Pregunta 2 – Matriu inversa

Donada la matriu $$M = \begin{pmatrix}2 & 1 & 0 \\ 1 & 3 & -1 \\0 & 1 & 2 \end{pmatrix}$$

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a) Calcula la matriu inversa $M^{-1}$ mitjançant el mètode de Gauss-Jordan. (1,5 punts)

Formem la matriu ampliada $[M | I_3]$:

$$\begin{pmatrix}
2 & 1 & 0 &| & 1 & 0 & 0 \\
1 & 3 & -1 &| & 0 & 1 & 0 \\
0 & 1 & 2 &| & 0 & 0 & 1 \\
\end{pmatrix}$$

Objectiu: Transformar la part esquerra en $I_3$ mitjançant operacions elementals.

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Pas 1: Intercanviem F1 i F2 per tenir 1 al pivot $1,1$:

$$R_1 \leftrightarrow R_2$$

$$\begin{array}{ccc|ccc}
1 & 3 & -1 & 0 & 1 & 0 \\
2 & 1 & 0 & 1 & 0 & 0 \\
0 & 1 & 2 & 0 & 0 & 1 \\
\end{array}$$

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Pas 2: Eliminem sota el pivot: $R_2 \leftarrow R_2 – 2R_1$

$$R_2 – 2R_1 = (2,1,0,1,0,0) – 2(1,3,-1,0,1,0) = (0, -5, 2, 1, -2, 0)$$

$$\begin{array}{ccc|ccc}
1 & 3 & -1 & 0 & 1 & 0 \\
0 & -5 & 2 & 1 & -2 & 0 \\
0 & 1 & 2 & 0 & 0 & 1 \\
\end{array}$$

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Pas 3: Fem que el pivot $2,2$ sigui 1: $R_2 \leftarrow -\frac{1}{5} R_2$

$$-\frac{1}{5}(0, -5, 2, 1, -2, 0) = \left(0, 1, -\frac{2}{5}, -\frac{1}{5}, \frac{2}{5}, 0\right)$$

$$\begin{array}{ccc|ccc}
1 & 3 & -1 & 0 & 1 & 0 \\
0 & 1 & -\frac{2}{5} & -\frac{1}{5} & \frac{2}{5} & 0 \\
0 & 1 & 2 & 0 & 0 & 1 \\
\end{array}$$

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Pas 4: Eliminem sota i sobre el pivot:

• $R_3 \leftarrow R_3 – R_2$
• $R_1 \leftarrow R_1 – 3R_2$

Fila 3:
$$R_3 – R_2 = (0,1,2,0,0,1) – \left(0,1,-\frac{2}{5},-\frac{1}{5},\frac{2}{5},0\right) = \left(0, 0, 2 + \frac{2}{5}, \frac{1}{5}, -\frac{2}{5}, 1\right) = \left(0, 0, \frac{12}{5}, \frac{1}{5}, -\frac{2}{5}, 1\right)$$

Fila 1:
$$R_1 – 3R_2 = (1,3,-1,0,1,0) – 3\left(0,1,-\frac{2}{5},-\frac{1}{5},\frac{2}{5},0\right) = (1, 0, -1 + \frac{6}{5}, \frac{3}{5}, 1 – \frac{6}{5}, 0) = \left(1, 0, \frac{1}{5}, \frac{3}{5}, -\frac{1}{5}, 0\right)$$

$$\begin{array}{ccc|ccc}
1 & 0 & \frac{1}{5} & \frac{3}{5} & -\frac{1}{5} & 0 \\
0 & 1 & -\frac{2}{5} & -\frac{1}{5} & \frac{2}{5} & 0 \\
0 & 0 & \frac{12}{5} & \frac{1}{5} & -\frac{2}{5} & 1 \\
\end{array}$$

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Pas 5: Fem pivot $3,3$ = 1: $R_3 \leftarrow \frac{5}{12} R_3$

$$\frac{5}{12} \left(0, 0, \frac{12}{5}, \frac{1}{5}, -\frac{2}{5}, 1\right) = \left(0, 0, 1, \frac{1}{12}, -\frac{1}{6}, \frac{5}{12}\right)$$

$$\begin{array}{ccc|ccc}
1 & 0 & \frac{1}{5} & \frac{3}{5} & -\frac{1}{5} & 0 \\
0 & 1 & -\frac{2}{5} & -\frac{1}{5} & \frac{2}{5} & 0 \\
0 & 0 & 1 & \frac{1}{12} & -\frac{1}{6} & \frac{5}{12} \\
\end{array}$$

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Pas 6: Eliminem sobre el pivot $3,3$:

• $R_1 \leftarrow R_1 – \frac{1}{5} R_3$
• $R_2 \leftarrow R_2 + \frac{2}{5} R_3$

Fila 1:
$$R_1 – \frac{1}{5} R_3 = \left(1, 0, \frac{1}{5}, \frac{3}{5}, -\frac{1}{5}, 0\right) – \frac{1}{5}\left(0, 0, 1, \frac{1}{12}, -\frac{1}{6}, \frac{5}{12}\right)
= \left(1, 0, 0, \frac{3}{5} – \frac{1}{60}, -\frac{1}{5} + \frac{1}{30}, -\frac{1}{12}\right)
= \left(1, 0, 0, \frac{36}{60} – \frac{1}{60}, -\frac{6}{30} + \frac{1}{30}, -\frac{1}{12}\right)
= \left(1, 0, 0, \frac{35}{60}, -\frac{5}{30}, -\frac{1}{12}\right)
= \left(1, 0, 0, \frac{7}{12}, -\frac{1}{6}, -\frac{1}{12}\right)$$

Fila 2:
$$R_2 + \frac{2}{5} R_3 = \left(0, 1, -\frac{2}{5}, -\frac{1}{5}, \frac{2}{5}, 0\right) + \frac{2}{5}\left(0, 0, 1, \frac{1}{12}, -\frac{1}{6}, \frac{5}{12}\right)
= \left(0, 1, 0, -\frac{1}{5} + \frac{2}{60}, \frac{2}{5} – \frac{2}{30}, \frac{2}{5} \cdot \frac{5}{12}\right)
= \left(0, 1, 0, -\frac{12}{60} + \frac{2}{60}, \frac{12}{30} – \frac{2}{30}, \frac{2}{12}\right)
= \left(0, 1, 0, -\frac{10}{60}, \frac{10}{30}, \frac{1}{6}\right)
= \left(0, 1, 0, -\frac{1}{6}, \frac{1}{3}, \frac{1}{6}\right)$$

$$\boxed{
\begin{array}{ccc|ccc}
1 & 0 & 0 & \dfrac{7}{12} & -\dfrac{1}{6} & -\dfrac{1}{12} \\
0 & 1 & 0 & -\dfrac{1}{6} & \dfrac{1}{3} & \dfrac{1}{6} \\
0 & 0 & 1 & \dfrac{1}{12} & -\dfrac{1}{6} & \dfrac{5}{12} \\
\end{array}
}$$

Conclusió:

$$\boxed{M^{-1} = \begin{pmatrix}
\dfrac{7}{12} & -\dfrac{1}{6} & -\dfrac{1}{12} \\
-\dfrac{1}{6} & \dfrac{1}{3} & \dfrac{1}{6} \\
\dfrac{1}{12} & -\dfrac{1}{6} & \dfrac{5}{12}
\end{pmatrix}}$$

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b) Verifica que $M \cdot M^{-1} = I_3$. (0,5 punts)

Calculem $M \cdot M^{-1}$:

$$M \cdot M^{-1} = \begin{pmatrix}
2 & 1 & 0 \\
1 & 3 & -1 \\
0 & 1 & 2
\end{pmatrix}
\begin{pmatrix}
\frac{7}{12} & -\frac{1}{6} & -\frac{1}{12} \\
-\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\
\frac{1}{12} & -\frac{1}{6} & \frac{5}{12}
\end{pmatrix}$$

Fila 1:

• $(1,1)): (2\cdot\frac{7}{12} + 1\cdot(-\frac{1}{6}) + 0 = \frac{14}{12} – \frac{1}{6} = \frac{7}{6} – \frac{1}{6} = 1$
• $(1,2)): (2\cdot(-\frac{1}{6}) + 1\cdot\frac{1}{3} + 0 = -\frac{1}{3} + \frac{1}{3} = 0$
• $(1,3)): (2\cdot(-\frac{1}{12}) + 1\cdot\frac{1}{6} + 0 = -\frac{1}{6} + \frac{1}{6} = 0$

Fila 2:

• $(2,1)): (1\cdot\frac{7}{12} + 3\cdot(-\frac{1}{6}) + (-1)\cdot\frac{1}{12} = \frac{7}{12} – \frac{3}{6} – \frac{1}{12} = \frac{7}{12} – \frac{6}{12} – \frac{1}{12} = 0$
• $(2,2)): (1\cdot(-\frac{1}{6}) + 3\cdot\frac{1}{3} + (-1)\cdot(-\frac{1}{6}) = -\frac{1}{6} + 1 + \frac{1}{6} = 1$
• ((2,3)): (1\cdot(-\frac{1}{12}) + 3\cdot\frac{1}{6} + (-1)\cdot\frac{5}{12} = -\frac{1}{12} + \frac{3}{6} – \frac{5}{12} = -\frac{1}{12} + \frac{6}{12} – \frac{5}{12} = 0)

Fila 3:

• $(3,1)): (0 + 1\cdot(-\frac{1}{6}) + 2\cdot\frac{1}{12} = -\frac{1}{6} + \frac{2}{12} = -\frac{2}{12} + \frac{2}{12} = 0$
• $(3,2)): (0 + 1\cdot\frac{1}{3} + 2\cdot(-\frac{1}{6}) = \frac{1}{3} – \frac{1}{3} = 0$
• $(3,3)): (0 + 1\cdot\frac{1}{6} + 2\cdot\frac{5}{12} = \frac{2}{12} + \frac{10}{12} = \frac{12}{12} = 1$

$$\therefore \quad M \cdot M^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I_3$$

Verificat.

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Resum de respostes:

• a)
  $$\boxed{M^{-1} = \begin{pmatrix}
  \frac{7}{12} & -\frac{1}{6} & -\frac{1}{12} \\
  -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\
  \frac{1}{12} & -\frac{1}{6} & \frac{5}{12}
  \end{pmatrix}}$$
• b) $M \cdot M^{-1} = I_3$ (verificat)

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