Resolución del ejercicio de Matrices
a) Halle la matriz A A A que satisface la ecuación P − 1 ⋅ A ⋅ P = J P^{-1} \cdot A \cdot P = J P − 1 ⋅ A ⋅ P = J . Para hallar la matriz A A A , debemos despejarla de la ecuación dada. Multiplicando por P P P por la izquierda y por P − 1 P^{-1} P − 1 por la derecha en ambos lados de la ecuación, obtenemos:
P ⋅ ( P − 1 ⋅ A ⋅ P ) ⋅ P − 1 = P ⋅ J ⋅ P − 1 P \cdot (P^{-1} \cdot A \cdot P) \cdot P^{-1} = P \cdot J \cdot P^{-1} P ⋅ ( P − 1 ⋅ A ⋅ P ) ⋅ P − 1 = P ⋅ J ⋅ P − 1 ( P ⋅ P − 1 ) ⋅ A ⋅ ( P ⋅ P − 1 ) = P ⋅ J ⋅ P − 1 (P \cdot P^{-1}) \cdot A \cdot (P \cdot P^{-1}) = P \cdot J \cdot P^{-1} ( P ⋅ P − 1 ) ⋅ A ⋅ ( P ⋅ P − 1 ) = P ⋅ J ⋅ P − 1 I ⋅ A ⋅ I = P ⋅ J ⋅ P − 1 I \cdot A \cdot I = P \cdot J \cdot P^{-1} I ⋅ A ⋅ I = P ⋅ J ⋅ P − 1 A = P ⋅ J ⋅ P − 1 A = P \cdot J \cdot P^{-1} A = P ⋅ J ⋅ P − 1 Primero, calculamos la inversa de la matriz P P P , P − 1 P^{-1} P − 1 .
P = ( 1 0 1 0 1 0 1 − 1 − 1 ) P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & -1 & -1 \end{pmatrix} P = 1 0 1 0 1 − 1 1 0 − 1 Calculamos el determinante de P P P :
det ( P ) = 1 ⋅ det ( 1 0 − 1 − 1 ) − 0 ⋅ det ( 0 0 1 − 1 ) + 1 ⋅ det ( 0 1 1 − 1 ) \det(P) = 1 \cdot \det\begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix} - 0 \cdot \det\begin{pmatrix} 0 & 0 \\ 1 & -1 \end{pmatrix} + 1 \cdot \det\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix} det ( P ) = 1 ⋅ det ( 1 − 1 0 − 1 ) − 0 ⋅ det ( 0 1 0 − 1 ) + 1 ⋅ det ( 0 1 1 − 1 ) det ( P ) = 1 ⋅ ( 1 ⋅ ( − 1 ) − 0 ⋅ ( − 1 ) ) − 0 + 1 ⋅ ( 0 ⋅ ( − 1 ) − 1 ⋅ 1 ) \det(P) = 1 \cdot (1 \cdot (-1) - 0 \cdot (-1)) - 0 + 1 \cdot (0 \cdot (-1) - 1 \cdot 1) det ( P ) = 1 ⋅ ( 1 ⋅ ( − 1 ) − 0 ⋅ ( − 1 )) − 0 + 1 ⋅ ( 0 ⋅ ( − 1 ) − 1 ⋅ 1 ) det ( P ) = 1 ⋅ ( − 1 ) + 1 ⋅ ( − 1 ) = − 1 − 1 = − 2 \det(P) = 1 \cdot (-1) + 1 \cdot (-1) = -1 - 1 = -2 det ( P ) = 1 ⋅ ( − 1 ) + 1 ⋅ ( − 1 ) = − 1 − 1 = − 2 Calculamos la matriz de cofactores de P P P :
C 11 = det ( 1 0 − 1 − 1 ) = − 1 C_{11} = \det\begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix} = -1 C 11 = det ( 1 − 1 0 − 1 ) = − 1 C 12 = − det ( 0 0 1 − 1 ) = 0 C_{12} = -\det\begin{pmatrix} 0 & 0 \\ 1 & -1 \end{pmatrix} = 0 C 12 = − det ( 0 1 0 − 1 ) = 0 C 13 = det ( 0 1 1 − 1 ) = − 1 C_{13} = \det\begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix} = -1 C 13 = det ( 0 1 1 − 1 ) = − 1 C 21 = − det ( 0 1 − 1 − 1 ) = − ( 0 − ( − 1 ) ) = − 1 C_{21} = -\det\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix} = -(0 - (-1)) = -1 C 21 = − det ( 0 − 1 1 − 1 ) = − ( 0 − ( − 1 )) = − 1 C 22 = det ( 1 1 1 − 1 ) = − 1 − 1 = − 2 C_{22} = \det\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = -1 - 1 = -2 C 22 = det ( 1 1 1 − 1 ) = − 1 − 1 = − 2 C 23 = − det ( 1 0 1 − 1 ) = − ( − 1 − 0 ) = 1 C_{23} = -\det\begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix} = -(-1 - 0) = 1 C 23 = − det ( 1 1 0 − 1 ) = − ( − 1 − 0 ) = 1 C 31 = det ( 0 1 1 0 ) = − 1 C_{31} = \det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = -1 C 31 = det ( 0 1 1 0 ) = − 1 C 32 = − det ( 1 1 0 0 ) = 0 C_{32} = -\det\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix} = 0 C 32 = − det ( 1 0 1 0 ) = 0 C 33 = det ( 1 0 0 1 ) = 1 C_{33} = \det\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 1 C 33 = det ( 1 0 0 1 ) = 1 La matriz de cofactores es:
C = ( − 1 0 − 1 − 1 − 2 1 − 1 0 1 ) C = \begin{pmatrix} -1 & 0 & -1 \\ -1 & -2 & 1 \\ -1 & 0 & 1 \end{pmatrix} C = − 1 − 1 − 1 0 − 2 0 − 1 1 1 La matriz adjunta de P P P es la traspuesta de la matriz de cofactores:
adj ( P ) = C T = ( − 1 − 1 − 1 0 − 2 0 − 1 1 1 ) \text{adj}(P) = C^T = \begin{pmatrix} -1 & -1 & -1 \\ 0 & -2 & 0 \\ -1 & 1 & 1 \end{pmatrix} adj ( P ) = C T = − 1 0 − 1 − 1 − 2 1 − 1 0 1 Finalmente, la inversa de P P P es:
P − 1 = 1 det ( P ) adj ( P ) = 1 − 2 ( − 1 − 1 − 1 0 − 2 0 − 1 1 1 ) = ( 1 / 2 1 / 2 1 / 2 0 1 0 1 / 2 − 1 / 2 − 1 / 2 ) P^{-1} = \frac{1}{\det(P)} \text{adj}(P) = \frac{1}{-2} \begin{pmatrix} -1 & -1 & -1 \\ 0 & -2 & 0 \\ -1 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 & 1/2 \\ 0 & 1 & 0 \\ 1/2 & -1/2 & -1/2 \end{pmatrix} P − 1 = det ( P ) 1 adj ( P ) = − 2 1 − 1 0 − 1 − 1 − 2 1 − 1 0 1 = 1/2 0 1/2 1/2 1 − 1/2 1/2 0 − 1/2 Ahora, calculamos A = P ⋅ J ⋅ P − 1 A = P \cdot J \cdot P^{-1} A = P ⋅ J ⋅ P − 1 . Primero, calculamos el producto P ⋅ J P \cdot J P ⋅ J :
P ⋅ J = ( 1 0 1 0 1 0 1 − 1 − 1 ) ( 2 1 0 0 2 0 0 0 − 1 ) P \cdot J = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & -1 & -1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix} P ⋅ J = 1 0 1 0 1 − 1 1 0 − 1 2 0 0 1 2 0 0 0 − 1 P ⋅ J = ( ( 1 ⋅ 2 + 0 ⋅ 0 + 1 ⋅ 0 ) ( 1 ⋅ 1 + 0 ⋅ 2 + 1 ⋅ 0 ) ( 1 ⋅ 0 + 0 ⋅ 0 + 1 ⋅ ( − 1 ) ) ( 0 ⋅ 2 + 1 ⋅ 0 + 0 ⋅ 0 ) ( 0 ⋅ 1 + 1 ⋅ 2 + 0 ⋅ 0 ) ( 0 ⋅ 0 + 1 ⋅ 0 + 0 ⋅ ( − 1 ) ) ( 1 ⋅ 2 + ( − 1 ) ⋅ 0 + ( − 1 ) ⋅ 0 ) ( 1 ⋅ 1 + ( − 1 ) ⋅ 2 + ( − 1 ) ⋅ 0 ) ( 1 ⋅ 0 + ( − 1 ) ⋅ 0 + ( − 1 ) ⋅ ( − 1 ) ) ) P \cdot J = \begin{pmatrix} (1\cdot 2 + 0\cdot 0 + 1\cdot 0) & (1\cdot 1 + 0\cdot 2 + 1\cdot 0) & (1\cdot 0 + 0\cdot 0 + 1\cdot (-1)) \\ (0\cdot 2 + 1\cdot 0 + 0\cdot 0) & (0\cdot 1 + 1\cdot 2 + 0\cdot 0) & (0\cdot 0 + 1\cdot 0 + 0\cdot (-1)) \\ (1\cdot 2 + (-1)\cdot 0 + (-1)\cdot 0) & (1\cdot 1 + (-1)\cdot 2 + (-1)\cdot 0) & (1\cdot 0 + (-1)\cdot 0 + (-1)\cdot (-1)) \end{pmatrix} P ⋅ J = ( 1 ⋅ 2 + 0 ⋅ 0 + 1 ⋅ 0 ) ( 0 ⋅ 2 + 1 ⋅ 0 + 0 ⋅ 0 ) ( 1 ⋅ 2 + ( − 1 ) ⋅ 0 + ( − 1 ) ⋅ 0 ) ( 1 ⋅ 1 + 0 ⋅ 2 + 1 ⋅ 0 ) ( 0 ⋅ 1 + 1 ⋅ 2 + 0 ⋅ 0 ) ( 1 ⋅ 1 + ( − 1 ) ⋅ 2 + ( − 1 ) ⋅ 0 ) ( 1 ⋅ 0 + 0 ⋅ 0 + 1 ⋅ ( − 1 )) ( 0 ⋅ 0 + 1 ⋅ 0 + 0 ⋅ ( − 1 )) ( 1 ⋅ 0 + ( − 1 ) ⋅ 0 + ( − 1 ) ⋅ ( − 1 )) P ⋅ J = ( 2 1 − 1 0 2 0 2 − 1 1 ) P \cdot J = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 2 & 0 \\ 2 & -1 & 1 \end{pmatrix} P ⋅ J = 2 0 2 1 2 − 1 − 1 0 1 Finalmente, calculamos A = ( P ⋅ J ) ⋅ P − 1 A = (P \cdot J) \cdot P^{-1} A = ( P ⋅ J ) ⋅ P − 1 :
A = ( 2 1 − 1 0 2 0 2 − 1 1 ) ( 1 / 2 1 / 2 1 / 2 0 1 0 1 / 2 − 1 / 2 − 1 / 2 ) A = \begin{pmatrix} 2 & 1 & -1 \\ 0 & 2 & 0 \\ 2 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1/2 & 1/2 & 1/2 \\ 0 & 1 & 0 \\ 1/2 & -1/2 & -1/2 \end{pmatrix} A = 2 0 2 1 2 − 1 − 1 0 1 1/2 0 1/2 1/2 1 − 1/2 1/2 0 − 1/2 A = ( ( 2 ⋅ 1 / 2 + 1 ⋅ 0 − 1 ⋅ 1 / 2 ) ( 2 ⋅ 1 / 2 + 1 ⋅ 1 − 1 ⋅ ( − 1 / 2 ) ) ( 2 ⋅ 1 / 2 + 1 ⋅ 0 − 1 ⋅ ( − 1 / 2 ) ) ( 0 ⋅ 1 / 2 + 2 ⋅ 0 + 0 ⋅ 1 / 2 ) ( 0 ⋅ 1 / 2 + 2 ⋅ 1 + 0 ⋅ ( − 1 / 2 ) ) ( 0 ⋅ 1 / 2 + 2 ⋅ 0 + 0 ⋅ ( − 1 / 2 ) ) ( 2 ⋅ 1 / 2 − 1 ⋅ 0 + 1 ⋅ 1 / 2 ) ( 2 ⋅ 1 / 2 − 1 ⋅ 1 + 1 ⋅ ( − 1 / 2 ) ) ( 2 ⋅ 1 / 2 − 1 ⋅ 0 + 1 ⋅ ( − 1 / 2 ) ) ) A = \begin{pmatrix} (2\cdot 1/2 + 1\cdot 0 - 1\cdot 1/2) & (2\cdot 1/2 + 1\cdot 1 - 1\cdot (-1/2)) & (2\cdot 1/2 + 1\cdot 0 - 1\cdot (-1/2)) \\ (0\cdot 1/2 + 2\cdot 0 + 0\cdot 1/2) & (0\cdot 1/2 + 2\cdot 1 + 0\cdot (-1/2)) & (0\cdot 1/2 + 2\cdot 0 + 0\cdot (-1/2)) \\ (2\cdot 1/2 - 1\cdot 0 + 1\cdot 1/2) & (2\cdot 1/2 - 1\cdot 1 + 1\cdot (-1/2)) & (2\cdot 1/2 - 1\cdot 0 + 1\cdot (-1/2)) \end{pmatrix} A = ( 2 ⋅ 1/2 + 1 ⋅ 0 − 1 ⋅ 1/2 ) ( 0 ⋅ 1/2 + 2 ⋅ 0 + 0 ⋅ 1/2 ) ( 2 ⋅ 1/2 − 1 ⋅ 0 + 1 ⋅ 1/2 ) ( 2 ⋅ 1/2 + 1 ⋅ 1 − 1 ⋅ ( − 1/2 )) ( 0 ⋅ 1/2 + 2 ⋅ 1 + 0 ⋅ ( − 1/2 )) ( 2 ⋅ 1/2 − 1 ⋅ 1 + 1 ⋅ ( − 1/2 )) ( 2 ⋅ 1/2 + 1 ⋅ 0 − 1 ⋅ ( − 1/2 )) ( 0 ⋅ 1/2 + 2 ⋅ 0 + 0 ⋅ ( − 1/2 )) ( 2 ⋅ 1/2 − 1 ⋅ 0 + 1 ⋅ ( − 1/2 )) A = ( 1 − 1 / 2 1 + 1 + 1 / 2 1 + 1 / 2 0 2 0 1 + 1 / 2 1 − 1 − 1 / 2 1 − 1 / 2 ) A = \begin{pmatrix} 1 - 1/2 & 1 + 1 + 1/2 & 1 + 1/2 \\ 0 & 2 & 0 \\ 1 + 1/2 & 1 - 1 - 1/2 & 1 - 1/2 \end{pmatrix} A = 1 − 1/2 0 1 + 1/2 1 + 1 + 1/2 2 1 − 1 − 1/2 1 + 1/2 0 1 − 1/2 A = ( 1 / 2 5 / 2 3 / 2 0 2 0 3 / 2 − 1 / 2 1 / 2 ) A = \begin{pmatrix} 1/2 & 5/2 & 3/2 \\ 0 & 2 & 0 \\ 3/2 & -1/2 & 1/2 \end{pmatrix} A = 1/2 0 3/2 5/2 2 − 1/2 3/2 0 1/2 b) Compruebe que A 3 = P ⋅ J 3 ⋅ P − 1 A^3 = P \cdot J^3 \cdot P^{-1} A 3 = P ⋅ J 3 ⋅ P − 1 . Sabemos del apartado anterior que A = P ⋅ J ⋅ P − 1 A = P \cdot J \cdot P^{-1} A = P ⋅ J ⋅ P − 1 . Podemos calcular A 2 A^2 A 2 y A 3 A^3 A 3 usando esta relación.
A 2 = A ⋅ A = ( P ⋅ J ⋅ P − 1 ) ⋅ ( P ⋅ J ⋅ P − 1 ) A^2 = A \cdot A = (P \cdot J \cdot P^{-1}) \cdot (P \cdot J \cdot P^{-1}) A 2 = A ⋅ A = ( P ⋅ J ⋅ P − 1 ) ⋅ ( P ⋅ J ⋅ P − 1 ) Dado que la multiplicación de matrices es asociativa y P − 1 ⋅ P = I P^{-1} \cdot P = I P − 1 ⋅ P = I (matriz identidad):
A 2 = P ⋅ J ⋅ ( P − 1 ⋅ P ) ⋅ J ⋅ P − 1 A^2 = P \cdot J \cdot (P^{-1} \cdot P) \cdot J \cdot P^{-1} A 2 = P ⋅ J ⋅ ( P − 1 ⋅ P ) ⋅ J ⋅ P − 1 A 2 = P ⋅ J ⋅ I ⋅ J ⋅ P − 1 A^2 = P \cdot J \cdot I \cdot J \cdot P^{-1} A 2 = P ⋅ J ⋅ I ⋅ J ⋅ P − 1 A 2 = P ⋅ J 2 ⋅ P − 1 A^2 = P \cdot J^2 \cdot P^{-1} A 2 = P ⋅ J 2 ⋅ P − 1 Ahora, calculamos A 3 A^3 A 3 :
A 3 = A 2 ⋅ A = ( P ⋅ J 2 ⋅ P − 1 ) ⋅ ( P ⋅ J ⋅ P − 1 ) A^3 = A^2 \cdot A = (P \cdot J^2 \cdot P^{-1}) \cdot (P \cdot J \cdot P^{-1}) A 3 = A 2 ⋅ A = ( P ⋅ J 2 ⋅ P − 1 ) ⋅ ( P ⋅ J ⋅ P − 1 ) A 3 = P ⋅ J 2 ⋅ ( P − 1 ⋅ P ) ⋅ J ⋅ P − 1 A^3 = P \cdot J^2 \cdot (P^{-1} \cdot P) \cdot J \cdot P^{-1} A 3 = P ⋅ J 2 ⋅ ( P − 1 ⋅ P ) ⋅ J ⋅ P − 1 A 3 = P ⋅ J 2 ⋅ I ⋅ J ⋅ P − 1 A^3 = P \cdot J^2 \cdot I \cdot J \cdot P^{-1} A 3 = P ⋅ J 2 ⋅ I ⋅ J ⋅ P − 1 A 3 = P ⋅ J 3 ⋅ P − 1 A^3 = P \cdot J^3 \cdot P^{-1} A 3 = P ⋅ J 3 ⋅ P − 1 De este modo, queda comprobado que A 3 = P ⋅ J 3 ⋅ P − 1 A^3 = P \cdot J^3 \cdot P^{-1} A 3 = P ⋅ J 3 ⋅ P − 1 .
