有人可以检查我的代码吗？ [英] Can someone check my code please?
问题描述
大家好，我是这些论坛的新手，实际上是C ++的新手。我正在研究一个项目，如果我能够比那些对C ++有更好理解的人检查我的代码，我将非常感激。
我想要的是什么要做的代码就是要求一个数字，在2到9之间，然后根据给定的数字显示一个矩阵并显示其行列式。我不知道如何编程找到决定因素，所以我通过简单地在别处找到决定因素来解决这个问题。我打算上一个9x9矩阵，虽然我仍然坚持找到6x6矩阵向上的行列式。
这是我的代码：
Hello everyone, i am new to these forums and in fact new to C++. I am working on a project, and i would be most grateful if i could have my code checked by people who have a better understanding of C++ than i curently do.
What i intend this code to do is ask for a number off you, between 2 and 9, it then displays a matrix according to the number given and displays its determinant. I do not know how to program finding the determinant though, so i have got round this by simply finding the determinant elsewhere. I intend to go up a 9x9 matrix, although i am stuck on finding the determinant for 6x6 matrices upwards.
Here is my code:
推荐答案
对于任何非奇异方阵A，它可以分解为L * U == P * A
其中L是一个下三角矩阵，其对角线上的所有一个，U是一个上部
三角矩阵，P是一个行置换矩阵。
因为det（L * U）== det（P * A）和det（P）是+ - 1;如果已经执行了一个奇数的行数交换，则为负数，否则为正数。
三角矩阵的行列式是对角线上元素的乘积。 det（L）== 1
（对角线上只有一个，所以det（A）== det（U）* s，其中s == + -1。
分解被称为低层分解具有部分旋转''。
如果分解失败（部分对角线上全部为零），则b / b矩阵A的行列式为零。
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亲切的问候，
Jos
For any non-singular square matrix A it can be decomposed as L*U == P * A
where L is a lower triangular matrix with all ones on its diagonal, U is an upper
triangular matrix and P is a row permutation matrix.
Since det(L*U) == det(P * A) and det(P) is +- 1; it is negative if an odd number
of row swaps have been performed, positive otherwise. The determinant of a
triangular matrix is the product of its elements on the diagonal. det(L) == 1
(there are only ones on its diagonal, so det(A) == det(U)*s where s == +-1.
The decomposition is called a ''lower-upper decomposition with partial pivoting''.
If the decomposition fails (all zeros on the partial diagonal), the determinant of
matrix A is zero.
kind regards,
Jos
对于任何非奇异方阵A它可以被分解为L * U == P * A
其中L是一个下三角矩阵，其对角线上的所有一个，U是一个上层
三角矩阵和P是一个行置换矩阵。
For any non-singular square matrix A it can be decomposed as L*U == P * A
where L is a lower triangular matrix with all ones on its diagonal, U is an upper
triangular matrix and P is a row permutation matrix.
这仍然会让你首先找到LU分解的问题。我不知道这将是多么容易。我只会使用辅助因子扩展（也就是扩展未成年人）将大矩阵的行列式减少为一系列决定因素，连续较小的矩阵（未成年人）乘以其辅助因子。最终，你最终会得到2x2矩阵行列式的总和。
This still leaves you with the problem of finding the LU decomposition in the first place. I don''t know how easy that would be. I would just use a cofactor expansion (a.k.a. expansion by minors) to reduce the determinant of the large matrix into a sum of determinants successively smaller matrices (minors) multiplied by their cofactor. Eventually, you end up with a sum of 2x2 matrix determinants.
这仍然会让你在找到LU分解的问题第一名。我不知道会有多容易。
This still leaves you with the problem of finding the LU decomposition in the first place. I don''t know how easy that would be.
这非常简单：只需高斯消除下三角矩阵即可。
足够。交换两行只会改变行列式的符号。
亲切的问候，
Jos
It''s extremely easy: just Gaussian elimination of the lower triangular matrix is
enough. Swapping two rows only changes the sign of the determinant.
kind regards,
Jos
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