题目
将向量beta表示为其余向量的线性组合 beta=(1,2,1,1)^T, alpha_1=(1,1,1,1)^T, alpha_2=(1,1,-1,-1)^T, alpha_3=(1,-1,1,-1)^T, alpha_4=(1,-1,-1,1)^TA. beta=(5)/(4)alpha_1-(1)/(4)alpha_2-(1)/(4)alpha_3-(1)/(4)alpha_4B. beta=(5)/(4)alpha_1+(1)/(4)alpha_2-(1)/(4)alpha_3-(1)/(4)alpha_4C. beta=(5)/(4)alpha_1+(1)/(4)alpha_2-(5)/(4)alpha_3-(1)/(4)alpha_4D. beta=(5)/(4)alpha_1+(1)/(4)alpha_2-(1)/(4)alpha_3+(1)/(4)alpha_4
将向量$\beta$表示为其余向量的线性组合 $\beta=(1,2,1,1)^T$, $\alpha_1=(1,1,1,1)^T$, $\alpha_2=(1,1,-1,-1)^T$, $\alpha_3=(1,-1,1,-1)^T$, $\alpha_4=(1,-1,-1,1)^T$
A. $\beta=\frac{5}{4}\alpha_1-\frac{1}{4}\alpha_2-\frac{1}{4}\alpha_3-\frac{1}{4}\alpha_4$
B. $\beta=\frac{5}{4}\alpha_1+\frac{1}{4}\alpha_2-\frac{1}{4}\alpha_3-\frac{1}{4}\alpha_4$
C. $\beta=\frac{5}{4}\alpha_1+\frac{1}{4}\alpha_2-\frac{5}{4}\alpha_3-\frac{1}{4}\alpha_4$
D. $\beta=\frac{5}{4}\alpha_1+\frac{1}{4}\alpha_2-\frac{1}{4}\alpha_3+\frac{1}{4}\alpha_4$
题目解答
答案
B. $\beta=\frac{5}{4}\alpha_1+\frac{1}{4}\alpha_2-\frac{1}{4}\alpha_3-\frac{1}{4}\alpha_4$
解析
步骤 1：建立方程组
将向量 $\beta = (1, 2, 1, 1)^T$ 表示为向量 $\alpha_1 = (1, 1, 1, 1)^T$, $\alpha_2 = (1, 1, -1, -1)^T$, $\alpha_3 = (1, -1, 1, -1)^T$, $\alpha_4 = (1, -1, -1, 1)^T$ 的线性组合，即求解方程组： \[ \begin{cases} x_1 + x_2 + x_3 + x_4 = 1 \\ x_1 + x_2 - x_3 - x_4 = 2 \\ x_1 - x_2 + x_3 - x_4 = 1 \\ x_1 - x_2 - x_3 + x_4 = 1 \end{cases} \]
步骤 2：解方程组
解上述方程组，得到：$x_1 = \frac{5}{4}$, $x_2 = \frac{1}{4}$, $x_3 = -\frac{1}{4}$, $x_4 = -\frac{1}{4}$。
步骤 3：表示向量 $\beta$
根据解得的系数，将向量 $\beta$ 表示为：$\beta = \frac{5}{4} \alpha_1 + \frac{1}{4} \alpha_2 - \frac{1}{4} \alpha_3 - \frac{1}{4} \alpha_4$。
将向量 $\beta = (1, 2, 1, 1)^T$ 表示为向量 $\alpha_1 = (1, 1, 1, 1)^T$, $\alpha_2 = (1, 1, -1, -1)^T$, $\alpha_3 = (1, -1, 1, -1)^T$, $\alpha_4 = (1, -1, -1, 1)^T$ 的线性组合，即求解方程组： \[ \begin{cases} x_1 + x_2 + x_3 + x_4 = 1 \\ x_1 + x_2 - x_3 - x_4 = 2 \\ x_1 - x_2 + x_3 - x_4 = 1 \\ x_1 - x_2 - x_3 + x_4 = 1 \end{cases} \]
步骤 2：解方程组
解上述方程组，得到：$x_1 = \frac{5}{4}$, $x_2 = \frac{1}{4}$, $x_3 = -\frac{1}{4}$, $x_4 = -\frac{1}{4}$。
步骤 3：表示向量 $\beta$
根据解得的系数，将向量 $\beta$ 表示为：$\beta = \frac{5}{4} \alpha_1 + \frac{1}{4} \alpha_2 - \frac{1}{4} \alpha_3 - \frac{1}{4} \alpha_4$。