Bernard Tran
07/03/2023 · Primary School
Determinación de los productos de dos matrices En los ejercicios 17 a 30, halle (a) \( A B \) y (b) \( B A \) (si están definidas). 17. \( A=\left[\begin{array}{rr}1 & 2 \\ 4 & 2\end{array}\right], \quad B=\left[\begin{array}{rr}2 & -1 \\ -1 & 8\end{array}\right] \) 18. \( A=\left[\begin{array}{rr}2 & -2 \\ -1 & 4\end{array}\right], \quad B=\left[\begin{array}{rr}4 & 1 \\ 2 & -2\end{array}\right] \)
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Para resolver los ejercicios, vamos a calcular los productos de las matrices \( A \) y \( B \) en ambos órdenes, \( A B \) y \( B A \), si están definidos.
### Ejercicio 17
Dadas las matrices:
\[
A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -1 \\ -1 & 8 \end{bmatrix}
\]
#### (a) Calcular \( A B \)
\[
A B = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 8 \end{bmatrix}
\]
Calculamos cada elemento de la matriz resultante:
- Elemento (1,1):
\[
1 \cdot 2 + 2 \cdot (-1) = 2 - 2 = 0
\]
- Elemento (1,2):
\[
1 \cdot (-1) + 2 \cdot 8 = -1 + 16 = 15
\]
- Elemento (2,1):
\[
4 \cdot 2 + 2 \cdot (-1) = 8 - 2 = 6
\]
- Elemento (2,2):
\[
4 \cdot (-1) + 2 \cdot 8 = -4 + 16 = 12
\]
Por lo tanto,
\[
A B = \begin{bmatrix} 0 & 15 \\ 6 & 12 \end{bmatrix}
\]
#### (b) Calcular \( B A \)
\[
B A = \begin{bmatrix} 2 & -1 \\ -1 & 8 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}
\]
Calculamos cada elemento de la matriz resultante:
- Elemento (1,1):
\[
2 \cdot 1 + (-1) \cdot 4 = 2 - 4 = -2
\]
- Elemento (1,2):
\[
2 \cdot 2 + (-1) \cdot 2 = 4 - 2 = 2
\]
- Elemento (2,1):
\[
-1 \cdot 1 + 8 \cdot 4 = -1 + 32 = 31
\]
- Elemento (2,2):
\[
-1 \cdot 2 + 8 \cdot 2 = -2 + 16 = 14
\]
Por lo tanto,
\[
B A = \begin{bmatrix} -2 & 2 \\ 31 & 14 \end{bmatrix}
\]
### Resumen del Ejercicio 17
\[
A B = \begin{bmatrix} 0 & 15 \\ 6 & 12 \end{bmatrix}, \quad B A = \begin{bmatrix} -2 & 2 \\ 31 & 14 \end{bmatrix}
\]
---
### Ejercicio 18
Dadas las matrices:
\[
A = \begin{bmatrix} 2 & -2 \\ -1 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 1 \\ 2 & -2 \end{bmatrix}
\]
#### (a) Calcular \( A B \)
\[
A B = \begin{bmatrix} 2 & -2 \\ -1 & 4 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & -2 \end{bmatrix}
\]
Calculamos cada elemento de la matriz resultante:
- Elemento (1,1):
\[
2 \cdot 4 + (-2) \cdot 2 = 8 - 4 = 4
\]
- Elemento (1,2):
\[
2 \cdot 1 + (-2) \cdot (-2) = 2 + 4 = 6
\]
- Elemento (2,1):
\[
-1 \cdot 4 + 4 \cdot 2 = -4 + 8 = 4
\]
- Elemento (2,2):
\[
-1 \cdot 1 + 4 \cdot (-2) = -1 - 8 = -9
\]
Por lo tanto,
\[
A B = \begin{bmatrix} 4 & 6 \\ 4 & -9 \end{bmatrix}
\]
#### (b) Calcular \( B A \)
\[
B A = \begin{bmatrix} 4 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ -1 & 4 \end{bmatrix}
\]
Calculamos cada elemento de la matriz resultante:
- Elemento (1,1):
\[
4 \cdot 2 + 1 \cdot (-1) = 8 - 1 = 7
\]
- Elemento (1,2):
\[
4 \cdot (-2) + 1 \cdot 4 = -8 + 4 = -4
\]
- Elemento (2,1):
\[
2 \cdot 2 + (-2) \cdot (-1) = 4 + 2 = 6
\]
- Elemento (2,2):
\[
2 \cdot (-2) + (-2) \cdot 4 = -4 - 8 = -12
\]
Por lo tanto,
\[
B A = \begin{bmatrix} 7 & -4 \\ 6 & -12 \end{bmatrix}
\]
### Resumen del Ejercicio 18
\[
A B = \begin{bmatrix} 4 & 6 \\ 4 & -9 \end{bmatrix}, \quad B A = \begin{bmatrix} 7 & -4 \\ 6 & -12 \end{bmatrix}
\]
Quick Answer
### Ejercicio 17
\[
A B = \begin{bmatrix} 0 & 15 \\ 6 & 12 \end{bmatrix}, \quad B A = \begin{bmatrix} -2 & 2 \\ 31 & 14 \end{bmatrix}
\]
### Ejercicio 18
\[
A B = \begin{bmatrix} 4 & 6 \\ 4 & -9 \end{bmatrix}, \quad B A = \begin{bmatrix} 7 & -4 \\ 6 & -12 \end{bmatrix}
\]
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