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Sagot :
¡Claro! Vamos a resolver ambos apartados paso a paso.
Dadas las matrices:
[tex]\[ A = \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} \][/tex]
### Parte a) Calcular [tex]\( -2A + 3B \)[/tex]
1. Multiplicamos la matriz [tex]\(A\)[/tex] por [tex]\(-2\)[/tex]:
[tex]\[ -2A = -2 \cdot \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -12 & 4 \\ -4 & -8 \end{pmatrix} \][/tex]
2. Multiplicamos la matriz [tex]\(B\)[/tex] por 3:
[tex]\[ 3B = 3 \cdot \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} -9 & 0 \\ -6 & 6 \end{pmatrix} \][/tex]
3. Sumamos las matrices resultantes:
[tex]\[ -2A + 3B = \begin{pmatrix} -12 & 4 \\ -4 & -8 \end{pmatrix} + \begin{pmatrix} -9 & 0 \\ -6 & 6 \end{pmatrix} = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
Así que,
[tex]\[ -2A + 3B = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
### Parte b) Calcular [tex]\( \frac{1}{2} A \times B \)[/tex] (Multiplicación de matrices)
El producto de dos matrices [tex]\(A \times B\)[/tex] se obtiene multiplicando cada fila de [tex]\(A\)[/tex] por cada columna de [tex]\(B\)[/tex]:
1. Multiplicamos [tex]\(A\)[/tex] por [tex]\(B\)[/tex]:
[tex]\[ A \times B = \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} \times \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} \][/tex]
Para la posición (1,1):
[tex]\[ (6 \cdot -3) + (-2 \cdot -2) = -18 + 4 = -14 \][/tex]
Para la posición (1,2):
[tex]\[ (6 \cdot 0) + (-2 \cdot 2) = 0 - 4 = -4 \][/tex]
Para la posición (2,1):
[tex]\[ (2 \cdot -3) + (4 \cdot -2) = -6 - 8 = -14 \][/tex]
Para la posición (2,2):
[tex]\[ (2 \cdot 0) + (4 \cdot 2) = 0 + 8 = 8 \][/tex]
Entonces,
[tex]\[ A \times B = \begin{pmatrix} -14 & -4 \\ -14 & 8 \end{pmatrix} \][/tex]
2. Multiplicamos la matriz resultante por [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \frac{1}{2} \times \begin{pmatrix} -14 & -4 \\ -14 & 8 \end{pmatrix} = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
Así que,
[tex]\[ \frac{1}{2} A \times B = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
### Resumen Final
a)
[tex]\[ -2A + 3B = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
b)
[tex]\[ \frac{1}{2} A \times B = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
Dadas las matrices:
[tex]\[ A = \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} \][/tex]
### Parte a) Calcular [tex]\( -2A + 3B \)[/tex]
1. Multiplicamos la matriz [tex]\(A\)[/tex] por [tex]\(-2\)[/tex]:
[tex]\[ -2A = -2 \cdot \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} -12 & 4 \\ -4 & -8 \end{pmatrix} \][/tex]
2. Multiplicamos la matriz [tex]\(B\)[/tex] por 3:
[tex]\[ 3B = 3 \cdot \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} = \begin{pmatrix} -9 & 0 \\ -6 & 6 \end{pmatrix} \][/tex]
3. Sumamos las matrices resultantes:
[tex]\[ -2A + 3B = \begin{pmatrix} -12 & 4 \\ -4 & -8 \end{pmatrix} + \begin{pmatrix} -9 & 0 \\ -6 & 6 \end{pmatrix} = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
Así que,
[tex]\[ -2A + 3B = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
### Parte b) Calcular [tex]\( \frac{1}{2} A \times B \)[/tex] (Multiplicación de matrices)
El producto de dos matrices [tex]\(A \times B\)[/tex] se obtiene multiplicando cada fila de [tex]\(A\)[/tex] por cada columna de [tex]\(B\)[/tex]:
1. Multiplicamos [tex]\(A\)[/tex] por [tex]\(B\)[/tex]:
[tex]\[ A \times B = \begin{pmatrix} 6 & -2 \\ 2 & 4 \end{pmatrix} \times \begin{pmatrix} -3 & 0 \\ -2 & 2 \end{pmatrix} \][/tex]
Para la posición (1,1):
[tex]\[ (6 \cdot -3) + (-2 \cdot -2) = -18 + 4 = -14 \][/tex]
Para la posición (1,2):
[tex]\[ (6 \cdot 0) + (-2 \cdot 2) = 0 - 4 = -4 \][/tex]
Para la posición (2,1):
[tex]\[ (2 \cdot -3) + (4 \cdot -2) = -6 - 8 = -14 \][/tex]
Para la posición (2,2):
[tex]\[ (2 \cdot 0) + (4 \cdot 2) = 0 + 8 = 8 \][/tex]
Entonces,
[tex]\[ A \times B = \begin{pmatrix} -14 & -4 \\ -14 & 8 \end{pmatrix} \][/tex]
2. Multiplicamos la matriz resultante por [tex]\( \frac{1}{2} \)[/tex]:
[tex]\[ \frac{1}{2} \times \begin{pmatrix} -14 & -4 \\ -14 & 8 \end{pmatrix} = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
Así que,
[tex]\[ \frac{1}{2} A \times B = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
### Resumen Final
a)
[tex]\[ -2A + 3B = \begin{pmatrix} -21 & 4 \\ -10 & -2 \end{pmatrix} \][/tex]
b)
[tex]\[ \frac{1}{2} A \times B = \begin{pmatrix} -7 & -2 \\ -7 & 4 \end{pmatrix} \][/tex]
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