对于一个随机数组rand（4）进行lu(rand(4))分解，为什么得到的不是一个上三角和一个下三角矩阵。
\u5728MATLAB\u4e2d\uff0c\u5df2\u77e5a=rand(3,4),a(1,2)=rand\u7684\u610f\u601da=rand(3,4) % \u751f\u6210\u4e00\u4e2a3\u884c4\u5217\u7684\u968f\u673a\u6570\u7ec4\uff0c\u5176\u4e2d\u6bcf\u4e2a\u5143\u7d20\u57280-1\u4e4b\u95f4a(1,2)=rand % \u5c06a\u6570\u7ec4\u7684\u7b2c\u4e00\u884c\u7b2c\u4e8c\u5217\u4f4d\u7f6e\u7684\u5143\u7d20\u8bbe\u4e3a\u4e00\u4e2a0-1\u4e4b\u95f4\u7684\u968f\u673a\u6570
array_rand\u4f20\u4e0d\u540c\u7684\u53c2\u6570\u5f97\u5230\u7684\u8fd4\u56de\u503c\u7c7b\u578b\u662f\u4e0d\u540c\u7684\u54e6\u3002
<?php// \u5f53\u7b2c\u4e8c\u4e2a\u53c2\u6570\u5927\u4e8e1\u65f6,array_rand\u53ea\u4f1a\u591a\u4e2a\u968f\u673a\u5143\u7d20\u7684\u7d22\u5f15\u7ec4\u6210\u7684\u6570\u7ec4,\u9700\u8981\u904d\u5386\u8be5\u8fd4\u56de\u503c\u518d\u53d6\u5143\u7d20$input = array( "1" , "2" , "3" , "4" , "5" );$rand_keys = array_rand ( $input , 3 );echo $input [ $rand_keys [ 0 ]] . "\n" ;echo $input [ $rand_keys [ 1 ]] . "\n" ;echo $input [ $rand_keys [ 2 ]] . "\n" ;// \u5f53\u7b2c\u4e8c\u4e2a\u53c2\u6570\u4e3a1\u6216\u8005\u4e0d\u4f20\u7b2c\u4e8c\u4e2a\u53c2\u6570\u65f6,array_rand\u53ea\u4f1a\u8fd4\u56de\u4e00\u4e2a\u6570\u5b57(\u968f\u673a\u5143\u7d20\u7684\u7d22\u5f15)$rand_key = array_rand ( $input , 1 );//\u8fd9\u91cc\u76f4\u63a5\u83b7\u53d6\u5373\u53efecho $input[$rand_key];
[L2,U2,P]=lu(x)
[L3,U3,P,Q] = lu(X)
MATLAB中[L1,U1]=lu(x)的结果：
L是下三角的置换矩阵即L1=p*L2，U是上三角阵。Matlab中LU分解采用高斯消元法，结果是不唯一的，只要[L1,U1]=lu(x)满足L1*U1=x, [L2,U2,P]=lu(x)满足L2*U2=p*x，[L3,U3,P,Q] = lu(X)满足 L3*U3= P*X*Q就行了
绛旓細L鏄笅涓夎鐨勭疆鎹㈢煩闃靛嵆L1=p*L2锛孶鏄笂涓夎闃点侻atlab涓璍U鍒嗚В閲囩敤楂樻柉娑堝厓娉曪紝缁撴灉鏄笉鍞竴鐨勶紝鍙[L1,U1]=lu(x)婊¤冻L1*U1=x, [L2,U2,P]=lu(x)婊¤冻L2*U2=p*x锛孾L3,U3,P,Q] = lu(X)婊¤冻 L3*U3= P*X*Q灏辫浜 鍙傝冭祫鏂欙細http://www.ilovematlab.cn/thread-71643-1-1.htm...
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