handbook of linear algebra

", stdin); -Freopen ("Bzoj_1013.out","W", stdout); the #endif * intN; $scanf"%d",n);Panax Notoginseng for(intI=1; i1; i++) - for(intj=1; j) thescanf"%LF",a[i][j]); + A for(intI=1; i){ the for(intj=1; j){ +a[i][n+1]-=a[i][j]*a[i][j]-a[i+1][j]*a[i+1][j]; -a[i][j]=2* (a[i+1][j]-a[i][j]); $ } $ } - - Solve (n); the - for(intI=1; i)Wuyiprintf"%.3LF", a[i][n+1]); theprintf"\ n"); - return 0; Wu}

matrix A and e just can reflect these 3 elementary transformations.The proof that the determinant is equal to the original determinant:This problem is very simple, but we should be able to realize the meaning of this theorem, it makes the row transformation and column transformation has the equivalence, that is, the application of the transformation of the row is applicable to the column.A brief proving process: defining determinant A and writing out its transpose matrix a^t.The A determinant i

Based on the previous chapters, we can easily draw the concept of eigenvectors and eigenvalues.First we know that the product of a and n dimensional vector v of n x n matrices will get an n-dimensional vector, then we now find that, after calculating u=av, the resulting vector u is collinear with V, that is, vector v is multiplied by matrix A to get the vector u "stretched" with respect to vector V, which satisfies the following equation:Av =λv=uSo here we call λ the eigenvalues of matrix A, and

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

is divided into two steps: Judging the legality of two matrices; The K line of the A matrix is extracted and multiplied by the column I of the b Matrix, and the first column I of the target matrix is obtained. The following two kinds of writing are the above ideas, the first one faster, occupy less memory, the second closer to People's thinking (the second folding please expand).Template Matrixoperator* (matrix//operator Overloading * Overloading for point multiplication{ /

The MIT Challenge My friend Scott Young recently finished a astounding Feat:he completed all the courses in MIT's fabled computer science C Urriculum, from Linear Algebra to Theory of computation, in less than one year. More importantly, he does it all in his own, watching the lectures online and evaluating himself using the actual exams. (see Scott's FAQ page for the details about how he ran this challenge

I. Five Representation Methods of Matrix Multiplication 1. General Form 2. Multiply the matrix and column vector 3. Multiply a matrix and a row vector 4. Multiply Matrix Blocks Ii. Matrix Inversion For phalanx, left inverse = Right Inverse Multiply the original matrix by its inverse matrix to obtain the Unit Matrix Several methods to determine whether it is reversible:1. the determinant is 02. the columns of the matrix are linear combinations of

in the European coordinate system are obtained after the X vector (also a coordinate point in the M coordinate) is left multiplied by m in the Custom space. Space coordinates are converted. If the implementation of the European coordinate transformation to the M coordinate system, can be on both sides of the same time left multiplied by a m of the inverse matrix M-1, (M-1) * m * x = (M-1) * B is X = (M-1) * B. After B is used, X can be obtained, and then the coordinate of point X in the M coord

- voidInsert (Char*R) { About intLen=strlen (s), p=RT; $ for(intI=0; i) - if(Ch[p][id (S[i])) -p=Ch[p][id (S[i]); - Else AP=ch[p][id (S[i])]=++CNT; +tag[p]=true; the } - $ voidBuild () { thequeueint>Q; the for(intI=0;i4; i++) the if(Ch[rt][i]) thefail[ch[rt][i]]=Rt,q.push (Ch[rt][i]); - Else inch[rt][i]=RT; the the while(!Q.empty ()) { About intx=Q.front (); Q.pop (); th

[find (1)]!=1){ thememset (Vis,0,sizeof(VIS)); num=0; the for(intI=2; i1; i++){94 if(Vis[find (i)])Continue; theVis[find (i)]=++num; the } thenum=0;98 for(intI=2; i1; i++) Aboutnum=num*Ten+Vis[find (i)]; -a.mat[id[num]][id[mem[t]]]+=1;101 }102 }103 }104 return; the }106 107 intMain () {108 #ifndef Online_judge109Freopen ("count.in","R", stdin); theFreopen ("Count.out","W", stdout);111 #endif thesc

] ofRec; V:Array[0..2002000] ofBoolean; N,m,i,j,k,l,st,ed,ww,top,tar,ans,x:longint;functionmin (aa,bb:longint): Longint;begin ifAa Thenexit (AA); exit (BB);End;procedureAdd (st,ed,ww:longint);beginInc (top); A[TOP].S:=St; A[TOP].E:=Ed; A[TOP].W:=ww; A[top].next:=B[st]; B[ST]:=top;End;functionBfs:boolean;varHead,tail,x,u:longint; Y:rec;beginFillchar (v,sizeof (v), false); Tail:=1; head:=0; d[st]:=1; V[ST]:=true; q[1]:=St; whileHead Do beginInc (head); x:=Q[head]; U:=B[x]; whileU>0 Do begi

1.The calculate the slope:the covariance of X and Y divided by the variance of X　　From NumPy import CoVslope_density = CoV (wine_quality["quality"],wine_quality["density"]) [0,1]/wine_quality["Density"].var () #cov ( X, y) is the function from NumPy, which returns a 2*2 Metric,.var () is Pandas function.2.To get the INTERCEPT:B = Y-ax (x and Y is the mean value of each column)Intercept_density = wine_quality["Quality"].mean ()-wine_quality["Density"].mean () * (Calc_slope (wine_quality[) Density

[Linear algebra] matrix addition 1 # Include 2 Using Namespace STD; 3 4 5 Int Main () 6 { 7 Int Matrixa [ 100 ] [ 100 ]; // Matrixa 8 Int Matrixb [ 100 ] [ 100 ]; // Matrixb 9 Int Plusresult [ 100 ] [ 100 ]; // Matrixa + matrixb = plusresult (this is a maxtrix) 10 Int M, N; 11 Cout " Enter the required and number of rows and columns in the matrix. " 12 Ci

1. Ways to get a column of another column with max/min values:A. Most_bars_country = flags["name"][flags["Bars"].idxmax ()]B. bars_sorted = flags.sort_values ("Bars", ascending=[0])Most_bars_country = bars_sorted["Name"].iloc[0]2. The probability of a certain value in a column:orange_probability = flags[flags["Orange"]==1].shape[0]/flags.shape[0]3. The calculate combination by using factorial:　　　Import Mathdef find_outcome_combinations (N, k): # Calculate The numerator of our formula.Numerator =

Compute Solution of Ax=b (X=XP+XN)Rank rR=M Solutions ExistR=n Solutions UniqueExampleIf we want to solve the equation, what conditions does b1,b2,b3 need to meet? The observation matrix shows that the third line is the first two rows and so the B1+B2=B3Solvability Condition on B:Ax=b is solvable when B was in C (A)If a combination of Rows of a gives zero row, then the same combination of entries of B must give 0Assuming that the above matrix becomes:To find complete solution to ax=b:1.Xp (parti

1. Get the R value and the P value between the dataset:R_fta_pts,p_value = Pearsonr (nba_stats["pts"],nba_stats["FTA"])R_stl_pf,p_value = Pearsonr (nba_stats["STL"],nba_stats["PF"]) # It'll return R value and P value.2. The function of getting convariance form the data set, the Convariance is the value of this measure how much both variables correlated with all other. If one changes to bigger, the other changes to bigger. Which said these, variables is corresponse. Here is the function of gettin

[Article] the full text of this series of articles on the wall is reprinted. Every language has libraries, besides the big. net libraries, F # has two own: the Core, which is shipped with Visual Studio 2010, and the PowerPack, which is an external library developed by MSR Cambridge and Visual Studio Team. notice that the code quality in PowerPack is actually quite high, it is put outside the Core library because it is evolving fast. once stable, they may be put into the Core. Our concern is mat

BZOJ 3996 TJOI2015 linear algebra network stream, bzojtjoi2015 Given N then n Matrix B And one 1 limit n Line vector C , Find one 1 limit n 01 matrix A , Make (A × B − C) × Max (A × B − C) × AT = A × B × AT −c × We can consider N Items, each item does not select the corresponding A Where each location is 1 Or 0 Then the row vector

the determinant of the time will give a detailed proof.The other part is the general algorithm used when solving the inverse matrix of 3 order and above.First we give a lemma:Theorem 1: if n x n matrix A is reversible, then for any r^n vector B, the solution to the matrix equation ax = b is only present.Proof: existence, in this matrix equation is multiplied by the inverse matrix of a, then there is x = a^-1 B. Uniqueness, combined with the properties of the inverse matrix uniqueness mentioned