Hilton Campos
02/12/2024 · Primary School
\( A=\left(\begin{array}{ccc}1 & -4 & -4 \\ 2 & 2 & 0 \\ -1 & 1 & 2\end{array}\right) ; B=\left(\begin{array}{cc}6 & -1 \\ 4 & 5 \\ 3 & -4\end{array}\right) ; C=\left(\begin{array}{ccc}3 & 4 & 6 \\ 1 & -2 & 0\end{array}\right) ; D=\left(\begin{array}{ccc}6 & 7 & 0 \\ 3 & -2 & 4 \\ 0 & -1 & 0\end{array}\right) \) Realice las operaciones algebralcas correspondientes segan \( U= \) \( (2 B) \cdot(4 C)+(A-5 D)^{T} M \) obtenga la matriz \( U \).
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Tutor-Verified Answer
Step-by-step Solution
Para calcular la matriz \( U \) según la expresión dada:
\[
U = (2B) \cdot (4C) + (A - 5D)^{T} M
\]
Primero, calculamos \( 2B \):
\[
B = \begin{pmatrix} 6 & -1 \\ 4 & 5 \\ 3 & -4 \end{pmatrix}
\]
\[
2B = 2 \cdot \begin{pmatrix} 6 & -1 \\ 4 & 5 \\ 3 & -4 \end{pmatrix} = \begin{pmatrix} 12 & -2 \\ 8 & 10 \\ 6 & -8 \end{pmatrix}
\]
Ahora, calculamos \( 4C \):
\[
C = \begin{pmatrix} 3 & 4 & 6 \\ 1 & -2 & 0 \end{pmatrix}
\]
\[
4C = 4 \cdot \begin{pmatrix} 3 & 4 & 6 \\ 1 & -2 & 0 \end{pmatrix} = \begin{pmatrix} 12 & 16 & 24 \\ 4 & -8 & 0 \end{pmatrix}
\]
Ahora, realizamos el producto \( (2B) \cdot (4C) \). Notamos que \( 2B \) es una matriz de \( 3 \times 2 \) y \( 4C \) es una matriz de \( 2 \times 3 \), por lo que el resultado será una matriz de \( 3 \times 3 \).
Calculamos el producto:
\[
(2B) \cdot (4C) = \begin{pmatrix} 12 & -2 \\ 8 & 10 \\ 6 & -8 \end{pmatrix} \cdot \begin{pmatrix} 12 & 16 & 24 \\ 4 & -8 & 0 \end{pmatrix}
\]
Calculamos cada elemento de la matriz resultante:
1. Fila 1, Columna 1:
\[
12 \cdot 12 + (-2) \cdot 4 = 144 - 8 = 136
\]
2. Fila 1, Columna 2:
\[
12 \cdot 16 + (-2) \cdot (-8) = 192 + 16 = 208
\]
3. Fila 1, Columna 3:
\[
12 \cdot 24 + (-2) \cdot 0 = 288 + 0 = 288
\]
4. Fila 2, Columna 1:
\[
8 \cdot 12 + 10 \cdot 4 = 96 + 40 = 136
\]
5. Fila 2, Columna 2:
\[
8 \cdot 16 + 10 \cdot (-8) = 128 - 80 = 48
\]
6. Fila 2, Columna 3:
\[
8 \cdot 24 + 10 \cdot 0 = 192 + 0 = 192
\]
7. Fila 3, Columna 1:
\[
6 \cdot 12 + (-8) \cdot 4 = 72 - 32 = 40
\]
8. Fila 3, Columna 2:
\[
6 \cdot 16 + (-8) \cdot (-8) = 96 + 64 = 160
\]
9. Fila 3, Columna 3:
\[
6 \cdot 24 + (-8) \cdot 0 = 144 + 0 = 144
\]
Por lo tanto, el resultado de \( (2B) \cdot (4C) \) es:
\[
(2B) \cdot (4C) = \begin{pmatrix} 136 & 208 & 288 \\ 136 & 48 & 192 \\ 40 & 160 & 144 \end{pmatrix}
\]
Ahora, calculamos \( A - 5D \):
\[
A = \begin{pmatrix} 1 & -4 & -4 \\ 2 & 2 & 0 \\ -1 & 1 & 2 \end{pmatrix}, \quad D = \begin{pmatrix} 6 & 7 &
Quick Answer
\[
U = \begin{pmatrix} 136 & 208 & 288 \\ 136 & 48 & 192 \\ 40 & 160 & 144 \end{pmatrix}
\]
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