explain gauss jordan elimination method for matrices as simplified as possible
212379
Here's an article about it:\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
117040
The techniques used to solve systems of linear equations (substitution and elimination) tend to do a pretty good job of solving quadratic systems.\n\nThis page does a pretty good job of demonstrating how:
439014
I just noticed that two rows are the same, right?  This is big, but I'll have to study up what it means.  Intuitively, I think that it means that the matrix has a null space, and thus and eigenvalue of zero.  \n\nor is it a type?\n\n\nRead up on it:\n\nhttp://en.wikipedia.org/wiki/Jordan_normal_form
113028
Lets call this matrix A.\n[ 2 0 4 ]\n[ 3 1 5 ]\n[ -1 1 -2 ] = A\n\nNow we need to solve the equation AV = I\nthen V is the inverse matrix.  Lets write this out\n\n[ 2 0 4 ] ` ` ` ` [ 1 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 1 0 ]\n[ -1 1 -2 ] V = [ 0 0 1 ]\n\nNow just multiply elementary matrices (row operations) to the left hand side of both sides of the equation until the left-hand entity is the identity.  Multiply the following matrix on the left-hand-side both sides [[1/2 0 0][0 1 0][0 0 -1]], that is replace the first row with 1/2 times the first row, and replace the third row with -1 times the third row:\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 ` 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\napply [[1 0 0][[-3 1 0][0 0 1]] both sides, that is replace second row with second row plus negative-three times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\nmultiply both sides on the left by the matrix [[1 0 0][0 1 0][ -1 0 1]], that replace third row with third row plus negative one times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 ` 1 0 ]\n[ 0 -1 0 ] V = [ -1/2 ` 0 -1 ]\n\nmultiply by [[1 0 0][0 0 -1][0 1 0]], that is place opposite of row 3 for row 2 and row 2 for row 3\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 1 -1 ] V = [ -3/2 1 0 ]\n\nnow apply [[1 0 0][0 1 0][ 0 -1 1]] both sides, that is replace third row with third row plus -1 times second row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 -1 ] V = [ -2 ` 1 -1 ]\n\napply [[ 1 0 2][0 1 0][0 0 -1]] both sides, replace row 1 with row 1 plus 2 times row 3, and replace row 3 with negative 1 times row 3\n\n[ 1 0 0 ] ` ` ` ` [ -3/2 2 -2 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 1 ] V = [ 2 -1 1 ]\n\nNow, since the left hand item is the identity, the inverse matrix is\n\n[ -3/2 2 -2 ]\n[ 1/2 0 1 ]\n[ 2 ` -1 1 ]\n\nThe product of all of the elementary operation matrices specified is also this matrix, and the reader can verity that AV = VA = I
29309
I recommend purlplemath.com.  i'm not sure how helpful it is overal, but i like how they explain matrix multiplication <http://www.purplemath.com/modules/mtrxmult.htm>.  you can also do a web search for the subject you have a problem with, there are plenty of sites.  and your teacher should love to help you.
554392
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
426810
The singular value decomposition for a matrix A writes A as a product (hanger)(stretcher)(aligner). \n \n\nIt's an amazing and useful fact that every m x n matrix has a singular value decomposition. \n\n-------------------\nIn linear algebra, the singular value decomposition (SVD) is an important factorization of a rectangular real or complex matrix, with several applications in signal processing and statistics.\n\nThe SVD can be seen as a generalization of the spectral theorem to arbitrary, not necessarily square, matrices. The spectral theorem says that normal matrices can be unitarily diagonalized using a basis of eigenvectors.\n\n---------------------------------\nSingular value decomposition takes a rectangular matrix of gene expression data (defined as A, where A is a n x p matrix) in which the n rows represents the genes, and the p columns represents the experimental conditions.\n\n---------------------------------\nA very powerful set of techniques dealing with sets of equations or matrices that are either singular or numerically very close to singular is the so-called singular value decomposition (SVD). SVD allows one to diagnose the problems in a given matrix and provides numerical answer as well.\n\nAny m x n matrix a (m >= n) can be written as the product of an m x n column-orthogonal matrix u, an n x n diagonal matrix with positive or zero elements, and the transpose of an n x n orthogonal matrix v:\n\n\nI hope that all these are useful, good luck!!!!!!!!!!!!
285206
At first I thought you were wanting the orthogonal matrices, but your example shows you are not. Orthogonal matrices are those whose inverse is their transpose. Yours are ones where the elements of the transpose are the inverse of the elements of the original.\n\nCertainly such a matrix has to have +-1 on the diagonal. Your matrix is singluar and any 2x2 matrix with your property and all 1 on the diagonal will be also. This means that one eigenvalue is 0 and the other is 2.\n\nI don't see anything special for 3x3 offhand, but I will look further.
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| . x . x |\n| -x . -x |\n\nwhere x is any real number and x ≠ 0 (The last part is only needed so that A will be non-zero). The transpose also works.\n\nMultiply it out yourself. The result has nothing to do with triangular matrices, so I also suggest that you do not merely cite special characteristics of such matrices in your answer.
381208
The inversion of a 3x3 matrix is very easy to solve, however the equation is fairly long. You can find it on wikipedia at http://en.wikipedia.org/wiki/Inverse_matrix#Inversion_of_3_x_3_matrices
500012
Well you can, in a sense.\nI don't know a SIMPLE way to do it, in the sense that there is a SIMPLE way to do matrix multiplication. But maybe the following explanation will be helpful:\n\nIf [A] x [B] = [C] (i.e., if the product of matrices A and B equals matrix C), then we can say that [C] / [B] = [A]. So in that sense, we are dividing C by B.\n\nBut what if we don't know matrix A. What if we know only [B] and [C]? How can we divide [C] by [B] to find [A]?\n\nThe answer is that we find the "inverse" of matrix B, and then MULTIPLY matrix C by this inverse matrix (which is shown as [B]^-1). We define [B]^-1 as the matrix for which [B] x [B]^-1 = the identity matrix. (The identity matrix is a square matrix that has 1's on its main diagonal and 0's everywhere else.)\n\nThe problem, then, becomes how to find the inverse of [B]. Fortunately, there are procedures for this, and you can probably find them on line using Google. Not all matrices can be inverted. But if B can be inverted, then you have:\n\n[C] / [B] = [C] x [B]^-1 = [A]\n\nNote that if we substitute [A] x [B] for [C], then this becomes;\n([A] x [B]) / [B] = [A] x [B] x [B]^-1 = [A] x identity matrix = [A]
186163
The size of a matrix multiplication is as follows:\n\n(nxm) (mxp)= (nxp)\n\n(1x2) (1x1) = 1x1\n\n(3x1) (1x3) = 3x3
382205
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
420696
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.
96517
sorry i don't know mathematica but you can solve it easily using matlab. Enter rthe coefficient matrix as follows for example,\nA=[1 2 3 4; 2 3 3 4;........]\nand the right hand side\nb=[ 2 3 4 4]\nonly for illustration. ";"indicates columns.\nA/b solves this system and gives the solution vector x.
35837
Are there any linear algebra program solvers available? I am sure there are. Just use one of those.\n\n--------------\n\nI did a quick look on the web. I think it is built in already. Here are a couple of insruction manuals. Good luck.
33621
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.)