How can matrices be used in your real life? and How are they applicable?
954567
Check out Markov Chain's.  You will find many uses for matrices.
1111281
AB is only defined if A has the same number of columns as B has rows.  So two matrices with different dimensions can be multiplied if that constraint holds.
471695
They are a set of Matricies that map one set of {x,y,z} to another set of {x',y',z'} through a rotation angle.  In oder to use them you need to know your starting set of coordiantes and the three angles through which you want to rotate.\n\nOne such set of angles that forms a perfectly good rotation basis are the Euler angles.\n\nIf you know the rotations, you can actually write the whole transformation as one matrix (usefull if you have to apply it over and over again.
102340
I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
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"
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)).
1027338
(Remark: the fact that this is supposed to work with any field K is a hint that our matrices - elements of GLn(K) - can't be too complicated!  In fact we can get away with using only 0s and 1s.)\n\nThe key is to think about the elements of G, call them g1, g2, ..., gn, also as permutations of the set { g1, g2, ..., gn } itself:  any element gi permutes { g1, g2, ..., gn } to { g1 gi, g2 gi, ..., gn gi } (verify that that really is a permutation of the set!).\n\nThis is the key to showing that every finite group of order n is isomorphic to a subgroup of Sn, the symmetric group of order n!.  But Sn itself is isomorphic to a subgroup of GLn(K), namely the subgroup of "permutation matrices", which have exactly one 1 in every row and column and 0s everywhere else.
245939
animations and simulations uses matrices extensively
693372
The field, be it finite or whatever, makes no difference. Use common methods of linear algebra you know. You should just remember to add, multiply, divide, etc in your field correctly.
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...)
564843
1) you can calculate distances and areas, which is very helpful in real life when buying ground\n\n2) you can calculate circles in triangles and rectangles, which is the twodimensional variant of dense packing of atoms in chemistry (crystals)\n\n3) you learn the angle functions, which you can use in physics for polar coordinats and optics and in astronomy for describing the positon of stars
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.)
912333
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]
77094
I am not sure what you mean by this. In C# you can multiply a vector by a matrix. If you consider a Point object a vector, there is a Matrix.TransformPoints function that will transform that point to a new point using the matrix. For example\n\nMatrix m = new Matrix(1, 0, 0, 1, 10, 10);\nPoint[] points = new Point[] {new Point(5, 5)};\nm.TransformPoints(points);\n\nThe result will be that the Point within the array should be 15, 15. Becasue that matrix translates by 10, 10. Note that the array is required because that is the only way to do a transformation on a Point.
818178
Think of a matrix as defining a linear transformation. Now imagine what happens to the unit sphere when that transformation is applied: it will get stretched and rotated into some type of ellipsoid. The norm of the matrix is the distance from the origin to the furthest point on that ellipsoid.
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
159293
Depends on what you are analyzing.  Matrix algebra is used extensively in communications and signal processing applications.  Not very much, though, in financial applications.