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Math01: 两种方法求下三角矩阵的逆

时间:2015-09-18 00:48:54      阅读:215      评论:0      收藏:0      [点我收藏+]

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方法一:单位矩阵消元

 1 clear;
 2 n = 500;
 3 A = zeros(n,n);
 4 for j = 1:n
 5     for i = j:n
 6         A(i,j) = (i + j)^2;
 7     end
 8 end
 9 C = A;
10 B = eye(n);
11 
12 for i = 1:(n-1)
13     B(i,:) = B(i,:)/A(i,i);
14     A(i,:) = A(i,:)/A(i,i);
15     for j = (i+1):n
16         B(j,:) = B(j,:) - B(i,:)*A(j,i);
17         A(j,:) = A(j,:) - A(i,:)*A(j,i);
18     end
19 end
20 B(n,:) = B(n,:)/A(n,n);
21 A(n,:) = A(n,:)/A(n,n);

方法二:分块矩阵迭代

\left(\begin{matrix}A&C\\0&B\end{matrix}\right)^{-1}=\left(\begin{matrix}A^{-1}&-A^{-1}CB^{-1}\\0&B^{-1}\end{matrix}\right)


\left(\begin{matrix}A&0\\C&B\end{matrix}\right)^{-1}=\left(\begin{matrix}A^{-1}&0\\-B^{-1}CA^{-1}&B^{-1}\end{matrix}\right)

 1 clear;
 2 n = 500;
 3 A = zeros(n,n);
 4 for j = 1:n
 5     for i = j:n
 6         A(i,j) = (i + j)^2;
 7     end
 8 end
 9 
10 inv_p = 1/A(1,1);
11 for k = 2:n
12     c = A(k,1:(k-1));
13     zero = zeros((k-1),1);
14     inv_q = [inv_p, zero; -(1/A(k,k))*c*inv_p, 1/A(k,k)];
15     inv_p = inv_q;
16 end

第二种方法速度更快

Math01: 两种方法求下三角矩阵的逆

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原文地址:http://www.cnblogs.com/toorist/p/4818046.html

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