What is it in math?\nAnd please simple language. I did a search and found complicated explanaitions for a matrices.
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A MATRIX is a set of numbers written in brackets.  It has dimensions which are read row by column.\nEXAMPLE:  [1   0]  Its dimensions are 1 by 2 because it has 1 row and 2 columns.\nYou can add matrices that have the same dimensions together, and subtract in the same way.\nEXAMPLE:  [3   6]   +  [2   7]  =  [5   13]\nYou can also multiply them together, but that's a little bit complicated to explain on a computer...you can't divide them.\nI don't really know how else to explain it without going into other complicated subjects.
414161
The infinity norm is just the maximum absolute row sum - ie take each row, add together the absolute value of each element. Take the maximum over all your sums.\n\nThe reason its important(ish) is because it is one of the 'extreme' norms. Every matrix norm is "between" the 1-norm (which is the same thing, but for columns) and the infinity-norm.\nFor example, the square of the 2-norm is <= the product of the 1 norm and infinity-norm.
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I know what you are talking about.\n\nIts in Linear Algebra.\n\nThis system can be expressed quite compactly by writing Ax=b, where A is an m x n  matrix whose elements are the coefficients, aij, in the linear system. We use A to denote both the linear map and the matrix; note that A is linear map from R^n to R^m . The column vector of unknowns is x which belongs to R^n and the right-side column vector is b which belongs to R^n . When b=0, the system is said to be homogeneous.\n\nFirst consider the homogeneous system Ax=0. The solution set of this system is the null space of A, denoted null A. The question is whether or not the solution set includes more than the zero vector. We have seen that if A is not injective (one-to-one), then the null space contains more than {0} . Furthermore, we have seen that if m<n, then A cannot be injective. Putting these observations together, we see that a homogeneous system with more unknowns than equations (n>m) must have nonzero solutions. This makes sense: there are not enough equations (constraints) to determine a solution uniquely.\n\nOops you asked about "nonhomogenus solutions". We will get there now. :)\n\nNow consider the nonhomogeneous system Ax=b. In order for a solution to exist, b must be a linear combination of the columns of A (with weights given by the components of x). Using familiar terminology, this means that b must lie in the range of A = R^m. Introducing new terminology that is used with matrices, this means that b must lie in the column space of A, which is the span of the columns of A. If , then there is at least one solution for all choices of b belong to R^m. This means that if A is surjective (onto), then there are solutions for all . Conversely, if n<m, then A is not surjective and no solution exists for some choices of b. We conclude that for a nonhomogeneous system with more equations than unknowns (m>n), no solution exists for some choices of b. This also makes sense: it is possible that there are too many equations (constraints) to determine a solution. \n\nWhat about m=n?\nA special case arises when m=n. In this case, if the columns of A are linearly independent, they span R^m and all b belong to R^m  lie in the column space of A (including b=0). Thus, all vectors b admit a solution, x, whose components give the weights in expressing b as a linear combination of the columns of A. Because this representation is unique, the system Ax=b has a unique solution for all b belong to R^m , in the case that m=n and the columns of A are linearly independent. \n\nThus, \n\nAx=b has solutions for ALL b. It does NOT depend on the subspace of R^n.\n\nRegards
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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.
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One basic idea for analyzing a mathematical object is to look at morphisms into 'simple' objects of that type.\n\nFor example, a group G operating on a set S is a homomorphism of G into the symmetric group Sym(S). We regard the symmetric groups as 'simple' objects that we can use to understand other groups.\n\nA group representation is a homomorphism of the group G into the group gl(V) of invertible linear transformations on some vector space V. The good thing about this is that you get all of linear algebra to help you understand the group gl(V) and its subgroups, and hence to understand the group G. The homomorphism is called a representation of G.\n\nAmong the questions that can be asked:\n1. Are there any invariant subspaces of V? (an invariant subspace is one that is preserved by every linear transformation in the image of G).\n2. Can the vector space V be written as a direct sum of invariant subspaces in some nice way?\n3. Are there enough representations to distinguish all elements of the group?\n4. What happens if you look at representations in vector spaces over different base fields?\n\nIf your group is a Lie group (a differentiable manifold that is also a group), the you can look at smooth homomorphisms. It turns out the there is a vast amount of geometry that appears. This geometry is closely related to problems in particle physics.\n\nThis is just the beginning of the study of representations. There is much,much more. Yes, it is fun!
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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.
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Google the word algebra help.  Purple math is great, but there are other sites also.  I am sorry I do not have the names at the moment.
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Everything that has a beginning has an end. - The Matrix.\n\nThe only questions are when and how.  The rest is just details.
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A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
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http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3206\n\nhttp://mathforum.org/library/drmath/view/51899.html\n\nhttp://www-staff.mcs.uts.edu.au/~don/pubs/solving.html\n\nhttp://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=3206
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If R := I - BA (:= means "equals by def.") then, by rearranging (BA = I - R, so A-1 B-1 = (I - R)-1), we get \n\nA-1 = (I - R)-1 B\n\nand the matrix (I - R)-1 can be expanded in a geometric series (just as, for numbers, 1/(1 - r) = I + r + r^2 +...), that is, \n\n(I - R)-1 = I + R + R^2 + ...\n\nbecause B approximates A-1, so R is close to 0. (More precisely, the expansion holds if each eigenvalue of R is less than 1 in absolute value.) Putting the two together, and now writing the inverse properly as A^(-1) instead of A-1 (which might be mistaken for A - I), we get \n\nA^(-1) = (I + R + R^2 + R^3 + ...)B\n\nas was required. This formula is exact, so this is not actually an iterative process, by the way (an iterative process would consist in getting successive approximations to A-1 by repeating some transformation again and again).\nPls award Best Answer :))
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There's something missing in that hint, there must still be an 'n' in there somewhere.\n\nTo work out the general formula:\n\nThe number of 2*2 matrices is (n-1)(m-1). The number of 2*3 is (n-1)(m-2). The number of 2*4 is (n-1)(m-3)... and so on until you reach (n-1)(1) = (n-1)(m-(m-1))\n\nSo the total number of 2 * j matrices is\nSigma(j = 1) { (n-1)(m-j) }\n\nThen the number of 3 * j matrices is\nSigma(j = 1) { (n-2)(m-j) }\n\nAnd so on, up to\nSigma(j = 1) { (1)(m-j) }\n\nThe sum of all these gives the number of i * j matrices as:\n\nSigma(i=1 to n-1)Sigma(j=1 to m-1) {(n-i)(m-j)}
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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)).
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linear operator takes a vector and maps it into another vector while preserving addition and multiplication by a scalar:\n\nL(ax+y)=aL(x)+L(y)\nx,y are vectors, a is a scalar, L() is operator\n\nIt can be defined by a matrix, i.e. result of operator is given by multiplying that matrix by a vector.\n\ngo look it up on wikipedia.org or mathworld.wolfram.com
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Say a[i,j] are the elements of A. The definition of the determinant is:\n\ndetA = Sum over P: (-1)^P * a[1,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nwhere the sum is over all permutations P of (1,2,...,n) and (-1)^P is -1,+1 for odd and even permutations respectively. (P(i) is the image of i under P).\n\nNow let's say (for example) that matrix B is created multiplying the second row by a constant c and adding it to the first row:\n\na[1,i] --> a[1,i]+c*a[2,i]\n\nThen the determinant of this new matrix will be:\n\ndetB = Sum over P:\n(-1)^P * (a[1,P(1)]+c*a[2,P(1)]) * a[2,P(2)]*...* a[n,P(n)]\n\n= det A +\nc * Sum over P: (-1)^P * a[2,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nbut the last sum is zero, because for each permutation P there is exactly one other permutation differing only by a transposition T of (1,2)\n\nP' = PT , P = P'T\n\nand P and P' have opposite parity so the two terms generated by each of these n!/2 pairs cancel each other --> detB = detA
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1. Write down a couple of Jordan blocks.\n\n2. Write down their squares, cubes, maybe even a fourth power.\n\n3. Gaze at each set of identity matrix, J, J^2, J^3, ... until you see a simple linear combination of the matrices that sums to the all zero matrix.  That is your minimal polynomial.\n\n4. Generalize.\n\n5. Now it is obvious. :-)\n\nAlternative:\n\n1.  Google for some combination of:\n\nminimal polynomial Jordan block\n"minimal polynomial" "Jordan block"\nJordan block minimal polynomial\n"Jordan block" "minimal polynomial"\n\n2. Read answer.\n\n3. Too bad, it still isn't obvious. :-(
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The size of a matrix multiplication is as follows:\n\n(nxm) (mxp)= (nxp)\n\n(1x2) (1x1) = 1x1\n\n(3x1) (1x3) = 3x3