Questions: Sean B=[0 5; -4 -5; 2 1] y C=[0 -5; 2 -5; 1 7]. Hallar -6B-5C.

Transcript text: Sean $B=\left[\begin{array}{cc}0 & 5 \\ -4 & -5 \\ 2 & 1\end{array}\right]$ y $C=\left[\begin{array}{cc}0 & -5 \\ 2 & -5 \\ 1 & 7\end{array}\right]$. Hallar $-6 B-5 C$.

Solution

To solve the problem of finding $-6B - 5C$, we need to perform scalar multiplication on each matrix and then subtract the resulting matrices. First, multiply each element of matrix $B$ by $-6$ and each element of matrix $C$ by $-5$. Then, subtract the corresponding elements of the two resulting matrices to get the final matrix.

Paso 1: Definición de las matrices

Sean las matrices $ B $ y $ C $ definidas como: \[ B = \begin{bmatrix} 0 & 5 \\ -4 & -5 \\ 2 & 1 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & -5 \\ 2 & -5 \\ 1 & 7 \end{bmatrix} \]

Paso 2: Cálculo de $-6B$

Multiplicamos cada elemento de la matriz $ B $ por $-6$: \[ -6B = -6 \cdot \begin{bmatrix} 0 & 5 \\ -4 & -5 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -30 \\ 24 & 30 \\ -12 & -6 \end{bmatrix} \]

Paso 3: Cálculo de $-5C$

Multiplicamos cada elemento de la matriz $ C $ por $-5$: \[ -5C = -5 \cdot \begin{bmatrix} 0 & -5 \\ 2 & -5 \\ 1 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 25 \\ -10 & 25 \\ -5 & -35 \end{bmatrix} \]

Paso 4: Cálculo de $-6B - 5C$

Restamos las matrices obtenidas en los pasos anteriores: \[ -6B - 5C = \begin{bmatrix} 0 & -30 \\ 24 & 30 \\ -12 & -6 \end{bmatrix} - \begin{bmatrix} 0 & 25 \\ -10 & 25 \\ -5 & -35 \end{bmatrix} = \begin{bmatrix} 0 - 0 & -30 - 25 \\ 24 - (-10) & 30 - 25 \\ -12 - (-5) & -6 - (-35) \end{bmatrix} \] \[ = \begin{bmatrix} 0 & -55 \\ 34 & 5 \\ -7 & 29 \end{bmatrix} \]

Respuesta Final

La matriz resultante es: \[ \boxed{\begin{bmatrix} 0 & -55 \\ 34 & 5 \\ -7 & 29 \end{bmatrix}} \]

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