a) Resuelva la siguiente ecuación A ⋅ B ⋅ X ⋅ C = ( 1 0 0 0 1 0 ) A \cdot B \cdot X \cdot C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} A ⋅ B ⋅ X ⋅ C = ( 1 0 0 1 0 0 ) . Primero, calculamos el producto A ⋅ B A \cdot B A ⋅ B :
A ⋅ B = ( 1 − 1 1 − 2 1 0 ) ⋅ ( 0 − 1 1 0 − 1 2 ) = ( ( 1 ) ( 0 ) + ( − 1 ) ( 1 ) + ( 1 ) ( − 1 ) ( 1 ) ( − 1 ) + ( − 1 ) ( 0 ) + ( 1 ) ( 2 ) ( − 2 ) ( 0 ) + ( 1 ) ( 1 ) + ( 0 ) ( − 1 ) ( − 2 ) ( − 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 2 ) ) A \cdot B = \begin{pmatrix} 1 & -1 & 1 \\ -2 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} (1)(0)+(-1)(1)+(1)(-1) & (1)(-1)+(-1)(0)+(1)(2) \\ (-2)(0)+(1)(1)+(0)(-1) & (-2)(-1)+(1)(0)+(0)(2) \end{pmatrix} A ⋅ B = ( 1 − 2 − 1 1 1 0 ) ⋅ 0 1 − 1 − 1 0 2 = ( ( 1 ) ( 0 ) + ( − 1 ) ( 1 ) + ( 1 ) ( − 1 ) ( − 2 ) ( 0 ) + ( 1 ) ( 1 ) + ( 0 ) ( − 1 ) ( 1 ) ( − 1 ) + ( − 1 ) ( 0 ) + ( 1 ) ( 2 ) ( − 2 ) ( − 1 ) + ( 1 ) ( 0 ) + ( 0 ) ( 2 ) ) A ⋅ B = ( 0 − 1 − 1 − 1 + 0 + 2 0 + 1 + 0 2 + 0 + 0 ) = ( − 2 1 1 2 ) A \cdot B = \begin{pmatrix} 0-1-1 & -1+0+2 \\ 0+1+0 & 2+0+0 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ 1 & 2 \end{pmatrix} A ⋅ B = ( 0 − 1 − 1 0 + 1 + 0 − 1 + 0 + 2 2 + 0 + 0 ) = ( − 2 1 1 2 ) Denominamos P = A ⋅ B P = A \cdot B P = A ⋅ B . La ecuación se transforma en P ⋅ X ⋅ C = R P \cdot X \cdot C = R P ⋅ X ⋅ C = R , donde R = ( 1 0 0 0 1 0 ) R = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} R = ( 1 0 0 1 0 0 ) .
Para resolver para X X X , necesitamos las inversas de P P P y C C C . La matriz P P P es de 2 × 2 2 \times 2 2 × 2 y la matriz C C C es de 3 × 3 3 \times 3 3 × 3 . La matriz R R R es de 2 × 3 2 \times 3 2 × 3 . Para que la ecuación tenga sentido, las dimensiones de X X X deben ser 2 × 3 2 \times 3 2 × 3 (ya que ( P 2 × 2 ⋅ X 2 × 3 ⋅ C 3 × 3 ) 2 × 3 = R 2 × 3 (P_{2 \times 2} \cdot X_{2 \times 3} \cdot C_{3 \times 3})_{2 \times 3} = R_{2 \times 3} ( P 2 × 2 ⋅ X 2 × 3 ⋅ C 3 × 3 ) 2 × 3 = R 2 × 3 ). Calculamos la inversa de P = ( − 2 1 1 2 ) P = \begin{pmatrix} -2 & 1 \\ 1 & 2 \end{pmatrix} P = ( − 2 1 1 2 ) :
det ( P ) = ( − 2 ) ( 2 ) − ( 1 ) ( 1 ) = − 4 − 1 = − 5 \det(P) = (-2)(2) - (1)(1) = -4 - 1 = -5 det ( P ) = ( − 2 ) ( 2 ) − ( 1 ) ( 1 ) = − 4 − 1 = − 5 P − 1 = 1 − 5 ( 2 − 1 − 1 − 2 ) = ( − 2 / 5 1 / 5 1 / 5 2 / 5 ) P^{-1} = \frac{1}{-5} \begin{pmatrix} 2 & -1 \\ -1 & -2 \end{pmatrix} = \begin{pmatrix} -2/5 & 1/5 \\ 1/5 & 2/5 \end{pmatrix} P − 1 = − 5 1 ( 2 − 1 − 1 − 2 ) = ( − 2/5 1/5 1/5 2/5 ) Calculamos la inversa de C = ( 1 3 2 1 1 1 0 3 1 ) C = \begin{pmatrix} 1 & 3 & 2 \\ 1 & 1 & 1 \\ 0 & 3 & 1 \end{pmatrix} C = 1 1 0 3 1 3 2 1 1 :
det ( C ) = 1 ( 1 ⋅ 1 − 1 ⋅ 3 ) − 3 ( 1 ⋅ 1 − 1 ⋅ 0 ) + 2 ( 1 ⋅ 3 − 1 ⋅ 0 ) = 1 ( 1 − 3 ) − 3 ( 1 − 0 ) + 2 ( 3 − 0 ) = 1 ( − 2 ) − 3 ( 1 ) + 2 ( 3 ) = − 2 − 3 + 6 = 1 \det(C) = 1(1 \cdot 1 - 1 \cdot 3) - 3(1 \cdot 1 - 1 \cdot 0) + 2(1 \cdot 3 - 1 \cdot 0) \\ = 1(1-3) - 3(1-0) + 2(3-0) \\ = 1(-2) - 3(1) + 2(3) = -2 - 3 + 6 = 1 det ( C ) = 1 ( 1 ⋅ 1 − 1 ⋅ 3 ) − 3 ( 1 ⋅ 1 − 1 ⋅ 0 ) + 2 ( 1 ⋅ 3 − 1 ⋅ 0 ) = 1 ( 1 − 3 ) − 3 ( 1 − 0 ) + 2 ( 3 − 0 ) = 1 ( − 2 ) − 3 ( 1 ) + 2 ( 3 ) = − 2 − 3 + 6 = 1 Calculamos la matriz de cofactores de C C C :
C o f ( C ) = ( + ( 1 − 3 ) − ( 1 − 0 ) + ( 3 − 0 ) − ( 3 − 6 ) + ( 1 − 0 ) − ( 3 − 0 ) + ( 3 − 2 ) − ( 1 − 2 ) + ( 1 − 3 ) ) = ( − 2 − 1 3 3 1 − 3 1 1 − 2 ) Cof(C) = \begin{pmatrix} +(1-3) & -(1-0) & +(3-0) \\ -(3-6) & +(1-0) & -(3-0) \\ +(3-2) & -(1-2) & +(1-3) \end{pmatrix} = \begin{pmatrix} -2 & -1 & 3 \\ 3 & 1 & -3 \\ 1 & 1 & -2 \end{pmatrix} C o f ( C ) = + ( 1 − 3 ) − ( 3 − 6 ) + ( 3 − 2 ) − ( 1 − 0 ) + ( 1 − 0 ) − ( 1 − 2 ) + ( 3 − 0 ) − ( 3 − 0 ) + ( 1 − 3 ) = − 2 3 1 − 1 1 1 3 − 3 − 2 La adjunta de C C C es la traspuesta de la matriz de cofactores:
a d j ( C ) = C o f ( C ) T = ( − 2 3 1 − 1 1 1 3 − 3 − 2 ) adj(C) = Cof(C)^T = \begin{pmatrix} -2 & 3 & 1 \\ -1 & 1 & 1 \\ 3 & -3 & -2 \end{pmatrix} a d j ( C ) = C o f ( C ) T = − 2 − 1 3 3 1 − 3 1 1 − 2 La inversa de C C C es:
C − 1 = 1 det ( C ) a d j ( C ) = 1 1 ( − 2 3 1 − 1 1 1 3 − 3 − 2 ) = ( − 2 3 1 − 1 1 1 3 − 3 − 2 ) C^{-1} = \frac{1}{\det(C)} adj(C) = \frac{1}{1} \begin{pmatrix} -2 & 3 & 1 \\ -1 & 1 & 1 \\ 3 & -3 & -2 \end{pmatrix} = \begin{pmatrix} -2 & 3 & 1 \\ -1 & 1 & 1 \\ 3 & -3 & -2 \end{pmatrix} C − 1 = det ( C ) 1 a d j ( C ) = 1 1 − 2 − 1 3 3 1 − 3 1 1 − 2 = − 2 − 1 3 3 1 − 3 1 1 − 2 La ecuación P ⋅ X ⋅ C = R P \cdot X \cdot C = R P ⋅ X ⋅ C = R se resuelve como X = P − 1 ⋅ R ⋅ C − 1 X = P^{-1} \cdot R \cdot C^{-1} X = P − 1 ⋅ R ⋅ C − 1 .
Primero, calculamos P − 1 ⋅ R P^{-1} \cdot R P − 1 ⋅ R :
P − 1 ⋅ R = ( − 2 / 5 1 / 5 1 / 5 2 / 5 ) ⋅ ( 1 0 0 0 1 0 ) = ( − 2 / 5 1 / 5 0 1 / 5 2 / 5 0 ) P^{-1} \cdot R = \begin{pmatrix} -2/5 & 1/5 \\ 1/5 & 2/5 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -2/5 & 1/5 & 0 \\ 1/5 & 2/5 & 0 \end{pmatrix} P − 1 ⋅ R = ( − 2/5 1/5 1/5 2/5 ) ⋅ ( 1 0 0 1 0 0 ) = ( − 2/5 1/5 1/5 2/5 0 0 ) Finalmente, calculamos X = ( P − 1 ⋅ R ) ⋅ C − 1 X = (P^{-1} \cdot R) \cdot C^{-1} X = ( P − 1 ⋅ R ) ⋅ C − 1 :
X = ( − 2 / 5 1 / 5 0 1 / 5 2 / 5 0 ) ⋅ ( − 2 3 1 − 1 1 1 3 − 3 − 2 ) X = \begin{pmatrix} -2/5 & 1/5 & 0 \\ 1/5 & 2/5 & 0 \end{pmatrix} \cdot \begin{pmatrix} -2 & 3 & 1 \\ -1 & 1 & 1 \\ 3 & -3 & -2 \end{pmatrix} X = ( − 2/5 1/5 1/5 2/5 0 0 ) ⋅ − 2 − 1 3 3 1 − 3 1 1 − 2 X = ( ( − 2 / 5 ) ( − 2 ) + ( 1 / 5 ) ( − 1 ) + ( 0 ) ( 3 ) ( − 2 / 5 ) ( 3 ) + ( 1 / 5 ) ( 1 ) + ( 0 ) ( − 3 ) ( − 2 / 5 ) ( 1 ) + ( 1 / 5 ) ( 1 ) + ( 0 ) ( − 2 ) ( 1 / 5 ) ( − 2 ) + ( 2 / 5 ) ( − 1 ) + ( 0 ) ( 3 ) ( 1 / 5 ) ( 3 ) + ( 2 / 5 ) ( 1 ) + ( 0 ) ( − 3 ) ( 1 / 5 ) ( 1 ) + ( 2 / 5 ) ( 1 ) + ( 0 ) ( − 2 ) ) X = \begin{pmatrix} (-2/5)(-2)+(1/5)(-1)+(0)(3) & (-2/5)(3)+(1/5)(1)+(0)(-3) & (-2/5)(1)+(1/5)(1)+(0)(-2) \\ (1/5)(-2)+(2/5)(-1)+(0)(3) & (1/5)(3)+(2/5)(1)+(0)(-3) & (1/5)(1)+(2/5)(1)+(0)(-2) \end{pmatrix} X = ( ( − 2/5 ) ( − 2 ) + ( 1/5 ) ( − 1 ) + ( 0 ) ( 3 ) ( 1/5 ) ( − 2 ) + ( 2/5 ) ( − 1 ) + ( 0 ) ( 3 ) ( − 2/5 ) ( 3 ) + ( 1/5 ) ( 1 ) + ( 0 ) ( − 3 ) ( 1/5 ) ( 3 ) + ( 2/5 ) ( 1 ) + ( 0 ) ( − 3 ) ( − 2/5 ) ( 1 ) + ( 1/5 ) ( 1 ) + ( 0 ) ( − 2 ) ( 1/5 ) ( 1 ) + ( 2/5 ) ( 1 ) + ( 0 ) ( − 2 ) ) X = ( 4 / 5 − 1 / 5 + 0 − 6 / 5 + 1 / 5 + 0 − 2 / 5 + 1 / 5 + 0 − 2 / 5 − 2 / 5 + 0 3 / 5 + 2 / 5 + 0 1 / 5 + 2 / 5 + 0 ) X = \begin{pmatrix} 4/5-1/5+0 & -6/5+1/5+0 & -2/5+1/5+0 \\ -2/5-2/5+0 & 3/5+2/5+0 & 1/5+2/5+0 \end{pmatrix} X = ( 4/5 − 1/5 + 0 − 2/5 − 2/5 + 0 − 6/5 + 1/5 + 0 3/5 + 2/5 + 0 − 2/5 + 1/5 + 0 1/5 + 2/5 + 0 ) X = ( 3 / 5 − 5 / 5 − 1 / 5 − 4 / 5 5 / 5 3 / 5 ) X = \begin{pmatrix} 3/5 & -5/5 & -1/5 \\ -4/5 & 5/5 & 3/5 \end{pmatrix} X = ( 3/5 − 4/5 − 5/5 5/5 − 1/5 3/5 ) X = ( 3 / 5 − 1 − 1 / 5 − 4 / 5 1 3 / 5 ) X = \begin{pmatrix} 3/5 & -1 & -1/5 \\ -4/5 & 1 & 3/5 \end{pmatrix} X = ( 3/5 − 4/5 − 1 1 − 1/5 3/5 ) b) Halle las dimensiones de las matrices D D D y E E E para que tenga sentido la igualdad A ⋅ D = E ⋅ B A \cdot D = E \cdot B A ⋅ D = E ⋅ B . Las dimensiones de las matrices dadas son:
A A A : 2 × 3 2 \times 3 2 × 3
B B B : 3 × 2 3 \times 2 3 × 2
Para que el producto A ⋅ D A \cdot D A ⋅ D esté definido, el número de columnas de A A A debe ser igual al número de filas de D D D . Por lo tanto, si D D D tiene dimensiones r D × c D r_D \times c_D r D × c D , entonces r D = 3 r_D = 3 r D = 3 . La matriz resultante A ⋅ D A \cdot D A ⋅ D tendrá dimensiones 2 × c D 2 \times c_D 2 × c D . Para que el producto E ⋅ B E \cdot B E ⋅ B esté definido, el número de columnas de E E E debe ser igual al número de filas de B B B . Por lo tanto, si E E E tiene dimensiones r E × c E r_E \times c_E r E × c E , entonces c E = 3 c_E = 3 c E = 3 . La matriz resultante E ⋅ B E \cdot B E ⋅ B tendrá dimensiones r E × 2 r_E \times 2 r E × 2 . Para que la igualdad A ⋅ D = E ⋅ B A \cdot D = E \cdot B A ⋅ D = E ⋅ B tenga sentido, las matrices resultantes deben tener las mismas dimensiones. Es decir, ( A ⋅ D ) 2 × c D (A \cdot D)_{2 \times c_D} ( A ⋅ D ) 2 × c D debe ser igual a ( E ⋅ B ) r E × 2 (E \cdot B)_{r_E \times 2} ( E ⋅ B ) r E × 2 .
Esto implica que:
2 = r E 2 = r_E 2 = r E
c D = 2 c_D = 2 c D = 2 Por lo tanto, las dimensiones de D D D son 3 × 2 3 \times 2 3 × 2 (ya que r D = 3 r_D = 3 r D = 3 y c D = 2 c_D = 2 c D = 2 ).
Y las dimensiones de E E E son 2 × 3 2 \times 3 2 × 3 (ya que r E = 2 r_E = 2 r E = 2 y c E = 3 c_E = 3 c E = 3 ).
Verificación:
A 2 × 3 ⋅ D 3 × 2 → ( A ⋅ D ) 2 × 2 A_{2 \times 3} \cdot D_{3 \times 2} \rightarrow (A \cdot D)_{2 \times 2} A 2 × 3 ⋅ D 3 × 2 → ( A ⋅ D ) 2 × 2
E 2 × 3 ⋅ B 3 × 2 → ( E ⋅ B ) 2 × 2 E_{2 \times 3} \cdot B_{3 \times 2} \rightarrow (E \cdot B)_{2 \times 2} E 2 × 3 ⋅ B 3 × 2 → ( E ⋅ B ) 2 × 2
Las dimensiones son consistentes.