Càlcul de la matriu inversa

Calculeu la matriu inversa $A^{-1}$ de la matriu $$A = \begin{pmatrix}1 & 0 & -1 \\ 0 & 2 & 3 \\ 1 & -1 & 1\end{pmatrix}$$

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✅ PAS 1: Calcula el determinant de $A$

Fem servir la fórmula del determinant per a matrius $3 \times 3$:

$$\text{det}(A) = a(ei – fh) – b(di – fg) + c(dh – eg)$$

Assignem les lletres corresponents:

$$A = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix}
= \begin{pmatrix}
1 & 0 & -1 \\
0 & 2 & 3 \\
1 & -1 & 1
\end{pmatrix}$$

Ara substituïm:

$$\text{det}(A) = 1(2 \cdot 1 – 3 \cdot (-1)) – 0(0 \cdot 1 – 3 \cdot 1) + (-1)(0 \cdot (-1) – 2 \cdot 1)$$
$$= 1(2 + 3) – 0(\cdot) + (-1)(0 – 2)$$
$$= 5 + 2 = 7$$

✅ Determinant de $A$ és 7

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✅ PAS 2: Calcular la matriu de cofactors

Per cada element $a_{ij}$, eliminem la fila $i$ i la columna $j$, i calculem el menor (determinant de la submatriu $2 \times 2$). Després, apliquem el signe corresponent segons la matriu de signes:

$$\begin{pmatrix} + & – & + \\ – & + & – \\ + & – & +\end{pmatrix}$$

Anem a fer cada cofactor:

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📌 Primera fila:

• $C_{11} = +\left|\begin{matrix}2 & 3 \\ -1 & 1\end{matrix}\right| = (2)(1) – (3)(-1) = 2 + 3 = 5$
• $C_{12} = -\left|\begin{matrix}0 & 3 \\ 1 & 1\end{matrix}\right| = -((0)(1) – (3)(1)) = -(-3) = 3$
• $C_{13} = +\left|\begin{matrix}0 & 2 \\ 1 & -1\end{matrix}\right| = (0)(-1) – (2)(1) = 0 – 2 = -2$

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📌 Segona fila:

• $C_{21} = -\left|\begin{matrix}0 & -1 \\ -1 & 1\end{matrix}\right| = -((0)(1) – (-1)(-1)) = -(-1) = 1$
• $C_{22} = +\left|\begin{matrix}1 & -1 \\ 1 & 1\end{matrix}\right| = (1)(1) – (-1)(1) = 1 + 1 = 2$
• $C_{23} = -\left|\begin{matrix}1 & 0 \\ 1 & -1\end{matrix}\right| = -((1)(-1) – (0)(1)) = -(-1) = 1$

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📌 Tercera fila:

• $C_{31} = +\left|\begin{matrix}0 & -1 \\ 2 & 3\end{matrix}\right| = (0)(3) – (-1)(2) = 0 + 2 = 2$
• $C_{32} = -\left|\begin{matrix}1 & -1 \\ 0 & 3\end{matrix}\right| = -((1)(3) – (-1)(0)) = -(3 – 0) = -3$
• $C_{33} = +\left|\begin{matrix}1 & 0 \\ 0 & 2\end{matrix}\right| = (1)(2) – (0)(0) = 2$

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✅ PAS 3: Matriu de cofactors

$$\text{Cof}(A) =
\begin{pmatrix}
5 & 3 & -2 \\
-1 & 2 & 1 \\
2 & -3 & 2
\end{pmatrix}$$

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✅ PAS 4: Matriu adjunta = transposada de la matriu de cofactors

$$\text{adj}(A) =
\begin{pmatrix}
5 & 1 & 2 \\
3 & 2 & -3 \\
-2 & 1 & 2
\end{pmatrix}$$

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✅ PAS 5: Inversa de $A$

$$A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) = \frac{1}{7} \cdot
\begin{pmatrix}
5 & 1 & 2 \\
3 & 2 & -3 \\
-2 & 1 & 2
\end{pmatrix}$$

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✅ RESULTAT FINAL:

$$A^{-1} =
\begin{pmatrix}
\frac{5}{7} & \frac{1}{7} & \frac{2}{7} \\
\frac{3}{7} & \frac{2}{7} & -\frac{3}{7} \\
-\frac{2}{7} & \frac{1}{7} & \frac{2}{7}
\end{pmatrix}$$