四阶行列式展开问题 关于行列式展开问题
\u56db\u9636\u884c\u5217\u5f0f\u5c55\u5f00\u95ee\u98984\u9636\u4ee5\u4e0a(\u542b4\u9636)\u6ca1\u6709\u5bf9\u89d2\u7ebf\u6cd5\u5219
\u5b83\u7684\u5b8c\u5168\u5c55\u5f00\u5f0f\u542b 4! = 24 \u9879
\u592a\u590d\u6742\u5e76\u4e0d\u6613\u8bb0\u5fc6
\u6240\u4ee5\u4e0d\u5982\u6ca1\u6709.
\u8865\u5145\u95ee\u9898\u4e0d\u6613\u770b\u5230, \u6700\u597d\u7528\u8ffd\u95ee, \u767e\u5ea6hi\u4f1a\u63d0\u9192
\u542ba11a23 \u7684\u4e00\u822c\u9879\u4e3a (-1)^t(13ij) a11a23a3ia4j
i,j \u5206\u522b\u53d62,4
\u5f53i=2,j=4\u65f6, t(1324) = 1,
\u6240\u4ee5 \u6c42\u542ba11a23\u7684\u9879\u4e3a -a11a23a32a44, a11a23a34a42
\u8fd9\u4e2a\u884c\u5217\u5f0f\u5e94\u8be5\u662f-2\uff0c\u4f60\u7684\u662f\u6309\u5217\u5c55\u5f00\u3002\u9664\u975e\u4f60\u53d8\u6362\u51fa\u9519\uff0c\u6309\u5217\u5c55\u5f00\u4e0d\u4f1a\u9519
丢项了只需考虑非零项.
第1行若取a1, 有2项非零项: a1a2a3a4, a1b2b3a4
第1行若取b1, 有2项非零项: b1b2b3b4, b1a2a3b4
然后考虑正负号(逆序数)就行了
满意请采纳^_^
共有4*3*2*1=24项,你只有8项.
你理解错了行列式的计算方法, 不是只取斜线上的元素相乘再相加.
当然,三阶行列式可以这样做.
我觉得你应该仔细看一下定义.
希望可以帮到你.
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