Càlcul de la matriu inversa

Calculeu la matriu inversa$$A = \begin{pmatrix}5 & 3 & 1 \\ 1 & -3 & -2 \\ -5 & 2 & 1\end{pmatrix}$$

Pas 1: Calculem el determinant de la matriu modificada

Utilitzem la definició del determinant, expandint per la primera fila:

$$\det A = 5 \begin{vmatrix} -3 & -2 \\ 2 & 1 \end{vmatrix} – 3 \begin{vmatrix} 1 & -2 \\ -5 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & -3 \\ -5 & 2 \end{vmatrix}.$$

Calculem cada determinant de 2×2:

1. Primer terme:
   $$\begin{vmatrix} -3 & -2 \\ 2 & 1 \end{vmatrix} = (-3) \cdot 1 – (-2) \cdot 2 = -3 + 4 = 1,$$
   $$5 \cdot 1 = 5.$$
2. Segon terme:
   $$\begin{vmatrix} 1 & -2 \\ -5 & 1 \end{vmatrix} = 1 \cdot 1 – (-2) \cdot (-5) = 1 – 10 = -9,$$
   $$-3 \cdot (-9) = 27.$$
3. Tercer terme:
   $$\begin{vmatrix} 1 & -3 \\ -5 & 2 \end{vmatrix} = 1 \cdot 2 – (-3) \cdot (-5) = 2 – 15 = -13,$$
   $$1 \cdot (-13) = -13.$$

Ara sumem els termes:

$$\det A = 5 + 27 + (-13) = 5 + 27 – 13 = 19.$$

El determinant de la matriu és:

$$\det A = 19 \neq 0.$$

Com que el determinant és diferent de zero, la matriu $A$ segueix sent invertible, i podem calcular la seva inversa.

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Pas 2: Calculem la matriu inversa $A^{-1}$

La inversa es calcula amb la fórmula:

$$A^{-1} = \frac{1}{\det A} \cdot \text{adj}(A),$$

on $\text{adj}(A)$ és la matriu adjunta, que és la transposada de la matriu de cofactors.

Calculem els cofactors $A_{ij}$

Els cofactors es calculen com $(-1)^{i+j}$ multiplicat pel determinant del menor corresponent.

$$A_{11} = (-1)^{1+1} \begin{vmatrix} -3 & -2 \\ 2 & 1 \end{vmatrix} = 1 \cdot (-3 \cdot 1 – (-2) \cdot 2) = 1 \cdot (-3 + 4) = 1,$$

$$A_{12} = (-1)^{1+2} \begin{vmatrix} 1 & -2 \\ -5 & 1 \end{vmatrix} = -1 \cdot (1 \cdot 1 – (-2) \cdot (-5)) = -1 \cdot (1 – 10) = -1 \cdot (-9) = 9,$$

$$A_{13} = (-1)^{1+3} \begin{vmatrix} 1 & -3 \\ -5 & 2 \end{vmatrix} = 1 \cdot (1 \cdot 2 – (-3) \cdot (-5)) = 1 \cdot (2 – 15) = -13,$$

$$A_{21} = (-1)^{2+1} \begin{vmatrix} 3 & 1 \\ 2 & 1 \end{vmatrix} = -1 \cdot (3 \cdot 1 – 1 \cdot 2) = -1 \cdot (3 – 2) = -1,$$

$$A_{22} = (-1)^{2+2} \begin{vmatrix} 5 & 1 \\ -5 & 1 \end{vmatrix} = 1 \cdot (5 \cdot 1 – 1 \cdot (-5)) = 1 \cdot (5 + 5) = 10,$$

$$A_{23} = (-1)^{2+3} \begin{vmatrix} 5 & 3 \\ -5 & 2 \end{vmatrix} = -1 \cdot (5 \cdot 2 – 3 \cdot (-5)) = -1 \cdot (10 + 15) = -25,$$

$$A_{31} = (-1)^{3+1} \begin{vmatrix} 3 & 1 \\ -3 & -2 \end{vmatrix} = 1 \cdot (3 \cdot (-2) – 1 \cdot (-3)) = 1 \cdot (-6 + 3) = -3,$$

$$A_{32} = (-1)^{3+2} \begin{vmatrix} 5 & 1 \\ 1 & -2 \end{vmatrix} = -1 \cdot (5 \cdot (-2) – 1 \cdot 1) = -1 \cdot (-10 – 1) = 11,$$

$$A_{33} = (-1)^{3+3} \begin{vmatrix} 5 & 3 \\ 1 & -3 \end{vmatrix} = 1 \cdot (5 \cdot (-3) – 3 \cdot 1) = 1 \cdot (-15 – 3) = -18.$$

La matriu de cofactors és:

$$\text{Matriu de cofactors} = \begin{pmatrix}
1 & 9 & -13 \\
-1 & 10 & -25 \\
-3 & 11 & -18
\end{pmatrix}.$$

Matriu adjunta

La matriu adjunta $\text{adj}(A)$ és la transposada de la matriu de cofactors:

$$\text{adj}(A) = \begin{pmatrix}
1 & -1 & -3 \\
9 & 10 & 11 \\
-13 & -25 & -18
\end{pmatrix}.$$

Calculem la inversa

Com que $\det A = 19$, la inversa és:

$$A^{-1} = \frac{1}{\det A} \cdot \text{adj}(A) = \frac{1}{19} \begin{pmatrix}
1 & -1 & -3 \\
9 & 10 & 11 \\
-13 & -25 & -18
\end{pmatrix} = \begin{pmatrix}
\frac{1}{19} & -\frac{1}{19} & -\frac{3}{19} \\
\frac{9}{19} & \frac{10}{19} & \frac{11}{19} \\
-\frac{13}{19} & -\frac{25}{19} & -\frac{18}{19}
\end{pmatrix}.$$

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Resposta final

$$A = \begin{pmatrix}
5 & 3 & 1 \\
1 & -3 & -2 \\
-5 & 2 & 1
\end{pmatrix},$$

el seu determinant és:

$$\det A = 19,$$

i la seva inversa és:

$$A^{-1} = \begin{pmatrix}
\frac{1}{19} & -\frac{1}{19} & -\frac{3}{19} \\
\frac{9}{19} & \frac{10}{19} & \frac{11}{19} \\
-\frac{13}{19} & -\frac{25}{19} & -\frac{18}{19}
\end{pmatrix}.$$

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Pas 3: Comprovació

Per verificar que la inversa és correcta, multipliquem $A \cdot A^{-1}$ i comprovem que el resultat és la matriu identitat $I$:

\begin{equation}
I = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}.
\end{equation}

Multipliquem $A \cdot A^{-1}$

• Primera fila de $A$: $(5, 3, 1)$

Primera columna de (A^{-1}):

\begin{equation}
5 \cdot \frac{1}{19} + 3 \cdot \frac{9}{19} + 1 \cdot \left(-\frac{13}{19}\right) = \frac{5}{19} + \frac{27}{19} – \frac{13}{19} = \frac{19}{19} = 1,
\end{equation}

Segona columna de $A^{-1}$:

\begin{equation}
5 \cdot \left(-\frac{1}{19}\right) + 3 \cdot \frac{10}{19} + 1 \cdot \left(-\frac{25}{19}\right) = -\frac{5}{19} + \frac{30}{19} – \frac{25}{19} = \frac{0}{19} = 0,
\end{equation}

Tercera columna de $A^{-1}$:

\begin{equation}
5 \cdot \left(-\frac{3}{19}\right) + 3 \cdot \frac{11}{19} + 1 \cdot \left(-\frac{18}{19}\right) = -\frac{15}{19} + \frac{33}{19} – \frac{18}{19} = \frac{0}{19} = 0.
\end{equation}

• Segona fila de $A$: $(1, -3, -2)$

Primera columna de $A^{-1}$:

\begin{equation}
1 \cdot \frac{1}{19} + (-3) \cdot \frac{9}{19} + (-2) \cdot \left(-\frac{13}{19}\right) = \frac{1}{19} – \frac{27}{19} + \frac{26}{19} = \frac{0}{19} = 0,
\end{equation}

Segona columna de $A^{-1}$:

\begin{equation}
1 \cdot \left(-\frac{1}{19}\right) + (-3) \cdot \frac{10}{19} + (-2) \cdot \left(-\frac{25}{19}\right) = -\frac{1}{19} – \frac{30}{19} + \frac{50}{19} = \frac{19}{19} = 1,
\end{equation}

Tercera columna de $A^{-1}$:

\begin{equation}
1 \cdot \left(-\frac{3}{19}\right) + (-3) \cdot \frac{11}{19} + (-2) \cdot \left(-\frac{18}{19}\right) = -\frac{3}{19} – \frac{33}{19} + \frac{36}{19} = \frac{0}{19} = 0.
\end{equation}

• Tercera fila de $A$: $(-5, 2, 1)$

Primera columna de $A^{-1}$:

\begin{equation}
(-5) \cdot \frac{1}{19} + 2 \cdot \frac{9}{19} + 1 \cdot \left(-\frac{13}{19}\right) = -\frac{5}{19} + \frac{18}{19} – \frac{13}{19} = \frac{0}{19} = 0,
\end{equation}

Segona columna de (A^{-1}):

\begin{equation}
(-5) \cdot \left(-\frac{1}{19}\right) + 2 \cdot \frac{10}{19} + 1 \cdot \left(-\frac{25}{19}\right) = \frac{5}{19} + \frac{20}{19} – \frac{25}{19} = \frac{0}{19} = 0,
\end{equation}

Tercera columna de (A^{-1}):

\begin{equation}
(-5) \cdot \left(-\frac{3}{19}\right) + 2 \cdot \frac{11}{19} + 1 \cdot \left(-\frac{18}{19}\right) = \frac{15}{19} + \frac{22}{19} – \frac{18}{19} = \frac{19}{19} = 1.
\end{equation}

Resultat de la multiplicació:

\begin{equation}
A \cdot A^{-1} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix} = I.
\end{equation}

Verificació del determinant

Expandim per la tercera fila per confirmar el determinant:

\begin{equation}
\det A = (-5) \begin{vmatrix} 3 & 1 \\ -3 & -2 \end{vmatrix} – 2 \begin{vmatrix} 5 & 1 \\ 1 & -2 \end{vmatrix} + 1 \begin{vmatrix} 5 & 3 \ 1 & -3 \end{vmatrix},
\end{equation}

\begin{equation}
(-5) \cdot (3 \cdot (-2) – 1 \cdot (-3)) = (-5) \cdot (-6 + 3) = (-5) \cdot (-3) = 15,
\end{equation}

\begin{equation}
-2 \cdot (5 \cdot (-2) – 1 \cdot 1) = -2 \cdot (-10 – 1) = -2 \cdot (-11) = 22,
\end{equation}

\begin{equation}
1 \cdot (5 \cdot (-3) – 3 \cdot 1) = 1 \cdot (-15 – 3) = -18,
\end{equation}

\begin{equation}
\det A = 15 + 22 – 18 = 19.
\end{equation}

Coincideix amb el càlcul anterior.

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Conclusió

La matriu inversa està ben calculada, ja que (A \cdot A^{-1} = I), i el determinant també està verificat. La resposta final és:

\begin{equation}
A = \begin{pmatrix}
5 & 3 & 1 \\
1 & -3 & -2 \\
-5 & 2 & 1
\end{pmatrix},
\end{equation}

\begin{equation}
\det A = 19,
\end{equation}

\begin{equation}
A^{-1} = \begin{pmatrix}
\frac{1}{19} & -\frac{1}{19} & -\frac{3}{19} \
\frac{9}{19} & \frac{10}{19} & \frac{11}{19} \
-\frac{13}{19} & -\frac{25}{19} & -\frac{18}{19}
\end{pmatrix}.
\end{equation}