matrices precalculus

Matrix A-type N-orderreversible matricesThe equivalent conditions:1. A'sdeterminant not equal to 02. A'srank equals n, that is, A isFull RankMatrix3. Row of a (column)Vector Group Linear Independent4. Homogeneous equation setax=0 only 0 Solutions5. For any b belongs to RN (n is superscript, which represents vector space), ax=b always has a unique solution6. A is equivalent to the unit matrix7. A can be expressed as a product of several elementary matrices8. The column vectors of a can be used as

equations:Figure 2Based on the above equation, we can extend it. We make a full differential on each of the items in X , and the right side of the equation biases all q . The specific expansion is as follows, and we can convert the left-hand expansion of Figure 3 into the matrix form on the right:Figure 3Some conclusions can be found, such as that the line of a matrix is equal to the number of items in X , that is, the number of degrees of freedom of the end controller, and the matrix column eq

] * vplaneequation[3]; -}View Code1 //Given three points on a plane in counter clockwise order, calculate the unit normal2 voidGltgetnormalvector (ConstGLTVector3 vP1,ConstGLTVector3 vP2,ConstGLTVector3 vP3, GLTVector3 vnormal)3 {4 GLTVector3 vV1, vV2;5 6 gltsubtractvectors (vP2, vP1, vV1);7 gltsubtractvectors (vP3, vP1, vV2);8 9 gltvectorcrossproduct (vV1, vV2, vnormal);Ten Gltnormalizevector (vnormal); One}View CodeThe second type:The Createshadowmatrix function is the focus1 The s

assume that the line expression is as follows:Then calculate the error function:The error function e is obtained by the deviation of the coefficient, a, a, and the 0:The value of the coefficients, a, a, a, and a linear expression is obtained by the above formula:The curves obtained by the least squares are as follows:Linear algebraic approach to least squares: Similarly, the expression of a fitted line is set to:The purpose of fitting is to make the data points satisfy the above function expres

1 >> A=zeros (5,3); 2 >> b=[2; 3]; 3 >> A (:,2) =B; 4 Assignment dimension mismatch with subscript.How to solve the above problem?1>>a2 3A =4 5 0 0 06 0 0 07 0 0 08 0 0 09 0 0 0Ten One>>b A -b = - the 2 - 3You want to assign B to the second column of a, but there is an error in the direct assignment;How to assign the values of different dimensions to matrices in matlab

sub-array and is:max= "return 0;}Three.Development ResultsIv. Summary of the pairThe idea of discovering this program is very simple to think of, but in the process of writing Cheng the most troublesome is to calculate the number of cycles from the first several cycles, but another partner, can easily pass the problem. And in the final stage of development, the partners gave me a lot of suggestions such as the output of the maximum sub-matrix, although it is more troublesome for me to make chan

;nu; b->nu=a->mu; B->tu=a->Tu; /*the number of rows, columns, and elements of a sparse matrix*/ if(B->tu! =0)/*with a non-0 element, the conversion*/{Q=0; for(col=1; Col) {/*converting by a sequence of columns*/ 　　 for(p=1; P/*Scan the entire ternary group table*/ 　　 if(a->data[p].j==Col) {B->data[q].i= a->DATA[P].J; B->data[q].j= a->data[p].i; B->data[q].v= a->data[p].v; Q++; }/*if*/}}/*if (b->tu>0)*/ returnB/*returns a pointer to the transpose matrix*/} /*TransM1*/

will be OK.In the usual process we do not need to consider the problem of the main sequence of the column, because we calculate the matrix (such as through LookAt, or GLM related functions obtained by the matrix) and shader in the required format is consistent, so we can use the incoming, There is only one case where we need to consider the column main or row main order relationship of the Matrix, that is, the GLSL matrix into a custom memory block is, when the matrix is passed to the uniform b

Transpose of the MatrixSwap the rows and columns of the original matrix, that is, the data on [I][j] and [j][i] positions.650) this.width=650; "title=" Picture 1.png "src=" http://s5.51cto.com/wyfs02/M01/81/FF/ Wkiom1dg3jdyhtjnaabwv61ugxk712.png "alt=" Wkiom1dg3jdyhtjnaabwv61ugxk712.png "/>Program code:#include Operation Result:1 0 0 2 0 00 0 0 0 0 03 0 0 4 0 00 0 0 0 0 05 0 0 6 0 0Quick Transpose650) this.width=650; "title=" Picture 2.png "src=" http://s5.51cto.com/wyfs02/M02/81/FE/ Wkiol1dg3bm

In Python, there are two common methods of implementing multidimensional arrays or matrices:Built-in list method and numpy Scientific calculation package method.The following is an example of creating a 10*10 matrix or multidimensional array, and initializing it to 0, the program is as follows:#Method 1:listARR1 = [[0]*10 forIinchRange (10)]arr1[0][0]= 1Print "Method 1:\n", ARR1ARR2= [[0 forIinchRange (10)] forIinchRange (10)]arr2[0][0]= 1Printarr2#Wrong simplified versionA = [[0]*10]*10A[0][0]=

Defines a Matrix class to implement addition and multiplication of matrices. # Include Running result:

Problem Description:Given a 2D board and a word, find if the word exists in the grid.The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells is those horizontally or V Ertically neighboring. The same letter cell is used more than once.For example,Given board =[ [' A ', ' B ', ' C ', ' e '], [' s ', ' F ', ' C ', ' s '], [' A ', ' D ', ' e ', ' e ']]Word = "ABCCED" , returns true ,Word = "SEE" , returns true ,Word = "ABCB" , returns false .Algorithm analy

"); $ for(i=0; i) $ { - for(j=0; j) - { theprintf"%3d", B[i][j]); - }Wuyiprintf"\ n"); the } - //array Multiplication Wuprintf"\ n matrix a multiplied by matrix B results to: \ n"); - for(i=0; i) About { $ for(j=0; j) - { -C[I][J] =0; - for(k=0; k) AC[I][J] = c[i][j]+a[i][k]*B[k][j]; + } the } -printf"\ n"); $ //values in array C are displayed in a branch the for(i=0; i) the { the f

=New float*[n-1]; for(intI=0; i1; i++) {Temp[i]=New float[N-1]; } for(intI=0; i) { for(intj=0; j) { for(intk=0; k1; k++) { for(intm=0; m1; m++) { intflag1=0; intFlag2=0; if(k0; Elseflag1=1; if(M0; ElseFlag2=1; TEMP[K][M]=array[k+flag1][m+Flag2]; } } intflag3=-1; if((i+j)%2==0) flag3=1; Resultmarin[j][i]=(float) Flag3*value (temp,n-1)/resultsum; } } returnResultmarin;}voidOu

Principle Analysis:This section describes the transpose of the Matrix. The transpose of the matrix will change the row and column elements of the matrix, that is, the first column of the second row (with C21, after the same) and the first row of the second column (C12) element swap position, the original C31 and C13 Exchange. namely CIJ and cji Exchange.(Fill in the illustration here)C + + language:The first thing we think about is to remove the column J of row I from Row J, which is simple enou

adjacency vertex for access */}/* Depth traversal operation of adjacency matrix */ void Dfstraverse (mgraph G) { int i; for (i = 0; i Visited[i] = FALSE; /* Initial all vertex states are not visited status */ for (i = 0; i if (!visited[i])/* Dfs is invoked on an unreachable vertex, if connected graph, only once */ D FS (G, i); } /* Breadth traversal algorithm for adjacency Matrix */void Bfstraverse (Mgraph G){int I, J;Queue Q;for (i = 0; i Visited[i] = FALSE;Initqueue (Q);/* Initializes an

matrix of the largest sub-matrix X1, and then move Make up a new matrix and then find the maximal sub-matrix X2 ... Move one column to the right, forming a new matrix and finding the maximum sub-matrix xn. Compares the size of the X1,X2......,XN and outputs the maximum value to the position. Iv. Source Code#include 　　Five, run six, knot to summarize this topic has been done many times, can say has been done rotten. But each time the subject requirements are different, the teacher always on the

numerical solution is obtained that is close enough to the eigenvalues of the Matrix.Again, the computer can only solve the numerical solution and cannot find the analytic solution. def eig(self, A) : ifisNone : return forin range(020) : (Q, R) = self.qr_decomposition(A) A = self.multiply(R, Q) return AOn the diagonal is the eigenvalues of the Matrix.After finding the eigenvalues, according to a x span class= "Mo" id= "mathjax-s

//Transpose of the matrix//swaps a two-dimensional array row and column elements into another two-dimensional array. //For example//a=1 2 3//4 5 6//b=1 4//2 5//3 6//The procedure is as follows:#include using namespacestd;intMain () {inta[2][3]={{1,2,3},{4,5,6}}; intb[3][2],i,j; cout"Array A:"Endl; for(i=0;i2; i++) { for(j=0;j3; j + +) {cout" "; B[j][i]=A[i][j]; } coutEndl; } cout"Array B:"Endl; for(i=0; i2; i++) { for(j=0; j1; j + +) cout" "; coutEndl; } System ("PAUSE");

+ ")" + ",");}System.out.println ();}}Main function testpublic static void Main (string[] args) {TODO auto-generated Method Stubint a[][] = new int[][] {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};int b[][] = new int[][] {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};int c[][] = new INT[3][3];System.out.println ("Matrix A:");PrintArray (a);System.out.println ("A matrix after the interchange of rows and columns:");Swaprowcloumn (a);PrintArray (a);System.out.println ("Matrix B:");PrintArray (b);System.out.println ("A ma