八阶行列式怎么展开?
八阶行列式怎么展开?
有道程序题,要求八阶行列式的值,我把它展开如下
bh(0) = ah(0, 0) * ah(1, 1) * ah(2, 2) * ah(3, 3) * ah(4, 4) * ah(5, 5) * ah(6, 6) * ah(7, 7) - ah(7, 0) * ah(6, 1) * ah(5, 2) * ah(4, 3) * ah(3, 4) * ah(2, 5) * ah(1, 6) * ah(0, 7)
bh(1) = ah(1, 0) * ah(2, 1) * ah(3, 2) * ah(4, 3) * ah(5, 4) * ah(6, 5) * ah(7, 6) * ah(0, 7) - ah(6, 0) * ah(5, 1) * ah(4, 2) * ah(3, 3) * ah(2, 4) * ah(1, 5) * ah(0, 6) * ah(7, 7)
bh(2) = ah(2, 0) * ah(3, 1) * ah(4, 2) * ah(5, 3) * ah(6, 4) * ah(7, 5) * ah(0, 6) * ah(1, 7) - ah(5, 0) * ah(4, 1) * ah(3, 2) * ah(2, 3) * ah(1, 4) * ah(0, 5) * ah(7, 6) * ah(6, 7)
bh(3) = ah(3, 0) * ah(4, 1) * ah(5, 2) * ah(6, 3) * ah(7, 4) * ah(0, 5) * ah(1, 6) * ah(2, 7) - ah(4, 0) * ah(3, 1) * ah(2, 2) * ah(1, 3) * ah(0, 4) * ah(7, 5) * ah(6, 6) * ah(5, 7)
bh(4) = ah(4, 0) * ah(5, 1) * ah(6, 2) * ah(7, 3) * ah(0, 4) * ah(1, 5) * ah(2, 6) * ah(3, 7) - ah(3, 0) * ah(2, 1) * ah(1, 2) * ah(0, 3) * ah(7, 4) * ah(6, 5) * ah(5, 6) * ah(4, 7)
bh(5) = ah(5, 0) * ah(6, 1) * ah(7, 2) * ah(0, 3) * ah(1, 4) * ah(2, 5) * ah(3, 6) * ah(4, 7) - ah(2, 0) * ah(1, 1) * ah(0, 2) * ah(7, 3) * ah(6, 4) * ah(5, 5) * ah(4, 6) * ah(3, 7)
bh(6) = ah(6, 0) * ah(7, 1) * ah(0, 2) * ah(1, 3) * ah(2, 4) * ah(3, 5) * ah(4, 6) * ah(5, 7) - ah(1, 0) * ah(0, 1) * ah(7, 2) * ah(6, 3) * ah(5, 4) * ah(4, 5) * ah(3, 6) * ah(2, 7)
bh(7) = ah(7, 0) * ah(0, 1) * ah(1, 2) * ah(2, 3) * ah(3, 4) * ah(4, 5) * ah(5, 6) * ah(6, 7) - ah(0, 0) * ah(7, 1) * ah(6, 2) * ah(5, 3) * ah(4, 4) * ah(3, 5) * ah(2, 6) * ah(1, 7)
result_ok = bh(0) + bh(1) + bh(2) + bh(3) + bh(4) + bh(5) + bh(6) + bh(7)
Text2.Text = result_ok
为什么不对?那怎么展开呀?
如果是展开就展开的公式降阶就是了·······麻烦的话!直接简化成上(下)三角阵来解嘛!