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§1 . 1 二阶与三阶行列式

§1 . 1 二阶与三阶行列式. 一、二元线性方程组与二阶行列式. 二、三阶行列式. 补充例题. 首页. 上页. 返回. 下页. 结束. 铃. a 11 x 1  a 12 x 2  b 1. . 用消元法解二元线性方程组. 得. a 21 x 1  a 22 x 2  b 2. 一、二元线性方程组与二阶行列式. 提示 :. [ a 11 x 1 + a 12 x 2 = b 1 ].  a 22 . a 11 a 22 x 1 + a 12 a 22 x 2 = b 1 a 22 .

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§1 . 1 二阶与三阶行列式

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  1. §1.1 二阶与三阶行列式 一、二元线性方程组与二阶行列式 二、三阶行列式 补充例题 首页 上页 返回 下页 结束 铃

  2. a11x1a12x2b1  用消元法解二元线性方程组 得 a21x1a22x2b2 一、二元线性方程组与二阶行列式 提示: [a11x1+a12x2=b1] a22 a11a22x1+a12a22x2=b1a22  [a21x1+a22x2=b2] a12 a12a21x1+a12a22x2=a12b2  (a11a22-a12a21)x1=b1a22-a12b2  下页

  3. a11x1a12x2b1  用消元法解二元线性方程组 得 a21x1a22x2b2 一、二元线性方程组与二阶行列式 提示: [a11x1+a12x2=b1] a21 a11a21x1+a12a21x2=b1a21  [a21x1+a22x2=b2] a11 a11a21x1+a11a22x2=a11b2  (a11a22-a12a21) x2=a11b2-b1a21  下页

  4. a11x1a12x2b1  用消元法解二元线性方程组 得 a21x1a22x2b2 a11 a21 a12 a22 我们用符号 表示代数和a11a22a12a21  b1 b2 a12 a22 a11 a21 b1 b2 x1 ————  x2 ————  a11 a21 a12 a22 a11 a21 a12 a22 一、二元线性方程组与二阶行列式 这样就有 下页

  5. a11 a21 a12 a22 我们用 表示代数和a11a22a12a21 并称它为二阶行 列式 a11 a21 a12 a22 一、二元线性方程组与二阶行列式 行列式中的相关术语 行列式的元素、行、列、主对角线、副对角线 对角线法则 二阶行列式是主对角线上两元素之积减去的副对角线上二元素之积所得的差 =a11a22 a12a21 下页

  6. 例1求解二元线性方程组 a11 a21 a12 a22 =a11a22 a12a21 解 由于 下页

  7. a11x1a12x2a13x3b1 方程组 的解为 a21x1a22x2a23x3b2 a31x1a32x2a33x3b3 D=a11a22a33a12a23a31a13a21a32a11a23a32a12a21a33a13a22a31 其中 D1=b1a22a33a12a23b3a13b2a32b1a23a32a12b2a33a13a22b3 D2=a11b2a33b1a23a31a13a21b3a11a23b3b1a21a33a13b2a31 D3=a11a22b3a12b2a31b1a21a32a11b2a32a12a21b3b1a22a31 a11 a21 a31 a12 a22 a32 a13 a23 a33 为了便于记忆和计算 我们用符号 表示代数和 二、三阶行列式 a11a22a33a12a23a31a13a21a32a11a23a32a12a21a33a13a22a31 下页

  8. a11 a21 a31 a12 a22 a32 a13 a23 a33 我们用符号 表示代数和 二、三阶行列式 a11a22a33a12a23a31a13a21a32a11a23a32a12a21a33a13a22a31 并称它为三阶行列式 行列式中的相关术语 行列式的元素、行、列、主对角线、副对角线 对角线法则 a11a22a33a12a23a31a13a21a32 a11a23a32a12a21a33a13a22a31 下页

  9. -4 1 -2 1 -2 -3 2 2 4 例2计算三阶行列式 D= a11a22a33a12a23a31a13a21a32 a11a23a32a12a21a33a13a22a31 按对角线法则 有 解 D 12(2) 21(3) (4)(2)4 114 2(2)(2) (4)2(3) 46324824 14 下页

  10. 1 2 4 1 3 9 1 x x2 例3求解方程 =0  a11a22a33a12a23a31a13a21a32 a11a23a32a12a21a33a13a22a31 解 方程左端的三阶行列式 D3x24x189x2x212 x25x6 由x25x60解得 x2或x3 结束

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