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Sagot :
Claro, vamos a realizar la verificación de que [tex]\( AB \)[/tex] es simétrica para las dos parejas de matrices [tex]\( A \)[/tex] y [tex]\( B \)[/tex].
### Parte (a)
Dadas las matrices:
[tex]\[ A = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \][/tex]
[tex]\[ B = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \][/tex]
Paso 1: Calculemos [tex]\( AB \)[/tex].
[tex]\[ AB = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} = \begin{array}{cc} (2 \cdot 3 + 0 \cdot 0) & (2 \cdot 0 + 0 \cdot -10) \\ (0 \cdot 3 + 5 \cdot 0) & (0 \cdot 0 + 5 \cdot -10) \end{array} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Paso 2: Calculemos [tex]\( B^{\top} A^{\top} \)[/tex].
Primero tomamos las transpuestas:
[tex]\[ A^{\top} = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \][/tex] (como A es simétrica, A = [tex]\( A^{\top} \)[/tex])
[tex]\[ B^{\top} = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \][/tex] (como B es simétrica, B = [tex]\( B^{\top} \)[/tex])
Luego calculamos el producto:
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} = \begin{array}{cc} (3 \cdot 2 + 0 \cdot 0) & (3 \cdot 0 + 0 \cdot 5) \\ (0 \cdot 2 + -10 \cdot 0) & (0 \cdot 0 + -10 \cdot 5) \end{array} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Paso 3: Comparación
Comparando los resultados de [tex]\( AB \)[/tex] y [tex]\( B^{\top} A^{\top} \)[/tex]:
[tex]\[ AB = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Los dos son iguales, por lo que [tex]\( AB \)[/tex] es simétrica.
### Parte (b)
Dadas las matrices:
[tex]\[ A = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \][/tex]
[tex]\[ B = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \][/tex]
Paso 1: Calculemos [tex]\( AB \)[/tex].
[tex]\[ AB = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} = \begin{array}{cc} (0 \cdot 1 + -5 \cdot 2) & (0 \cdot 2 + -5 \cdot 0) \\ (-5 \cdot 1 + 1 \cdot 2) & (-5 \cdot 2 + 1 \cdot 0) \end{array} = \begin{array}{cc} -10 & 0 \\ -3 & -10 \end{array} \][/tex]
Paso 2: Calculemos [tex]\( B^{\top} A^{\top} \)[/tex].
Primero las transpuestas:
[tex]\[ A^{\top} = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \][/tex] (como A es simétrica, A = [tex]\( A^{\top} \)[/tex])
[tex]\[ B^{\top} = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \][/tex] (como B es simétrica, B = [tex]\( B^{\top} \)[/tex])
Luego calculamos el producto:
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} = \begin{array}{cc} (1 \cdot 0 + 2 \cdot -5) & (1 \cdot -5 + 2 \cdot 1) \\ (2 \cdot 0 + 0 \cdot -5) & (2 \cdot -5 + 0 \cdot 1) \end{array} = \begin{array}{cc} -10 & -3 \\ 0 & -10 \end{array} \][/tex]
Paso 3: Comparación
Comparando los resultados de [tex]\( AB \)[/tex] y [tex]\( B^{\top} A^{\top} \)[/tex]:
[tex]\[ AB = \begin{array}{cc} -10 & 0 \\ -3 & -10 \end{array} \][/tex]
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} -10 & -3 \\ 0 & -10 \end{array} \][/tex]
Los dos no son iguales, por lo que [tex]\( AB \)[/tex] no es simétrica.
### Resumen
- En la parte (a), [tex]\( AB \)[/tex] resultó simétrica.
- En la parte (b), [tex]\( AB \)[/tex] no resultó simétrica.
### Parte (a)
Dadas las matrices:
[tex]\[ A = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \][/tex]
[tex]\[ B = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \][/tex]
Paso 1: Calculemos [tex]\( AB \)[/tex].
[tex]\[ AB = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} = \begin{array}{cc} (2 \cdot 3 + 0 \cdot 0) & (2 \cdot 0 + 0 \cdot -10) \\ (0 \cdot 3 + 5 \cdot 0) & (0 \cdot 0 + 5 \cdot -10) \end{array} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Paso 2: Calculemos [tex]\( B^{\top} A^{\top} \)[/tex].
Primero tomamos las transpuestas:
[tex]\[ A^{\top} = \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \][/tex] (como A es simétrica, A = [tex]\( A^{\top} \)[/tex])
[tex]\[ B^{\top} = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \][/tex] (como B es simétrica, B = [tex]\( B^{\top} \)[/tex])
Luego calculamos el producto:
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 3 & 0 \\ 0 & -10 \end{array} \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} = \begin{array}{cc} (3 \cdot 2 + 0 \cdot 0) & (3 \cdot 0 + 0 \cdot 5) \\ (0 \cdot 2 + -10 \cdot 0) & (0 \cdot 0 + -10 \cdot 5) \end{array} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Paso 3: Comparación
Comparando los resultados de [tex]\( AB \)[/tex] y [tex]\( B^{\top} A^{\top} \)[/tex]:
[tex]\[ AB = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 6 & 0 \\ 0 & -50 \end{array} \][/tex]
Los dos son iguales, por lo que [tex]\( AB \)[/tex] es simétrica.
### Parte (b)
Dadas las matrices:
[tex]\[ A = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \][/tex]
[tex]\[ B = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \][/tex]
Paso 1: Calculemos [tex]\( AB \)[/tex].
[tex]\[ AB = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} = \begin{array}{cc} (0 \cdot 1 + -5 \cdot 2) & (0 \cdot 2 + -5 \cdot 0) \\ (-5 \cdot 1 + 1 \cdot 2) & (-5 \cdot 2 + 1 \cdot 0) \end{array} = \begin{array}{cc} -10 & 0 \\ -3 & -10 \end{array} \][/tex]
Paso 2: Calculemos [tex]\( B^{\top} A^{\top} \)[/tex].
Primero las transpuestas:
[tex]\[ A^{\top} = \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} \][/tex] (como A es simétrica, A = [tex]\( A^{\top} \)[/tex])
[tex]\[ B^{\top} = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \][/tex] (como B es simétrica, B = [tex]\( B^{\top} \)[/tex])
Luego calculamos el producto:
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} 1 & 2 \\ 2 & 0 \end{array} \begin{array}{cc} 0 & -5 \\ -5 & 1 \end{array} = \begin{array}{cc} (1 \cdot 0 + 2 \cdot -5) & (1 \cdot -5 + 2 \cdot 1) \\ (2 \cdot 0 + 0 \cdot -5) & (2 \cdot -5 + 0 \cdot 1) \end{array} = \begin{array}{cc} -10 & -3 \\ 0 & -10 \end{array} \][/tex]
Paso 3: Comparación
Comparando los resultados de [tex]\( AB \)[/tex] y [tex]\( B^{\top} A^{\top} \)[/tex]:
[tex]\[ AB = \begin{array}{cc} -10 & 0 \\ -3 & -10 \end{array} \][/tex]
[tex]\[ B^{\top} A^{\top} = \begin{array}{cc} -10 & -3 \\ 0 & -10 \end{array} \][/tex]
Los dos no son iguales, por lo que [tex]\( AB \)[/tex] no es simétrica.
### Resumen
- En la parte (a), [tex]\( AB \)[/tex] resultó simétrica.
- En la parte (b), [tex]\( AB \)[/tex] no resultó simétrica.
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