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Sagot :
Claro, vamos a resolver los problemas dados paso a paso.
Primero, se nos dan las matrices:
[tex]\[ A = \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} \][/tex]
### i) Calcular [tex]\( 2A - B \)[/tex]
Primero, calculamos [tex]\( 2A \)[/tex]:
[tex]\[ 2A = 2 \times \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times -1 & 2 \times 0 \\ 2 \times -3 & 2 \times 2 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ -6 & 4 \end{pmatrix} \][/tex]
Luego, restamos la matriz [tex]\( B \)[/tex] de [tex]\( 2A \)[/tex]:
[tex]\[ 2A - B = \begin{pmatrix} -2 & 0 \\ -6 & 4 \end{pmatrix} - \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} = \begin{pmatrix} -2 - (-1) & 0 - (-2) \\ -6 - (-3) & 4 - (-5) \end{pmatrix} \][/tex]
Realizamos las restas en cada elemento:
[tex]\[ 2A - B = \begin{pmatrix} -2 + 1 & 0 + 2 \\ -6 + 3 & 4 + 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
Por lo tanto, la respuesta a [tex]\( 2A - B \)[/tex] es:
[tex]\[ \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
### ii) Calcular [tex]\( AB \)[/tex]
Para multiplicar las matrices [tex]\( A \)[/tex] y [tex]\( B \)[/tex], utilizamos la regla de multiplicación de matrices:
[tex]\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{pmatrix} \][/tex]
Sustituimos los elementos de las matrices [tex]\( A \)[/tex] y [tex]\( B \)[/tex]:
[tex]\[ AB = \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} = \begin{pmatrix} (-1 \cdot -1 + 0 \cdot -3) & (-1 \cdot -2 + 0 \cdot -5) \\ (-3 \cdot -1 + 2 \cdot -3) & (-3 \cdot -2 + 2 \cdot -5) \end{pmatrix} \][/tex]
Realizamos los productos y sumas:
[tex]\[ AB = \begin{pmatrix} 1 + 0 & 2 + 0 \\ 3 - 6 & 6 - 10 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
Por lo tanto, la respuesta a [tex]\( AB \)[/tex] es:
[tex]\[ \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
### Resumen de resultados:
i) [tex]\( 2A - B \)[/tex]:
[tex]\[ \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
ii) [tex]\( AB \)[/tex]:
[tex]\[ \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
Primero, se nos dan las matrices:
[tex]\[ A = \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} \][/tex]
### i) Calcular [tex]\( 2A - B \)[/tex]
Primero, calculamos [tex]\( 2A \)[/tex]:
[tex]\[ 2A = 2 \times \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 \times -1 & 2 \times 0 \\ 2 \times -3 & 2 \times 2 \end{pmatrix} = \begin{pmatrix} -2 & 0 \\ -6 & 4 \end{pmatrix} \][/tex]
Luego, restamos la matriz [tex]\( B \)[/tex] de [tex]\( 2A \)[/tex]:
[tex]\[ 2A - B = \begin{pmatrix} -2 & 0 \\ -6 & 4 \end{pmatrix} - \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} = \begin{pmatrix} -2 - (-1) & 0 - (-2) \\ -6 - (-3) & 4 - (-5) \end{pmatrix} \][/tex]
Realizamos las restas en cada elemento:
[tex]\[ 2A - B = \begin{pmatrix} -2 + 1 & 0 + 2 \\ -6 + 3 & 4 + 5 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
Por lo tanto, la respuesta a [tex]\( 2A - B \)[/tex] es:
[tex]\[ \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
### ii) Calcular [tex]\( AB \)[/tex]
Para multiplicar las matrices [tex]\( A \)[/tex] y [tex]\( B \)[/tex], utilizamos la regla de multiplicación de matrices:
[tex]\[ AB = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22}b_{22} \end{pmatrix} \][/tex]
Sustituimos los elementos de las matrices [tex]\( A \)[/tex] y [tex]\( B \)[/tex]:
[tex]\[ AB = \begin{pmatrix} -1 & 0 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} -1 & -2 \\ -3 & -5 \end{pmatrix} = \begin{pmatrix} (-1 \cdot -1 + 0 \cdot -3) & (-1 \cdot -2 + 0 \cdot -5) \\ (-3 \cdot -1 + 2 \cdot -3) & (-3 \cdot -2 + 2 \cdot -5) \end{pmatrix} \][/tex]
Realizamos los productos y sumas:
[tex]\[ AB = \begin{pmatrix} 1 + 0 & 2 + 0 \\ 3 - 6 & 6 - 10 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
Por lo tanto, la respuesta a [tex]\( AB \)[/tex] es:
[tex]\[ \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
### Resumen de resultados:
i) [tex]\( 2A - B \)[/tex]:
[tex]\[ \begin{pmatrix} -1 & 2 \\ -3 & 9 \end{pmatrix} \][/tex]
ii) [tex]\( AB \)[/tex]:
[tex]\[ \begin{pmatrix} 1 & 2 \\ -3 & -4 \end{pmatrix} \][/tex]
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