How to find submatrices (mxm order) of a given matrix and to get them printed???
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.
858418
Click ON\nClick PRGM\nSelect NEW tab and hit ENTER\n\nTo select DISP and PROMPT:\n Hit PRGM when creating or editing a program,\n Select Tab I/O and you will see you selection\n\nTo get the arrow, click STO-> just to the left of 1\n\n\nThis is the Program: (or the one I use)\n\n:Disp "AX^2+BX+C"\n\n:Prompt A\n:Prompt B\n:Prompt C\n:(-B+*square root*(B^2-4AC))/(2A)->D\n:(-B-*square root*(B^2-4AC))/(2A)->E\n:Disp D\n:Disp E\n\nTo run program, go to the main screen, click PRGM, select your program, hit ENTER, input values for A, B, and C, hit ENTER and you result will show your two possible answers.
723293
you have to use the couchy schuartz method\nim nit sure on the spelling \nbasicly you take one vector (V1) and use it then you take the second vecter (U1) minous vi times(V1 interproduct U1)/(magnatude V1^2) and use that total as V2 then you do a simular thing for the third and fourth vector and up if you have them by subtrating V1<Ua,V1>/(mag V1)^2 plus V2<Ua,V2>/(mag V2)^2 .... till you have done that for each vector \nin other words every vector will have a term for all of the new previous vectors subtracted from it then finaly you take all the new vectors and devide them by their magnatude to get unit vectors and they will then all be orthOGANOL UNIT VECTORS
584514
Use 'Gaussian elimination' to obtain a basis for the span of matrix columns or rows.\n\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
737238
the 3 x 3 matrix A times the 3 x 3 matix B will yield a 3 x 3 matrix AB\n\nAB1,1 = (A1,1)(B1,1) + (A1,2)(B2,1) + (A1,3)(B3,1)\nAB1,1 = (-3)(-1) + (6)(0) + (2)(-5) = -7\n\nAB1,2 = (A1,1)(B1,2) + (A1,2)(B2,2) + (A1,3)(B3,2)\nAB1,2 = (-3)(8) + (6)(-3) + (2)(0) = -42\n\nAB2,1 = (A2,1)(B1,1) + (A2,2)(B2,1) + (A2,3)(B3,1)\nAB2,1 = (-4)(-1) + (8)(0) + (1)(-5) = -1\n\netc.
195858
Here is an easy way. Say you want to find the inverse of the 2x2 matrix\n\n4 1\n1 2\n\nTo type this matrix into a TI-89, you can type in a list of lists using the square brackets, like this:\n\n[[4,1],[1,2]]\n\nYou can also use semi colons to separate the rows instead of using sublists, which can be a bit easier to read, like this:\n\n[4,1;1,2]\n\nNow to find the inverse, just type ^(-1) onto the end of the matrix you typed in (and make sure you use the unary minus key, which is the gray (-) key on the number pad, not the black subtraction key which is on the right side of the calculator)\n\nAll put together, to find the inverse of the matrix above, you can type:\n\n[4,1;1,2]^(-1)\n\nand then push enter of course.\n\nAs another example, to find the inverse of the 3x3 matrix\n\n1 2 3\n5 4 8\n9 6 7\n\ntype in:\n\n[1,2,3;5,4,8;9,6,7]^(-1)
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
426810
search in www.matweb.com
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
1802
I haven't tried it, but check out MatPy (linked below).  Another possibility is Numerical Python (also linked below).
718910
ill use this as a key for a standard matrix:\n\na b\nc d\n\nin your coeff. matrix:\n\na=equation 1 x coeff\nb=eq.1 y coeff.\nc=eq.2 x coeff.\nd=eq.2 y coeff.\n\nso the coeff. matrix looks like\n\n3   11\n-1  4\n\nthe value of the determinant is always "ad-bc" for a 2 by 2, so yours youls be\n\n(3)(4)-(-1)(11) = 12+11 = 23\n\n\nand i dont know what you mean by adjoint, sorry
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
85187
I don't really know what you need help on....\n\nCheck out some of the sites below in the source section, or you could just post what in particular is troubling you...\n\n(I can sympathize if you're having trouble with Matrices though... those always have so many places to go wrong.. ugh.)\n\n(PPS: Wikipedia can be very useful for finding out more information, as you can tell from the number of links below.)
95774
The RANK of matrix A, an m x n matrix, is the dimension of the row spaces or the column space. Or, the number of basic variables in Ax = 0. \n\nUsually written as rank A.\n\nAlso, for m x n matrix A, the row space and the column space have the same dimension, \n\nwritten dim R(A) = dim C(A).
755982
Subscripts\nSubscripts are special postfix operators placed in square brackets ([ ]) after a matrix operand. Subscript operations have the general form \n\noperand[row,column] \nwhere \noperand \nis usually a matrix name, but it can also be an expression or literal. \n\n\nrow \nrefers to an expression, either scalar or vector, for selecting one or more rows from the operand. \ncolumn \nrefers to an expression, either scalar or vector, for selecting one or more columns from the operand. \nYou can use subscripts to \n\nrefer to a single element of a matrix \nrefer to an entire row or column of a matrix \nrefer to any submatrix contained within a matrix \nperform a reduction across rows or columns of a matrix \nIn expressions, subscripts have the same (high) precedence as the transpose postfix operator (). Note that when both row and column subscripts are used, they are separated by a comma. \n\nSelecting a Single Element\nYou can select a single element of a matrix in two ways. You can use two subscripts (row, column) to refer to its location, or you can use one subscript to look for the element down the rows. For example, referring to the coffee example used earlier, to find the element corresponding to the number of cups that Linda drank on Monday, you can use either of two statements. \nFirst, you can refer to the element by row and column location. In this case, you want the second row and first column. You can call this matrix c21. \n\n   >  print  coffee[rowname=names];\n\n           COFFEE\n           JENNY          4         2         2         3         2\n           LINDA          3         3         1         2         1\n           JIM            2         1         0         2         1\n           SAMUEL         5         4         4         3         4\n\n   >  c21=coffee[2,1];\n   >  print c21;\n\n                               C21\n                                 3\n\nYou can also look for the element down the rows. In this case, you refer to this element as the sixth element of COFFEE in row-major order. \n   >  c6=coffee[6];\n   >  print c6;\n\n                                C6\n                                 3\n\n\nSelecting a Row or Column\nTo refer to an entire row or column of a matrix, write the subscript with the row or column number, omitting the other subscript but not the comma. For example, to refer to the row of COFFEE that corresponds to Jim, you want the submatrix consisting of the third row and all columns: \n   >  jim=coffee[3,];\n   >  print jim;\n\n\n                JIM\n                  2         1         0         2         1\n\nIf you want the data for Friday, you know that the fifth column corresponds to Friday, so you want the submatrix consisting of the fifth column and all rows: \n   >  friday=coffee[,5];\n   >  print friday;\n\n                          FRIDAY\n                               2\n                               1\n                               1\n                               4\n\n\nSubmatrices\nYou refer to a submatrix by the specific rows and columns you want. Include within the brackets the rows you want, a comma, and the columns you want. For example, to create the submatrix of COFFEE consisting of the first and third rows and the second, third, and fifth columns, submit the following statements: \n   >  submat1=coffee[{1 3},{2 3 5}];\n   >  print submat1;\n\n                 SUBMAT1\n                       2         2         2\n                       1         0         1\n\nThe first vector, {1 3}, selects the rows, and the second vector, {2 3 5}, selects the columns. Alternately, you can create the vectors beforehand and supply their names as arguments. \n   >  rows={1 3};\n   >  cols={2 3 5};\n   >  submat1=coffee[rows,cols];\n\nYou can use index vectors generated by the index creation operator (:) in subscripts to refer to successive rows or columns. For example, to select the first three rows and last three colu
350177
Sometimes adding some indents will help figure out what's going on.  In your algorithm, you have a lot of nested if and while statements, so space them out like this first to see how each is related to each other (I'm using periods because Yahoo won't show spaces):\n\ni <- 0\nwhile (i < n) do\n...j <- i+1\n...m <- a[i]\n...while (j < n) do\n......if (m > a[j]) then\n.........m <- a[j]\n......end if\n......j <- j + 1\n...end while\n...b[i] <- m\n...i <- i + 1\nend while\n\nYour i is a counter that starts with 0, and your j is a counter that starts with 1.  j will always be more than i.\n\nThe "while (i < n) do" line limits this algorithm to 3 loops, since n is 3.\n\nDirectly after this statement, j is established as a counter, and m is defined to be the i-th element of your matrix.  The next while statement then compares m (the i-th element in the matrix) to the element after this (using j to read the next element).\n\nIf m is larger, then the if statement makes m take on the smaller value.  If m is smaller, nothing happens and j is indexed to the next value.  Then the process repeats itself.  If we get to the end of the matrix and m isn't replaced anymore, it becomes the i-th value in matrix B.  After this, i is indexed to the next element in matrix A and the entire process is repeated once again, until all of matrix A has been checked.\n\nThe result then is that through m, matrix A is sorted least to greatest and is written into matrix B, which is answer (d).
695328
(1) add row#1*6 from row#2, then from row#3\nthat is [–6+1*6, –47+6*6, -2+1*6] for row#2\nthat is [-6+1*6, -41+6*6, 2+1*6] for row#3\nthus:\n1……6……1       \n0…..-11…..4\n0……-5…..8\n(2) add row#2*(-5/11) from row#3\nthat is [0, -5+(-5)*(-5/11), 8+4*(-5/11)] for row#3 \nthus:\n1……6……1       \n0…...-11…..4\n0……0….68/11\nto check the result multiply main diagonal: 1*(-11)*(68/11)=-68