c程序求行列式的值的问题？ - ITeye问答
#include &stdio.h&
#include &math.h&
#include &stdlib.h&
/*a为行列式，n为行列式的阶*/
int f(int **a,int n)
int i,j,k;
/*子行列式的阶*/
len = n-1;
/*按照定义，初始化一个子行列式数组的空间*/
int **p = (int **)malloc(sizeof(int *)*len);
for(i=0;i&i++)
*(p+i) = (int *)malloc(len);
/*阶为2，按照定义计算*/
if( 2==n )
return a[0][0]*a[1][1]-a[1][0]*a[0][1];
for( k=0; k&n; k++)
for(i=0;i&i++)
for(j=0;j&j++)
*(*(p+i)+j) = a[k][j+1];/*初始化子行列式的值*/
*(*(p+i)+j) = a[k+1][j+1];
s += (int)pow(-1,i+j) * f(p,len);/*递归计算*/
void main()
int a[3][3] = {1,2,4,5,7,9,12,11,8};
printf("%d",f(p,3));
思路很清楚，就是按照定义求值，但调试总是出错，请大家指点下问题补充：一楼的童鞋，这里是二维数组，我想要体现维度问题补充：主要是要调试成功，并且有结果
采纳的答案
自己看吧：
#include &stdio.h&
#include &math.h&
#include &stdlib.h&
/*a为行列式，n为行列式的阶*/
int f(int **a,int n)
int i,j,k;
/*子行列式的阶*/
len = n-1;
/*按照定义，初始化一个子行列式数组的空间*/
int **p = (int **)malloc(sizeof(int *)*len);
for(i=0;i&i++)
p[i] = (int *)malloc(sizeof(int)*len);
/*阶为1，按照定义计算*/
if( 1==n )
return a[0][0];
for( k=0; k&n; k++)
for(i=0;i&i++)
for(j=0;j&j++)
p[i][j] = a[i][j+1];/*初始化子行列式的值*/
p[i][j] = a[i+1][j+1];
s += (int)pow(-1,k) * a[k][0]* f(p,len);/*递归计算*/
for(i=0;i&i++) free(*(p+i));
int main()
int a[3][3] = {1,2,4,5,7,9,12,11,8};
int **p = (int **)malloc(sizeof(int *)*3);
for(i=0;i&3;i++) p[i] = a[i];
printf("%d",f(p,3));
第十五行：
int **p = (int **)malloc(sizeof(int *)*len*len);
这样可以吗？
已解决问题
未解决问题From Wikipedia, the free encyclopedia
In , the Jacobian matrix (, ) is the
of all first-order
of a . When the matrix is a , both the matrix and its
are referred to as the Jacobian in literature.
Suppose f : Rn → Rm is a function which takes as input the vector x ∈ Rn and produces as output the vector f(x) ∈ Rm. Then the Jacobian matrix J of f is an m×n matrix, usually defined and arranged as follows:
{\displaystyle \mathbf {J} ={\begin{bmatrix}{\dfrac {\partial \mathbf {f} }{\partial x_{1}}}&\cdots &{\dfrac {\partial \mathbf {f} }{\partial x_{n}}}\end{bmatrix}}={\begin{bmatrix}{\dfrac {\partial f_{1}}{\partial x_{1}}}&\cdots &{\dfrac {\partial f_{1}}{\partial x_{n}}}\\\vdots &\ddots &\vdots \\{\dfrac {\partial f_{m}}{\partial x_{1}}}&\cdots &{\dfrac {\partial f_{m}}{\partial x_{n}}}\end{bmatrix}}}
or, component-wise:
{\displaystyle \mathbf {J} _{ij}={\frac {\partial f_{i}}{\partial x_{j}}}.}
This matrix, whose entries are functions of x, is also denoted by Df, Jf, and ?(f1,...,fm)/?(x1,...,xn). (Note that some literature defines the Jacobian as the
of the matrix given above.)
The Jacobian matrix is important because if the function f is
at a point x (this is a slightly stronger condition than merely requiring that all partial derivatives exist there), then the Jacobian matrix defines a
Rn → Rm, which is the best (pointwise)
of the function f near the point x. This linear map is thus the generalization of the usual notion of derivative, and is called the derivative or the differential of f at x.
If m = n, the Jacobian matrix is a square matrix, and its , a function of x1, …, xn, is the Jacobian determinant of f. It carries important information about the local behavior of f. In particular, the function f has locally in the neighborhood of a point x an
that is differentiable if and only if the Jacobian determinant is nonzero at x (see ). The Jacobian determinant also appears when changing the variables in
If m = 1, f is a
and the Jacobian matrix is reduced to a
of partial derivatives of f—i.e. the
These concepts are named after the
The Jacobian generalizes the
of a -valued function of multiple variables, which itself generalizes the derivative of a scalar-valued function of a single variable. In other words, the Jacobian for a scalar-valued
is the gradient and that of a scalar-valued function of single variable is simply its derivative. The Jacobian can also be thought of as describing the amount of "stretching", "rotating" or "transforming" that a transformation imposes locally. For example, if (x′, y′) = f(x, y) is used to transform an image, the Jacobian Jf(x, y), describes how the image in the neighborhood of (x, y) is transformed.
If a function is differentiable at a point, its derivative is given in coordinates by the Jacobian, but a function does not need to be differentiable for the Jacobian to be defined, since only the
are required to exist.
If p is a point in Rn and f is
at p, then its derivative is given by Jf(p). In this case, the
described by Jf(p) is the best
of f near the point p, in the sense that
{\displaystyle \mathbf {f} (\mathbf {x} )=\mathbf {f} (\mathbf {p} )+\mathbf {J} _{\mathbf {f} }(\mathbf {p} )(\mathbf {x} -\mathbf {p} )+o(\|\mathbf {x} -\mathbf {p} \|)}
for x close to p and where o is the
(for x → p) and ‖x - p‖ is the
between x and p. (See .)
Compare this to a
for a scalar function of a scalar argument, truncated to first order:
{\displaystyle f(x)=f(p)+f'(p)(x-p)+o(x-p).}
In a sense, both the
and Jacobian are ""—the former the first derivative of a scalar function of several variables, the latter the first derivative of a vector function of several variables.
The Jacobian of the gradient of a scalar function of several variables has a special name: the , which in a sense is the "" of the function in question.
A nonlinear map
{\displaystyle f\colon \mathbb {R} ^{2}\to \mathbb {R} ^{2}}
sends a small square to a distorted parallelogram, shown in red. The Jacobian at a point gives the best linear approximation of the square near that point, shown as a translucent white.
If m = n, then f is a function from Rn to itself and the Jacobian matrix is a . We can then form its , known as the Jacobian determinant. The Jacobian determinant is sometimes referred to as "the Jacobian".
The Jacobian determinant at a given point gives important information about the behavior of f near that point. For instance, the
near a point p ∈ Rn if the Jacobian determinant at p is non-zero. This is the . Furthermore, if the Jacobian determinant at p is , then f preserves orientation near p; if it is , f reverses orientation. The
of the Jacobian determinant at p gives us the factor by which the function f expands or shrinks
near p; this is why it occurs in the general .
The Jacobian determinant is used when making a
when evaluating a
of a function over a region within its domain. To accommodate for the change of coordinates the magnitude of the Jacobian determinant arises as a multiplicative factor within the integral. This is because the n-dimensional dV element is in general a
in the new coordinate system, and the n-volume of a parallelepiped is the determinant of its edge vectors.
The Jacobian can also be used to solve
or approximate solutions near an equilibrium point.
According to the , the
of the Jacobian matrix of an
is the Jacobian matrix of the inverse function. That is, if the Jacobian of the function f : Rn → Rn is continuous and nonsingular at the point p in Rn, then f is invertible when restricted to some neighborhood of p and
{\displaystyle \mathbf {J} _{\mathbf {f} ^{-1}}\circ \mathbf {f} ={\mathbf {J} _{\mathbf {f} }}^{-1}.}
Conversely, if the Jacobian determinant is not zero at a point, then the function is locally invertible near this point, that is, there is a
of this point in which the function is invertible.
The (unproved)
is related to global invertibility in the case of a polynomial function, that is a function defined by n
in n variables. It asserts that, if the Jacobian determinant is a non-zero constant (or, equivalently, that it does not have any complex zero), then the function is invertible and its inverse is a polynomial function.
If f : Rn → Rm is a , a critical point of f is a point where the
of the Jacobian matrix is not maximal. This means that the rank at the critical point is lower than the rank at some neighbour point. In other words, let k be the maximal dimension of the
contained in the image of f; then a point is critical if all
of rank k of f are zero.
In the case where 1 = m = n = k, a point is critical if the Jacobian determinant is zero.
Consider the function f : R2 → R2 given by
{\displaystyle \mathbf {f} (x,y)={\begin{bmatrix}x^{2}y\\5x+\sin y\end{bmatrix}}.}
Then we have
{\displaystyle f_{1}(x,y)=x^{2}y}
{\displaystyle f_{2}(x,y)=5x+\sin y}
and the Jacobian matrix of F is
{\displaystyle \mathbf {J} _{\mathbf {f} }(x,y)={\begin{bmatrix}{\dfrac {\partial f_{1}}{\partial x}}&{\dfrac {\partial f_{1}}{\partial y}}\\[1em]{\dfrac {\partial f_{2}}{\partial x}}&{\dfrac {\partial f_{2}}{\partial y}}\end{bmatrix}}={\begin{bmatrix}2xy&x^{2}\\5&\cos y\end{bmatrix}}}
and the Jacobian determinant is
{\displaystyle \det(\mathbf {J} _{\mathbf {f} }(x,y))=2xy\cos y-5x^{2}.}
The transformation from
(r, φ) to
(x, y), is given by the function F: R+ × [0, 2π) → R2 with components:
{\displaystyle {\begin{aligned}x&=r\cos \\\y&=r\sin \varphi .\end{aligned}}}
{\displaystyle \mathbf {J} (r,\varphi )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \varphi }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \varphi }}\end{bmatrix}}={\begin{bmatrix}\cos \varphi &-r\sin \varphi \\\sin \varphi &r\cos \varphi \end{bmatrix}}}
The Jacobian determinant is equal to r. This can be used to transform integrals between the two coordinate systems:
{\displaystyle \iint _{\mathbf {F} (A)}f(x,y)\,dx\,dy=\iint _{A}f(r\cos \varphi ,r\sin \varphi )\,r\,dr\,d\varphi .}
The transformation from
(r, θ, φ) to
(x, y, z), is given by the function F: R+ × [0, π] × [0, 2 π) → R3 with components:
{\displaystyle {\begin{aligned}x&=r\sin \theta \cos \\\y&=r\sin \theta \sin \\\z&=r\cos \theta .\end{aligned}}}
The Jacobian matrix for this coordinate change is
{\displaystyle \mathbf {J} _{\mathbf {F} }(r,\theta ,\varphi )={\begin{bmatrix}{\dfrac {\partial x}{\partial r}}&{\dfrac {\partial x}{\partial \theta }}&{\dfrac {\partial x}{\partial \varphi }}\\[1em]{\dfrac {\partial y}{\partial r}}&{\dfrac {\partial y}{\partial \theta }}&{\dfrac {\partial y}{\partial \varphi }}\\[1em]{\dfrac {\partial z}{\partial r}}&{\dfrac {\partial z}{\partial \theta }}&{\dfrac {\partial z}{\partial \varphi }}\end{bmatrix}}={\begin{bmatrix}\sin \theta \cos \varphi &r\cos \theta \cos \varphi &-r\sin \theta \sin \varphi \\\sin \theta \sin \varphi &r\cos \theta \sin \varphi &r\sin \theta \cos \varphi \\\cos \theta &-r\sin \theta &0\end{bmatrix}}.}
is r2 sin θ. As an example, since dV = dx dy dz this determinant implies that the
dV = r2 sin θ dr dθ dφ. Unlike for a change of , this determinant is not a constant, and varies with coordinates (r and θ).
The Jacobian matrix of the function F : R3 → R4 with components
{\displaystyle {\begin{aligned}y_{1}&=x_{1}\\y_{2}&=5x_{3}\\y_{3}&=4x_{2}^{2}-2x_{3}\\y_{4}&=x_{3}\sin x_{1}\end{aligned}}}
{\displaystyle \mathbf {J} _{\mathbf {F} }(x_{1},x_{2},x_{3})={\begin{bmatrix}{\dfrac {\partial y_{1}}{\partial x_{1}}}&{\dfrac {\partial y_{1}}{\partial x_{2}}}&{\dfrac {\partial y_{1}}{\partial x_{3}}}\\[1em]{\dfrac {\partial y_{2}}{\partial x_{1}}}&{\dfrac {\partial y_{2}}{\partial x_{2}}}&{\dfrac {\partial y_{2}}{\partial x_{3}}}\\[1em]{\dfrac {\partial y_{3}}{\partial x_{1}}}&{\dfrac {\partial y_{3}}{\partial x_{2}}}&{\dfrac {\partial y_{3}}{\partial x_{3}}}\\[1em]{\dfrac {\partial y_{4}}{\partial x_{1}}}&{\dfrac {\partial y_{4}}{\partial x_{2}}}&{\dfrac {\partial y_{4}}{\partial x_{3}}}\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&0&5\\0&8x_{2}&-2\\x_{3}\cos x_{1}&0&\sin x_{1}\end{bmatrix}}.}
This example shows that the Jacobian need not be a square matrix.
The Jacobian determinant of the function F : R3 → R3 with components
{\displaystyle {\begin{aligned}y_{1}&=5x_{2}\\y_{2}&=4x_{1}^{2}-2\sin(x_{2}x_{3})\\y_{3}&=x_{2}x_{3}\end{aligned}}}
{\displaystyle {\begin{vmatrix}0&5&0\\8x_{1}&-2x_{3}\cos(x_{2}x_{3})&-2x_{2}\cos(x_{2}x_{3})\\0&x_{3}&x_{2}\end{vmatrix}}=-8x_{1}{\begin{vmatrix}5&0\\x_{3}&x_{2}\end{vmatrix}}=-40x_{1}x_{2}.}
From this we see that F reverses orientation near those points where x1 and x2 the function is
invertible everywhere except near points where x1 = 0 or x2 = 0. Intuitively, if one starts with a tiny object around the point (1, 2, 3) and apply F to that object, one will get a resulting object with approximately 40 × 1 × 2 = 80 times the volume of the original one, with orientation reversed.
The Jacobian serves as a linearized
Consider a
of the form
{\displaystyle {\dot {\mathbf {x} }}=F(\mathbf {x} )}
{\displaystyle {\dot {\mathbf {x} }}}
is the (component-wise) derivative of
{\displaystyle \mathbf {x} }
with respect to the
{\displaystyle t}
(time), and
{\displaystyle F\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n}}
is differentiable. If
{\displaystyle F(\mathbf {x} _{0})=0}
{\displaystyle \mathbf {x} _{0}}
(also called a ). By the , the behavior of the system near a stationary point is related to the
{\displaystyle \mathbf {J} _{F}\left(\mathbf {x} _{0}\right)}
, the Jacobian of
{\displaystyle F}
at the stationary point. Specifically, if the eigenvalues all have real parts that are negative, then the system is stable near the stationary point, if any eigenvalue has a real part that is positive, then the point is unstable. If the largest real part of the eigenvalues is zero, the Jacobian matrix does not allow for an evaluation of the stability.
A square system of coupled nonlinear equations can be solved iteratively by . This method uses the Jacobian matrix of the system of equations.
Arrowsmith, D. K.; Place, C. M. (1992). . Dynamical Systems: Differential Equations, Maps, and Chaotic Behaviour. London: Chapman & Hall. pp. 77–81.  .
Hirsch, M Smale, Stephen (1974). Differential equations, dynamical systems and linear algebra.
(1996). Economic Dynamics (Third ed.). Berlin: Springer. pp. 305–330.  .
, ed. (2001) [1994], , , Springer Science+Business Media B.V. / Kluwer Academic Publishers,  
A more technical explanation of Jacobians【标准答案】已知四阶行列式D中第三行元素为（-1，2，0，1），它们的余子式依次分别为5，3，-7，4，则D的值等于()
电大职业技能实训平台答案 题目：已知四阶行列式D中第三行元素为（-1，2，0，1），它们的余子式依次分别为5，3，-7，4，则D的值等于()[A]5[B]-10[C]-15 【 收集整理了电大职业技能实训的大部分习题请使用功能查找相应习题】
》》》标准答案：C
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《经济数学基础12》
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设A 为4 阶矩阵,|A|=3,则其伴随矩阵A*的行列式|A*|=?
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|AA*|=|A||A*|=||A|E||; //现在都是数了,不是矩阵了,所以可以用代 数方法做了|A|=3是数,E是单位矩阵（也是上三角行列式）,那么||A|E|=3*3*3*3=81;//上三角行列式的计算,书上有写所以：|A*||A|=81,又因为|A|=3,所以除以掉下,就是27；
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|A*|=|A|^(n-1)=3^3=27 楼上的回答很搞3^3=9？！
A 为4 阶矩阵 |A|=3，所以 AA*=|A|E -> |A*|=|A|^(n-1)=3^3=27 (n=4)汗一下
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