Horton Horton
07/27/2023 · Senior High School

reducir Ia maTrig a una forma Triangular superior \( B=\left[\begin{array}{llll}1 & 3 & 2 & 1 \\ 3 & 5 & 3 & 2 \\ 3 & 6 & 3 & 2 \\ 6 & 4 & 5 & 3\end{array}\right] \)

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Para reducir la matriz \( B \) a una forma triangular superior, utilizaremos el método de eliminación de Gauss. La matriz inicial es: \[ B = \begin{bmatrix} 1 & 3 & 2 & 1 \\ 3 & 5 & 3 & 2 \\ 3 & 6 & 3 & 2 \\ 6 & 4 & 5 & 3 \end{bmatrix} \] ### Paso 1: Hacer ceros debajo del primer pivote El primer pivote es el elemento \( B[1,1] = 1 \). Usaremos este pivote para hacer ceros en las posiciones \( B[2,1] \), \( B[3,1] \) y \( B[4,1] \). - Para \( B[2,1] \): \[ R_2 \leftarrow R_2 - 3R_1 \implies R_2 = [3, 5, 3, 2] - 3[1, 3, 2, 1] = [0, -4, -3, -1] \] - Para \( B[3,1] \): \[ R_3 \leftarrow R_3 - 3R_1 \implies R_3 = [3, 6, 3, 2] - 3[1, 3, 2, 1] = [0, -3, -3, -1] \] - Para \( B[4,1] \): \[ R_4 \leftarrow R_4 - 6R_1 \implies R_4 = [6, 4, 5, 3] - 6[1, 3, 2, 1] = [0, -14, -7, -3] \] La matriz ahora es: \[ B = \begin{bmatrix} 1 & 3 & 2 & 1 \\ 0 & -4 & -3 & -1 \\ 0 & -3 & -3 & -1 \\ 0 & -14 & -7 & -3 \end{bmatrix} \] ### Paso 2: Hacer ceros debajo del segundo pivote El segundo pivote es \( B[2,2] = -4 \). Usaremos este pivote para hacer ceros en \( B[3,2] \) y \( B[4,2] \). - Para \( B[3,2] \): \[ R_3 \leftarrow R_3 - \frac{-3}{-4}R_2 \implies R_3 = [0, -3, -3, -1] - \frac{3}{4}[0, -4, -3, -1] = [0, 0, -\frac{9}{4}, -\frac{1}{4}] \] - Para \( B[4,2] \): \[ R_4 \leftarrow R_4 - \frac{-14}{-4}R_2 \implies R_4 = [0, -14, -7, -3] - \frac{14}{4}[0, -4, -3, -1] = [0, 0, \frac{7}{2}, \frac{1}{2}] \] La matriz ahora es: \[ B = \begin{bmatrix} 1 & 3 & 2 & 1 \\ 0 & -4 & -3 & -1 \\ 0 & 0 & -\frac{9}{4} & -\frac{1}{4} \\ 0 & 0 & \frac{7}{2} & \frac{1}{2} \end{bmatrix} \] ### Paso 3: Hacer ceros debajo del tercer pivote El tercer pivote es \( B[3,3] = -\frac{9}{4} \). Usaremos este pivote para hacer ceros en \( B[4,3] \). - Para \( B[4,3] \): \[ R_4 \leftarrow R_4 - \frac{\frac{7}{2}}{-\frac{9}{4}}R_3 \implies R_4 = [0, 0, \frac{7}{2}, \frac{1}{2}] - \frac{7}{2} \cdot \left(-\frac{4}{9}\right)[0, 0, -\frac{9}{4}, -\frac{1}{4}] = [0, 0, 0, 0] \] La

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Para reducir la matriz \( B \) a una forma triangular superior, se utilizó el método de eliminación de Gauss. La matriz se transformó de la siguiente manera: \[ B = \begin{bmatrix} 1 & 3 & 2 & 1 \\ 0 & -4 & -3 & -1 \\ 0 & 0 & -\frac{9}{4} & -\frac{1}{4} \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
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