First, the method of Lagrange multipliers
1, the popular interpretation
To a function: $ Z = f (x, y) $ How to find out its extreme point of it? With previous knowledge, simply, directly seeking its partial derivatives not to OK it?
If you say now that this function does add a constraint? Also say that if there is such a constraint $ 2xy 2yz + + = S $ 2zx , how to find that the function $ Z (x, y, z ) = xyz $ the maximum value of it?
Under such constraints, in the end what is the point we want?
If you say that we now have such a mountain, the height of this mountain is $ f (x, y) $ , which is a curve $ g (x, y) = C $. Curve embedded in the mountains, how do we find the lowest point of the curve of it?
To find the lowest point on the curve, first from the lowest contour number (0 piece) starts up. Third count, and finally contour intersection point of the curve (shown above). Because lower than the contours of this place is not within the constraints, so this is definitely the lowest point of the curve of this constraint.
Further constraint curve and contour intersection where possible, must be tangent . Because if it is intersected, as shown below, then there will be a portion of the curve in the region B, region B but lower than the contour line, this is not possible.
Tangent to two curves, which means that at this point in parallel to the normal line , that is a normal vector're just an arbitrary constant multiplier (taken as $ - \ lambda $): $ \ bigtriangledown f (x, y) = - \ lambda \ bigtriangledown g (x, y) $, where $ \ bigtriangledown $ denotes the partial derivative.
We can put on the right side to the left side of the formula, and the constant is then moved into the differential operator obtained: $ \ bigtriangledown F (X, Y) + \ the lambda \ bigtriangledown G (X, Y) = 0 $ .
Re-explain this formula, this is $ f (x, y) + \ lambda g (x, y) $ no requirement under the restriction of the extreme point. Briefly , the extremum is lower with constraints converted to extremum under unconstrained conditions.
2, using the method
Then we look at the Lagrange multiplier method using a specific sub law. Analytic function: $ z = f (x, y) $ condition $ \ varphi (x, y) = 0 $ under conditions of extreme value.
Since extreme value, that is, make the partial derivatives equal to zero.
Constructor $ F. (X, Y) = F (X, Y) + \ the lambda \ varphi (X, Y) $ , where $ \ $ the lambda is the Lagrangian multiplier. So, we can get the following expression
Solving this out (X, Y) coordinates of the extreme point is set by the equation above.
Lagrange multipliers generally used argument more than two conditions below.
Analytic function: $ u = f (x, y, z, t) $ condition $ 0 $ under \ varphi (x, y, z, t) = 0, \ psi (x, y, z, t) = a extremum.
Similarly Constructor: $ F (x, y, z, t) = f (x, y, z, t) + \ lambda _ {1} \ varphi (x, y, z, t) + \ lambda _ { 2} \ psi (x, y, z, t) $. Where $ \ lambda _ {1}, \ lambda _ {2} $ are Lagrangian multipliers, to solve the same extreme point coordinates 0 and constraint by the deflector.
3, example
Requests the function $ u = x ^ {3} y ^ {2} z $ the maximum constraint x + y + z = 12 in.
Similarly Constructor: $ F (x, y, z) = x ^ {3} y ^ {2} z + \ lambda (x + y + z-12) $. Then the partial derivatives, respectively, to obtain the following expression.
$\left\{\begin{matrix}
F_{x}'=3x^{2}y^{2}z+\lambda =0\\
F_{y}'=2x^{3}yz+\lambda =0\\
F_{z}'=x^{3}y^{2}+\lambda =0\\
x+y+z=12
\end{matrix}\right.$
求解上面的方程组可以得到唯一驻点（6，4，2），这样的话最大值$u_{max}=6^{3}\cdot 4^{2}\cdot 2=6912$。
二、行列式
1、二阶行列式
首先来看看二元线性方程组的求解：$\left\{\begin{matrix}a_{11}x_{1}+a_{12}x_{2}=b_{1}\\ a_{21}x_{1}+a_{22}x_{2}=b_{2}\end{matrix}\right.$
对上面这个方程组求解可得：$\begin{matrix}(a_{11}a_{22}-a_{12}a_{21})x_{1}=b_{1}a_{22}-a_{12}b_{2}\\ (a_{11}a_{22}-a_{12}a_{21})x_{2}=a_{11}b_{2}-b_{1}a_{21}\end{matrix}$。
当$a_{11}a_{22}-a_{12}a_{21}\neq 0$时方程组有唯一解：$x_{1}=\frac{b_{1}a_{22}-a_{12}b_{2}}{a_{11}a_{22}-a_{12}a_{21}},x_{2}=\frac{a_{11}b_{2}-b_{1}a_{21}}{a_{11}a_{22}-a_{12}a_{21}}$。
根据上面的解看起来好像有些规律呀
表达式$a_{11}a_{22}-a_{12}a_{21}$即为二阶行列式。
$D=\begin{vmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{vmatrix}=a_{11}a_{22}-a_{12}a_{21}$。其中aij(i=1,2;j=1,2)称为元素。i代表行标，j代表列标。
2、三阶行列式
二阶看起来挺容易就算出来了，三阶的呢？
3、例题
计算$D=\begin{vmatrix}1 & 2 & -4\\ -2 & 2 & 1\\ -3 & 4 & -2\end{vmatrix}$的行列式。
三、矩阵
1、何为矩阵
某航空公司在A，B，C，D四城市之间开辟了若干航线，如图所示表示了四城市间的航班图，如果从A到B有航班，则用带箭头的线连接 A 与B。
如果说我们用表格的形式来表示这种关系并且用1和0来表示城市之间是否联通。
何为矩阵：输入的数据就是矩阵，对数据做任何的操作都是矩阵的操作了。
矩阵的组成：矩阵是由行和列来组成的：
矩阵的特殊形式：行向量与列向量。$\begin{pmatrix}a_{1} & a_{2} & \cdots & a_{n}\end{pmatrix}$，$\begin{pmatrix}a_{1}\\ a_{2}\\ \vdots \\ a_{n}\end{pmatrix}$
2、行列式与矩阵的区别
方阵：行和列的数量一样就是方阵了，一般叫做n阶方阵。
下面介绍几种特殊的矩阵
同型矩阵和矩阵相等是一个事吗？
两个矩阵行列数相同的时候称为同型矩阵，例如$\begin{pmatrix}1 & 2\\ 2 & 3\end{pmatrix}$与$\begin{pmatrix}2 & 4\\ 1 & 2\end{pmatrix}$
在同型的前提下，并且各个元素相等，这就是矩阵相等了：
3、矩阵的基本运算
假如说有两个$M\times N$的矩阵$A=(a_{ij}),B=(b_{ij})$：
矩阵乘法的运算规律：
注意：矩阵的乘法是没有交换律的