can u give me smith's algorithm for polynomial matrices?
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.
1130286
There is a primer here:\nhttp://www.phon.ucl.ac.uk/courses/spsci/matlab/primer40.pdf\n\nand an online course here:\nhttp://www.phon.ucl.ac.uk/courses/spsci/matlab/\n\nI may just be reviewing them myself!
848509
http://mathworld.wolfram.com/johnsonsoli...
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.
319998
http://www.math.iitb.ac.in/resources/manuals/algorithms.html click d link to know. It may be helpful to u.
16666
it called the Cayley-Hamilton Theorem, type it in google and its proved in many places;\n\n\nhttp://delta.cs.cinvestav.mx/~mcintosh/comun/summer99/mcintosh/node9.html  <---this one is good :)\n\nLet B = x*I - A, where I is the identity matrix of the same size as A.  \nLet C be the classical adjoint of B, so that BC = (det B)I. Recall \nf(x) = det(B), by definition. We wish to show that f(A) = 0. It \nsuffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the \nk-th unit vector, so fix a value of k in that range.\n\nNow x = A implies that for all j,\n\n   x*e(j) = A*e(j),\n\n             n\n          = SUM A(i,j)*e(i),\n            i=1\n   \n      n                     n\n   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),\n     i=1                   i=1\n\n    n\n   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,\n   i=1\n\n    n\n   SUM B(i,j)*e(i) = 0,\n   i=1\n\n           n\n   C(j,k)*SUM B(i,j)*e(i) = 0,\n          i=1\n\n    n\n   SUM B(i,j)*C(j,k)*e(i) = 0.\n   i=1\n\nThen, summing over all j, reversing the order of summation, and using\nthe definition of the classical adjoint,\n\n    n   n\n   SUM SUM B(i,j)*C(j,k)*e(i) = 0,\n   j=1 i=1\n\n    n    n\n   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,\n   i=1  j=1\n\n    n\n   SUM [det(B)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(x)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(A)*delta(i,k)]*e(i) = 0,\n   i=1\n\n         n\n   f(A)*SUM [delta(i,k)]*e(i) = 0,\n        i=1\n\n   f(A)*e(k) = 0.\n\nThis proves that f(A) annihilates all the unit vectors, so must be\nthe zero matrix, and f(A) = 0.\n\n\nsecond one taken off the dr math website, but the link above might be easier to understand.
332712
The statement is true for :\nIf A is a skew-symmetric matrix, then A = - At (A transpose)\nSo (At)^(-1)= (A^(-1))t\nSo (-A)^(-1)= (A^(-1))t\nSo -1((A)^(-1)= (A^(-1))t\nSo A^(-1) is skew-symmetric
958486
Rs=Im*Rm/Is\n\nRs*Is = Im*Rm\nRm= (Rs*Is)/Im\n\nc=(8.85*10^-12)(K)(A/d)\nc*d = (8.85*10^-12)(K)(A)\nd= [8.85*10^-12)(K)(A)]/c\n\ny=mx+b\nb=y/(mx)\n\nx= (y-b)/m\n\nLm=(Lt-L1-L2)/2\n\nE=Q/C\n\n1/Rt - 1/R2 = 1/R1
210912
Put the left side coefficients into matrix [A],\n[6 2]\n[5 6]\nand the right side values into matrix [B]\n[-16]\n[..4]\nthen tell the calculator to multiply [A] inverse * [B]\nthe result is the solution matrix\n[x]\n[y]\n\nnote, the inverse key looks like x^-1.  \n\nsolve the second system the same way.  coefficient matrix will be 3x3, answer matrix 3x1.
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.)
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
734225
i just learned this in class... \nafter you enter your mass matix M,\nyour Dampening Matrix C\nand your Stiffness Matrix K\nyou take [K] - lamda [M]= (b) \nyou can type \n[Evals]=eigs(new)\nThese will also be the square of your natural frequencies.\nyour mode shapes are the eigvectors! fyi
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
348293
There are two seminar papers by Magnus Hestenes that should give you a good start:\n\n"A  Ternary Algebra with Applications to Matrices and Linear Transformation"\n\n"On a Ternary Algebra"
554827
Well, I suppose you want a program that computes the trace of a matrix. The trace is simply the sum of the elements in the diagonol in a matrix. I won't show it to you with OOP since there's no reason to.\n\nint trace(int matrix[][],int numrows,int numcols) {\n\nint tracetotal = 0;\n\nif (numcols != numrows) {\nstd::cout<<"Error: Can't compute trace. Not square matrix.&#92;n";\nexit(1);\n}\n\nfor (int i = 0; i < numrows; i++) {\ntracetotal+=matrix[i][i];\n}\n\nreturn tracetotal;\n}
211287
here you find a free download from a book (from the American Mathematical Society):\nhttp://www.ams.org/online_bks/coll17/\n\nThe wikipedia is also a good source\nhttp://en.wikipedia.org/wiki/Matrix_(mathematics)\n\nAnd here's an on-line calculator for matrix operations (it seems cool... I didn't try it)\nhttp://www.quickmath.com/www02/pages/modules/matrices/index.shtml\n\nAnd... mathworld\nhttp://mathworld.wolfram.com/Matrix.html\n\n\nGood work!\n(I'll download the online book myself: I didn't gave a look at it before giving you the link; the book is a little old (1937) but matrices haven't changed so much from that time on...)
78119
I guess you must factorize them.\nAll the polynomials you wrote are to be factorized by partial factorization.\nI think you know the general rule.\nA(x)P(x)+A(x)Q(x) + B(x)P(x) + B(x)Q(x) = \n---> factorize between the 1°-2° and 3°-4° <---\nA(x)[P(x) + Q(x)] + B(x)[P(x) + Q(x)]\nYou get the same binomial term [P(x) + Q(x)] which is a common factor. Factorize:\n[P(x) + Q(x)]*[A(x)+B(x)]\n\nFor example, in the first:\nA()=a\nB()=b\nP(x)=2x\nQ(y)=3y\nYou get \na(2x+3y)+b(2x+3y) = (2x+3y)(a+b)\n\nIn all the others you must rearrange the terms to use the above steps:\nin xy+32+8x+4y the repeating patterns are 8x + xy and 32 + 4y ....\n\nThink about it and good work