how are matrices used in mapping?
814015
Let us say you have a certain coordinate system C with the basis vectors:\nex= (1,0) and ey=(0,1)\n. ey↑\n. . .C →\n. . . . ex\nand you want to express points according to another coordinate system C' with the basis vectors\ne'x = (-1,0), and e'y =( 0,-1), the coordinates of the new basis vectors is according to C\n. ex'0(-1,0)\n←C'\n. . .↓ey'=(0,-1)\n\nThe relationship between the (x,y) in C and (x',y') in C' can be expressed by a Matrix mapping.\n. . . . . . .(-1 o) ..(x)\n(x',y') =  (0, -1) *(y) \n\nremember it is matrix multiplication. This was of cause a very simple example\n\nif we take a point (5,-1) in C its coordinates will be:\n\n. . . . . . . . . . . . . . (5)\n. . . (x',y') = (-1,0)*(-1) = (-5,1)\n. . . . . . . . .(0 ,-1)\n\naccording to the new basis ex',ey' in C'\nI hope its might have helped you \nthe matrix I have use is of cause \n\nE= (-1, 0)\n. . .(0 ,-1)\n\nI this simple situation I could also have used this very simple matrix = (-1,-1)\nand then the mapping would have looked like this:\n\n. . . . . .(-1)\n(x',y' )=(-1)*(x.y)\n\nAgain remember its matrix multiplication.
1069671
yep...scalars multiply vectors component by component just like you have shown.  you can also show it by plotting out the original vectors, scaling them appropriately, adding them, and finding the resukltant components.
415924
A matrix is an array of numbers that is usually studied in relation to solving systems of equations and more general problems in linear algebra. Familiarity with linear algebra is the gateway to higher mathematics; the matrix part of this is little more than advanced arithmetic.
551932
Assuming your matrix is of order nxn and you want submatrix of order mxm.\n\nThen there is (n-m+1)^2 submatrices.\n\nSo you can run this loop(Java code)\n\nfor(i=0;i==n-m;i++)\n{\nint k;\nfor(k=0;k==n-m;k++)\n{\nint p,q;\nfor(p=0;p<m;p++)\n{\nfor(q=0;q<m;q++)\n{\nprint("  ");\nprint(a[p][q]);\n}\nprintln();\n}\n}\n\n}\n\nTo print all the submatrices iterate over this procedure canging m from 1 to n.
37157
A covariance matrix is a useful tool in statistics. \n\nA matrix is an array of numbers, in the form of a rectangle, with several rows and several columns. A covariance matrix is a special kind of matrix that is used to find connections between variables. \n\nFor example, if you have N random variables, and you want to know how they are related to each other, then you can compute a covariance matrix. The covariance matrix for N random variables has N rows and N columns, that is, it is an N x N matrix. Note that the covariance matrix is a square matrix (it has the same number of rows and columns; in this case, N).\n \n\nThe diagonal elements of the covariance matrix give the variance of each of the random variables in turn; that is, they describe how much each of these random variables varies. \n\nThe off-diagonal elements describe how the random variable Xj corresponding to the jth row and the random variable Xk corresponding to the kth column behave. For example, if changes in Xj do not have any effect on Xk, then the covariance matrix value at row j and column k, C(j,k) will be zero: C(j,k)=0. If when Xj increases, Xk increases, then C(j,k)>0. If increasing Xj causes Xk to decrease, then C(j,k)<0.\n\nOf course, the details about how to compute the covariance matrix C(j,k) are slightly more complicated than this, but this gives you a sort of idea about what the covariance matrix means.
618738
I can't give you a definiton because I really didn't understand anything I read about matrices!!!  But here are a few links that may be helpful to you.
260574
Because it can be considered a "matrix"\n\n\nhttp://www.innvista.com/health/anatomy/connect.htm
1035025
A matrix is a rectangular array of numbers.  For our purposes, let z{m,n} stand for the number z in the mth row and nth column of a matrix.\n\nExample -- Consider the 2x2 matrix named ' A'.  Let A be composed of numbers z{1,1}=a,  z{1,2}=b,  z{2,1}=c, and z{2,2}=d.  \nOr more simply,   the matrix A is set up as\n| a   b |\n| c   d |\n\nBy definition:  If 'A' is square matrix, and if a matrix 'B' can be found such that  AB=BA=I, then A is said to be invertible, and B is called an inverse of A.\n\nNow, go back to our Example 2x2 matrix named 'A'.  If (ad - bc) is not equal to zero, then the inverse of A, called A^(-1), is equal to\n\n(1/ (ad-bc) ) times the matrix\n  |  d   (-b) |\n  | (-c)   a  |\nwhere z{1,1}=d,  z{1,2}= -b,  z{2,1}= -c, and z{2,2}=a.  \n\nOR\n\n |  (d/(ad-bc))     (-b/(ad-bc)) |\n |  (-c/(ad-bc))    (a/(ad-bc))  |\nwhere z{1,1} =(d/(ad-bc)),  z{1,2} =(-b/(ad-bc)),  z{2,1}=(-c/(ad-bc)), and z{2,2} =(a/(ad-bc)).
692261
You need to switch to the Projection Matrix when you are specifying the projections. \ne.g.\n glMatrixMode( GL_PROJECTION );\n glLoadIdentity();\n gluPerspective( 60.0, aspectRatio, 1.0, 30.0 );\n\n\nThe model view matrix is used for the rotation, scaling and translation stuff. So, you need to do glMatrixMode(GL_MODELVIEW) before you do your transformations.
949909
So Im guessing your have a transformation matrix.  \n\nYou have an initial state specified by matrix [A] and the linear transform specified by matrix [T].\n\nTo find the state after iteration 1 you multiply\n[A]x[T]=[y1]\n\nIf [y1] was your initial matrix you would multiply it by [T] to get the state after iteration.\n[y1]x[T]=[y2]\nby substitution\n([A] x [T]) x [T] = [y2]\nor \n[A] x [T]^2 = [y2]\n\nso what is T at the nth iteration?\n[A] x [T]^n = [yn]\n\nso what is the limit as n goes to inifinity?\n[A] x lim([T]^n,n->inf) = [y_inf]\n\nNumerical results break down, consider a rotation.  The limit as it goes to infinity is undefined.\nT(1,1)=sin(x)\nT(1,2)=cos(x)\nT(2,1)=-1*cos(x)\nT(2,2)=sin(x)\n\nWhen you square it you get\nT2(1,1)=sin(x)^2-cos(x)^2 = -cos(2*x)\nT2(1,2)=2*cos(x)*sin(x) = sin(2*x)\nT2(2,1)=-2*cos(x)*sin(x) = -sin(2*x)\nT2(2,2)=sin(x)^2-cos(x)^2 = -cos(2*x)\n\nWhen you cube it you get\nT3(1,1)=-sin(3*x)\nT3(1,2)=-cos(3*x)\nT3(2,1)=cos(3*x)\nT3(2,2)=-sin(3*x)\n\nWhen you raise it to the power of 100 (come on MatLab, dont die on me now, darnit) ... ahem.  When you raise it to the power of 10 (?)  aha\nT10(1,1)=-cos(10*x)\nT10(1,2)=sin(10*x)\nT10(2,1)=-sin(10*x)\nT10(2,2)=-cos(10*x)\n\nSee how that works?  Not bad huh?\n\nFor the pure rotation it never ends.\nWhat about a shrinking, a scaling by a matrix whose norm is less than 1? (note if you use 0.9, the norm is 1.8.)\n\nT(i,j)=0.45\nyields\nT2(i,j)=0.4050\nT3(i,j)=0.3645\nT10(i,j)=0.1743\nT1000(i,j)=0.1328*10^-4\n\nIt disappears to zero.\n\nintuitively you can also figure out that for a scaling by a matrix whose norm is larger than one the transform eventually goes to infinity as the limit is taken to infinity.\n\nyou could make a non-trigonometric rotation, like the sin(1/x)  Thats a fun expression.\n\nT(i,i)=sin(1/x) \nT(1,2)=cos(1/x)\nT(2,1)=-1*cos(1/x)\n\nAs x goes to infinity what happens?\nit becomes a reflection\nT_inf(i,i)=0\nT_inf(1,2)=1\nT_inf(2,1)=-1\n\nIm digressing.  Its a fun field with interesting applications.  Weather forecasting, deformation of solids, heat transfer, optimization, all use linear transformations to do their work.\n\nHave fun.
805688
U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
967181
depends on the variable, you could use a dot density map or a cartogram or a choropleth map, or evenanother type of map
387286
Karnaugh map:\n\nCheck out the following link which will help.
115853
#include <iostream>\n#include <vector>\nusing namespace std;\n#define rep(i, n) for((i) = 0; (i) < (n); (i)++)\n#define rep2(i, k, n) for((i) = k; (i) < (n); (i)++)\nvoid f21();\nint n;\ndouble mat[100][100];\nint main()\n{\nwhile(true)\n{\ncin >> n;\nif(!n) break;\nmemset(mat, 0, sizeof(mat));\nf21();\n}\nreturn 0;\n}\n\nvoid f21()\n{\ncout << "****************************************" << endl;\ncout << "*          INVERSE OF MATRIX           *" << endl;\ncout << "****************************************" << endl;\n\ncout << "ORDER OF THE MATRIX: " << n << endl;\ncout << "ENTER ELEMENT ROW-WISE" << endl;\nint m = 2*n;\nint i, j, k;\nrep(i, n) {\nrep(j, n) {\ncin >> mat[i][j];\ncout <<"&#92;t";\nprintf("%5.3lf", mat[i][j]);\n}\nrep2(j, n, m) {\nif(i == (j-n)){\nmat[i][j] = 1;\ncontinue;\n}\nmat[i][j] = 0;\n}\ncout << endl;\n}\n\ncout << endl << "INVERSE MATRIX: " << endl;\ndouble pivot, comm;\nrep(i, n) {\npivot = mat[i][i];\nif(mat[i][i] == 0){\ncout << "DIVISION BY ZERO." <<endl;\nreturn;\n}\nrep(j, m) mat[i][j] = mat[i][j]/pivot;\nrep(j, n) {\nif(i == j) continue;\ncomm = mat[j][i];\nrep2(k, i, m) {\nmat[j][k] = mat[j][k] - (comm*mat[i][k]);\n}\n}\n}\n\nrep(i, n) {\nrep2(j, n, m) {\ncout <<"&#92;t";\nprintf("%6.4lf", mat[i][j]);\n}\ncout << endl;\n}\ncout << endl;\n}\n\n/*This is writen in C++ this will output the inverse of a Matrix\ninput:\n4\n1 2 3 4\n2 1 4 2\n2 5 7 12\n1 -1 3 11\n\n5\n1.173 -0.634 0 0 0\n-0.539 1.173 -0.634 0 0\n0 -0.539 1.173 -0.634 0\n0 0 -0.539 1.173 -0.634\n0 0 0 -0.539 1.173\n\n0\n*/
889697
well gauss jordan is a easy method for matrix manipulations. if u use reverse triangulation also in conjunction with it . it wud be much more simpler.\n\n2x-y+3z=8\n-x+2y+z=4\n3x+y-z=0\n\nthis is set of equations given.write them in matrix form\n\n|| 2  -1  3 ||  ||x||     ||8||\n|| -1  2  1 ||  ||y||  = ||4|| \n|| 3   1  -1||  ||z||     ||0||\n\nA    X= B\nnow the equation is in the form of AX=B\n\nnow our concentration will be on matrices A and B\n\nour aim is to make the lower triangle elements equal to 0.\n\nnow follow the steps one by one. apply the calculations on both A and B  matrices.\n\nwe have 3 rows in matrix A and let us name them as R1 , R2 and R3.\n\nreplace R2 by R2/-1 and R3 by R3/3\n\nnow the matrices will be\n\n||2   -1    3    || ||x||   ||8 ||\n||1   -2   -1    || ||y||= ||-4||\n||1   1/3  -1/3 || ||z||   ||0 ||\nreplace R2=R1-2xR2\n||2   -1    3     || ||x||    ||8 ||\n||0    3    5     || ||y||=  ||16||\n||1   1/3  -1/3 ||  ||z||   ||0 ||\ncheck once what we have done here. \nevery element of R2 is calculated from the preceding element in that colum by the formula R1-2xR2.\n\nthe elelment of colum one is\n2-2x1=0\nthe element of colum 2 is -1-(2x-2)=3\nthe element of colum 3 is 3-(2x-1)=5\n\nlike wise \nreplace R3=R1-2xR3\n||2   -1    3      || ||x||     ||8 ||\n||0    3    5      || ||y|| =  ||16||\n||0   -5/3  11/3 || ||z||     ||8 ||\n\nnow we got the first colum zero in the lower triangle.\n\nnow the second element in the third row needs to be zeroed.\nreplace R3=R3x(-3/5) or  R3=R3/(-5/3)\n||2   -1    3      || ||x||   ||  8   ||\n||0    3    5      || ||y||= ||  16  ||\n||0    1  11/5   || ||z||   ||-24/5 ||\nreplace R2=R2/3\n||2   -1    3      || ||x||   ||  8   ||\n||0    1    5/3   || ||y||= || 16/3 ||\n||0    1  11/5   || ||z||   ||-24/5 ||\n\nnow replace R3=R3-R2\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\n\nnow apply the final steps of reverse triangulation technique.\nfrom the matrix equation, if again converted in to normal equations \n\n2x-y+3z   =8\n  y+5z/3  =16/3\n    8z/15 =-152/15\nnow z= (-152/15) divided by (8/15)\n= (-152/15)/(8/15) =-152/8=-19\nthere fore z=-19\nnow substitute z=-19 in equation 2\n\ny+5x(-19)/3=16/3\ny=(16/3)+(95/3)=111/3=37\nsubstitute the values of y and Z in equation 1\n\n2x-37+3x-19=8\n2x=8+37+57=102\nX=102/2=51\nfinally we got the values of x = 51, y=37 and z=-19\n\nactually u can continue the guass jordan method without applying the reverse triangulation method to get the values of x, y \n\nand z directy from the matrix for that after making the lower triangle zero we have to make the upper triangle elements also \n\nequal to zero.\n\nlet us see how to do that.\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\nreplace R3=R2/(8/15)\n||2   -1    3     || ||x||   ||  8     ||\n||0    1    5/3  || ||y||= || 16/3   ||\n||0    0    1     || ||z||   || -19    ||\nR2=R2-5xR3/3\n||2   -1    3    || ||x||  ||  8     ||\n||0    1    0    || ||y||= ||  37    ||\n||0    0    1    || ||z||  || -19    ||\n\nR1=R1/3\n||2/3   -1/3    1    || ||x||   ||  8/3   ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\n\nR1=R1-R3\n||2/3   -1/3    0    || ||x||   ||  65/3  ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\nR1=R1x-3\n||-2    1     0    || ||x||   ||  -65   ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1-R2\n||-2    0     0    || ||x||   ||  -102  ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1/(-2)\n||1    0     0    || ||x||  ||  51    ||\n||0       1     0    || ||y||=||  37    ||\n||0       0     1    || ||z||  || -19    ||\nnow we can diretly read the matrix into equations as\nx=51, y=37 and z=-19\n\n\nhope und
865560
A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
892094
Linear programming uses matrices, and it is used e.g. to optimize the manufacturing of certain goods. Engineering is another application, e.g. digital data storing and image compression. Economics is a third one, e.g. there are input-output models to model sectors of an economy. I'm sure there are a lot more.