & -1 \\ 4 & -2 & 3 \\ -1 & 3 & 0 \end{pmatrix}$$ Calculeu la matriu inversa pel mètode de Gauss-Jordan

Matriu $C$ i la matriu augmentada $[C\mid I]$:
$$\left[\begin{array}{ccc|ccc}
2 & 1 & -1 & 1 & 0 & 0\\
4 & -2 & 3 & 0 & 1 & 0\\
-1 & 3 & 0 & 0 & 0 & 1
\end{array}\right]$$

1. Posar pivot 1 a la primera fila: divideixo $R_1$ per (2):
$$R_1\leftarrow \tfrac{1}{2}R_1
\quad\Rightarrow\quad
\left[\begin{array}{ccc|ccc}
1 & \tfrac12 & -\tfrac12 & \tfrac12 & 0 & 0\\
4 & -2 & 3 & 0 & 1 & 0\\
-1 & 3 & 0 & 0 & 0 & 1
\end{array}\right]$$

2. Eliminar sota el primer pivot:
$$R_2\leftarrow R_2-4R_1,\qquad R_3\leftarrow R_3+R_1$$
Resulta:
$$\left[\begin{array}{ccc|ccc}
1 & \tfrac12 & -\tfrac12 & \tfrac12 & 0 & 0\\
0 & -4 & 5 & -2 & 1 & 0\\
0 & \tfrac{7}{2} & -\tfrac12 & \tfrac12 & 0 & 1
\end{array}\right]$$

3. Normalitzem el pivot de la segona fila: $R_2\leftarrow R_2/(-4)$:
$$\left[\begin{array}{ccc|ccc}
1 & \tfrac12 & -\tfrac12 & \tfrac12 & 0 & 0\\
0 & 1 & -\tfrac{5}{4} & \tfrac12 & -\tfrac14 & 0\\
0 & \tfrac{7}{2} & -\tfrac12 & \tfrac12 & 0 & 1
\end{array}\right]$$

4. Fem zeros a la columna 2 (a dalt i a baix):
$$R_1\leftarrow R_1-\tfrac12 R_2,\qquad R_3\leftarrow R_3-\tfrac{7}{2}R_2$$
Càlcul i resultat:
$$\left[\begin{array}{ccc|ccc}
1 & 0 & \tfrac{1}{8} & \tfrac{1}{4} & \tfrac{1}{8} & 0\\
0 & 1 & -\tfrac{5}{4} & \tfrac{1}{2} & -\tfrac{1}{4} & 0\\
0 & 0 & \tfrac{31}{8} & -\tfrac{5}{4} & \tfrac{7}{8} & 1
\end{array}\right]$$

5. Normalitzem el pivot de la tercera fila: $R_3\leftarrow R_3\cdot\frac{8}{31}$:
$$\left[\begin{array}{ccc|ccc}
1 & 0 & \tfrac{1}{8} & \tfrac{1}{4} & \tfrac{1}{8} & 0\\
0 & 1 & -\tfrac{5}{4} & \tfrac{1}{2} & -\tfrac{1}{4} & 0\\
0 & 0 & 1 & -\tfrac{10}{31} & \tfrac{7}{31} & \tfrac{8}{31}
\end{array}\right]$$

6. Eliminem la columna 3 a R1 i R2:
$$R_1\leftarrow R_1-\tfrac{1}{8}R_3,\qquad R_2\leftarrow R_2+\tfrac{5}{4}R_3$$
Fent les operacions s’arriba a:
$$\left[\begin{array}{ccc|ccc}
1 & 0 & 0 & \tfrac{9}{31} & \tfrac{3}{31} & -\tfrac{1}{31}\\
0 & 1 & 0 & \tfrac{3}{31} & \tfrac{1}{31} & \tfrac{10}{31}\\
0 & 0 & 1 & -\tfrac{10}{31} & \tfrac{7}{31} & \tfrac{8}{31}
\end{array}\right]$$

Ara hem obtingut $[I\mid C^{-1}]$. Per tant,
$$\boxed{,C^{-1}=\begin{pmatrix}
\frac{9}{31} & \frac{3}{31} & -\frac{1}{31}\\
\frac{3}{31} & \frac{1}{31} & \frac{10}{31}\\
-\frac{10}{31} & \frac{7}{31} & \frac{8}{31}
\end{pmatrix},}$$