题目

完成下面的解题过程(共10空):求解三元线性方程组x_{1)-2 x_(2)+x_(3)=-2 2 x_(1)+x_(2)-3 x_(3)=1 -x_(1)+x_(2)-x_(3)=0.解 方程组的系数行列式D=|1 & -2 & 1 2 & 1 & -3 -1 & 1 & -1|= ___ neq 0,而D_(1)=|-2 & 1 1 & -3 1 & -1|= ___ ,D_(3)=|1 & -2 & -2 2 & 1 & 1 -1 & 1 & 0|= ___ .故所求方程组的解为:x_(1)=(D_(1))/(D)= ___ ,x_(2)=(D_(2))/(D)= ___ ,x_(3)=(D_(3))/(D)= ___ .

完成下面的解题过程(共10空): 求解三元线性方程组$\left\{\begin{array}{c}x_{1}-2 x_{2}+x_{3}=-2 \\ 2 x_{1}+x_{2}-3 x_{3}=1 \\ -x_{1}+x_{2}-x_{3}=0\end{array}\right.$ 解 方程组的系数行列式 $D=\left|\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 1 & -1\end{array}\right|=$ ___ $\neq 0,$ 而 $D_{1}=\left|\begin{array}{ccc}-2 & 1 \\ 1 & -3 \\ 1 & -1\end{array}\right|=$ ___ ,$D_{2}=\left|\begin{array}{ccc}1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 0 & -1\end{array}\right|=$ ___ , $D_{3}=\left|\begin{array}{ccc}1 & -2 & -2 \\ 2 & 1 & 1 \\ -1 & 1 & 0\end{array}\right|=$ ___ . 故所求方程组的解为: $x_{1}=\frac{D_{1}}{D}=$ ___ ,$x_{2}=\frac{D_{2}}{D}=$ ___ ,$x_{3}=\frac{D_{3}}{D}=$ ___ .

题目解答

答案

计算系数行列式 $D$： \[ D = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 1 & -1 \end{vmatrix} = -5 \quad (\text{展开计算得}) \] 计算 $D_1, D_2, D_3$： \[ D_1 = \begin{vmatrix} -2 & -2 & 1 \\ 1 & 1 & -3 \\ 0 & 1 & -1 \end{vmatrix} = -5, \quad D_2 = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 0 & -1 \end{vmatrix} = -10, \quad D_3 = \begin{vmatrix} 1 & -2 & -2 \\ 2 & 1 & 1 \\ -1 & 1 & 0 \end{vmatrix} = -5 \] 解为： \[ x_1 = \frac{D_1}{D} = 1, \quad x_2 = \frac{D_2}{D} = 2, \quad x_3 = \frac{D_3}{D} = 1 \] **答案：** \[ \boxed{ \begin{array}{ccc} D = -5, & D_1 = -5, & D_2 = -10, & D_3 = -5, \\ x_1 = 1, & x_2 = 2, & x_3 = 1. \end{array} } \]

解析

本题考查克拉默法则的应用，即通过计算行列式求解三元线性方程组。解题核心在于：

  1. 计算系数行列式$D$，判断是否非零；
  2. 构造并计算$D_1, D_2, D_3$（分别用常数项替换系数矩阵的第1、2、3列）；
  3. 代入公式$x_i = \frac{D_i}{D}$求解各变量。

计算系数行列式$D$

$D = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 1 & -1 \end{vmatrix} = 1 \cdot (1 \cdot (-1) - (-3) \cdot 1) - (-2) \cdot (2 \cdot (-1) - (-3) \cdot (-1)) + 1 \cdot (2 \cdot 1 - 1 \cdot (-1)) = -5$

计算$D_1$（替换第1列为常数项）

$D_1 = \begin{vmatrix} -2 & -2 & 1 \\ 1 & 1 & -3 \\ 0 & 1 & -1 \end{vmatrix} = -2 \cdot (1 \cdot (-1) - (-3) \cdot 1) - (-2) \cdot (1 \cdot (-1) - (-3) \cdot 0) + 1 \cdot (1 \cdot 1 - 1 \cdot 0) = -5$

计算$D_2$（替换第2列为常数项）

$D_2 = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -3 \\ -1 & 0 & -1 \end{vmatrix} = 1 \cdot (1 \cdot (-1) - (-3) \cdot 0) - (-2) \cdot (2 \cdot (-1) - (-3) \cdot (-1)) + 1 \cdot (2 \cdot 0 - 1 \cdot (-1)) = -10$

计算$D_3$（替换第3列为常数项）

$D_3 = \begin{vmatrix} 1 & -2 & -2 \\ 2 & 1 & 1 \\ -1 & 1 & 0 \end{vmatrix} = 1 \cdot (1 \cdot 0 - 1 \cdot 1) - (-2) \cdot (2 \cdot 0 - 1 \cdot (-1)) + (-2) \cdot (2 \cdot 1 - 1 \cdot (-1)) = -5$

求解变量

$x_1 = \frac{D_1}{D} = \frac{-5}{-5} = 1, \quad x_2 = \frac{D_2}{D} = \frac{-10}{-5} = 2, \quad x_3 = \frac{D_3}{D} = \frac{-5}{-5} = 1$