still stuck with co-ordinate change matrices?
816222
(a .. b .. c)( 1 ) . (5 - x + 0x²)\n(d .. e .. f)( x ) = (6 + x + 0x²)\n(g .. h .. i)( x² ) . (1 + 3x + x²)\n\nSo, by equating coefficients:\na = 5, b = -1, c = 0\nd = 6, e = 1, f = 0\ng = 1, h = 5, i = 1\n\nSo the coordinate change matrix is\n\n( 5 . -1 . 0 )\n( 6 .. 1 . 0 )\n( 1 .. 5 . 1 )
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*/
732131
ha! I did this today in Algebra!!\n\nOkay, basically you take each of the coordinates and plug them into this fomula\n\n\nplusorminus 1/2  det| x1 y1 1|\n                                 | x2 y2 1|\n                                 | x3 y3 1|         solve for 1/2 of this detemanant just like you would solve for a 3 x 3 matrix.  You answer of course must always be positive.\n\nfor some reason, the two lines under plusorminus... got moved over, they should fall directly under |x1 y1 1|
1802
I haven't tried it, but check out MatPy (linked below).  Another possibility is Numerical Python (also linked below).
451843
If you have correlation matrix A, then A(i,j) is the covariance between i and j, I think.  The matrix should be symmetric across its main diagonal.
304446
hey!! chek out this website...\nBUT Y DO U WANT TO USE MATLAB FOR SEPARATING EVEN AND ODD?? THERE ARE OTHER METHODS...
499636
Look here:\nhttp://www.posc.org/Specifications/Epicentre_V30/DataModel/LogicalDictionary/StandardValues/projected_2d_coordinate_system.html\n\n\nhttp://www.ldeo.columbia.edu/res/pi/MB-System/html/mbmosaic.html
1062782
you should use matlab.
398921
To make an augmented matrix, just take your coefficients and put them in a matrix, then draw a vertical line and place the constants on the right side, like this:\n\n| -2 0 9 | 2 |\n| 0 -2 9 | 22 |\n| 3 5 -2 | 10 |\n\nFill in 0's for any variables that aren't there (i.e. there's no y in the first equation, so there's a 0 in its place in the matrix).
67077
Here ya go: http://www.longitudebooks.com/find/p/12027/mcms.html
726789
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!
1116867
try the conversion table. europe uses metrics
201111
You need to measure the distance between the closely spaced lines (J values).  Trans and cis J values will be different
1017997
The equations are wrongor insufficient. Post the correct data if you want me to help/ Alternately, you could also message me.
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.
953641
I can only say so much because those equations are too long for two in the morning. Not that it matters; I just won't be retyping them. But you certainly need a bunch of random things for this. First, you need a variable -- a char, probably, since you don't care about any input beyond that. I don't know what the compare thing is going to output, but that'll come.\n\nI suggest an array for the old and new charts.\n\nGeez, what's the syntax? int oldchart[11], or something? I'm getting too used to Java. Anyway. Since you're going in increments of 5 from 0 to 50, you'll have a 0, 5, 10, ... 45, 50 -- that is, 11 different speeds, right? So you'll need an array for the old chart and an array for the new chart, both with a size of 11. Do you get where I'm going?\n\nCalculating the chart takes a bit more thinking. We'll ignore the actual process of getting input for the moment and just deal with filling it in. I'll suggest a method, and explain afterwards:\n\nint speed;\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   oldchart[i] = <calculate wind chill>;\n}\n\nThe same thing could be applies to the newchart array. The index, i, will go from 0 to 10, matching the indexes of the array, right? Since you're going by increments of 5, the speed for slot i will be i*5. That takes you from 0 to 50.\n\nSo after you pull that off for the old and new, you've got charts saved in the variables. Printing them is kind of similar to what happened above -- you use a loop to output what's in the array, using i*5 for the speed.\n\nNow, granted, you could combine this and the above step so you wouldn't need the arrays at all--you'd just recalculate and print every time. But I find it easier to organize my thoughts by going a way that deals with one issue at a time... It depends on what you prefer at this point.\n\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   cout << "Speed: "<< speed << " Wind Chill: " << oldarray[i] << endl;\n}\n\nLet's see...\n\nDisplaying the menu is easy--just use cout << "your menu stuff" << endl;\nInput is a bit stranger, I think. For something like this, just use a cin and check afterwards if it's right. Um, let's see... I'll give this to you in pseudocode--you should be able to work out the actual stuff.\n\nwhile ( the user doesn't want to exit )\n//this is just a logic thing -- if the user is done, then what?\n{\n     take the user's input;\n\n     if the input is 'o' or 'O'\n     {\n          calculate and print the old chart\n     }\n\n     else if the input is 'n' or 'N'\n     {\n          calculate and print the new chart\n     }\n\n     else if the input is 'c' or 'C'\n     {\n          calculate and print both the old and new charts (?) (whatever compare does)\n     }\n\n     else if the user wants to exit\n     {\n          end the program (usu. with return 0 or something of that nature)\n     }\n\n     else the user's input was bad\n}\n\nMind that for this to go well, you'd have to mind a few issues -- as I wrote above, if the user doesn't enter what you expect them to. Also, you'd have to deal with a case where the user decided to print charts before entering data into the old or new charts. For that, you should give the temperature a default value, like 0, so the charts can still be calculated.\n\nYou can see that I wrote in a calculation for every print, which is what I was saying before with it not being necessary. You could technically use the for loop I gave for printing but just replace the reference to the array with the calculation for the wind chill.\n\n...Don't know if I covered all of it, but I should be getting to bed. ^^;
1126894
This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).