题(18分)分）用施密特正交化方法将下列向量组标准正交化alpha_(1)=(1,0,-1,1)^T,alpha_(2)=(1,-1,0,1)^T,alpha_(3)=(-1,1,1,0)^T.

1. **正交化** - $\beta_1 = \alpha_1 = (1, 0, -1, 1)^T$ - $\beta_2 = \alpha_2 - \frac{\alpha_2^T\beta_1}{\beta_1^T\beta_1}\beta_1 = \left(\frac{1}{3}, -1, \frac{2}{3}, \frac{1}{3}\right)^T$ - $\beta_3 = \alpha_3 - \frac{\alpha_3^T\beta_1}{\beta_1^T\beta_1}\beta_1 - \frac{\alpha_3^T\beta_2}{\beta_2^T\beta_2}\beta_2 = \left(-\frac{1}{5}, \frac{3}{5}, \frac{3}{5}, \frac{4}{5}\right)^T$ 2. **单位化** - $\eta_1 = \frac{\beta_1}{$\beta_1$} = \frac{1}{\sqrt{3}}(1, 0, -1, 1)^T$ - $\eta_2 = \frac{\beta_2}{$\beta_2$} = \frac{1}{\sqrt{15}}(1, -3, 2, 1)^T$ - $\eta_3 = \frac{\beta_3}{$\beta_3$} = \frac{1}{\sqrt{35}}(-1, 3, 3, 4)^T$ **答案：** \[ \boxed{ \begin{aligned} &\eta_1 = \frac{1}{\sqrt{3}}(1, 0, -1, 1)^T, \\ &\eta_2 = \frac{1}{\sqrt{15}}(1, -3, 2, 1)^T, \\ &\eta_3 = \frac{1}{\sqrt{35}}(-1, 3, 3, 4)^T. \end{aligned} } \]