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[综合讨论] 求解方程的问题

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发表于 2010-12-15 10:00 | 显示全部楼层 |阅读模式

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A=[1;2;3];    B=[4;5;6];    C=[7;8;9]
A+t×B=C
求t (t是一个实数;A、B、C是实数矩阵)

怎么求解t啊?用solve不能做啊。
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发表于 2010-12-15 10:50 | 显示全部楼层
这个移项不就解了吗?
发表于 2010-12-15 12:34 | 显示全部楼层
最小二乘即可
 楼主| 发表于 2010-12-15 14:56 | 显示全部楼层
最小二乘法?用这个lsqnonlin命令?
移项之后,分解成若干个小方程求解吗?我想用个有效率点的方法。
发表于 2010-12-15 15:14 | 显示全部楼层
A=[1;2;3]; B=[4;5;6]; C=[7;8;9];
B\(C-A)

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 楼主| 发表于 2010-12-15 18:12 | 显示全部楼层
\
       

Backslash or matrix left division. If A is a square matrix, A\B is roughly the same as inv(A)*B, except it is computed in a different way. If A is an n-by-n matrix and B is a column vector with n components, or a matrix with several such columns, then X = A\B is the solution to the equation AX = B. A warning message is displayed if A is badly scaled or nearly singular. See the reference page for mldivide for more information.

       

If A is an m-by-n matrix with m ~= n and B is a column vector with m components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to the under- or overdetermined system of equations AX = B. The effective rank, k, of A is determined from the QR decomposition with pivoting (see Algorithm for details). A solution X is computed that has at most k nonzero components per column. If k < n, this is usually not the same solution as pinv(A)*B, which is the least squares solution with the smallest norm .

谢谢啦。

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