Implement the divide-and-conquer algorithm for finding an element K of nn matrix M in which the elements of e
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I would like to help, but having a complete question to answer would make it a bit easier.
865560
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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d=nt2/p\npd=2nt\np=2nt/d
1802
I haven't tried it, but check out MatPy (linked below).  Another possibility is Numerical Python (also linked below).
837785
hi\np=(qm)/n\ny=x+(b/m) or x+(bq/np)
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
1103228
You have to measure the approximate number or operations for each algorithm.  Say one is O(n) and another is O(2n).  \n\ncheck out some sites about algorithm complexity\n\nhttp://www.cs.ucd.ie/staff/jcarthy/home/complexity.html\nhttp://en.wikipedia.org/wiki/Big_O_notation
433211
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
717242
I apologize for the notation in advance.\n\nCholesky factorization assumes that a matrix is symmetric and positive definite.  Symmetric is easy to determine.  Since Cholesky is fairly efficient, it can be used to determine if in fact a symmetric matrix is positive definite.  \n\nThe gist of the algorithm is this:  let's assume that we can write a symmetric positive definite matrix A as\nA=U'*U (' means transpose), where U is upper triangular.  To think about this, let's write the matrix A in blocks:\n\nA=(a_11 A_1,A'_1 A_22)\n\nwhere a_11 is the 1,1 element of A, A_1 is the remainder of row 1 of A, and A_22 is the 2,2 cofactor of A.  Similarly write U as \n\nU=(u_11 U_1,0 U_22)\n\nThen U'*U=A gives us\n\nu_11*u_11=a_11\nu_11*U_1=A_1\nU'_1*U_1 + U'_22*U_22 = A_22\n\nThis doesn't look like it gets us that far, but notice that U'_22*U_22 is a symmetric positive definite matrix, to which the preceding steps of the algorithm can be applied.  That's the way Cholesky works; proceed as above until done.  \n\nQuick example:\nA=(1 2 3,2 8 8,3 8 11)\nThen u_11=sqrt(1)=1\nU_1=(1/u_11)*A_1=(1/1)*(2 3)=(2 3) (this is a row vector)\n\nU'_22*U_22 = A_22 - (U'_1*U_1)\n=(8 8,8 11)-(2 3)'*(2 3)=(4 2,2 2)\n\nRepeat with this new matrix:\nu_22=sqrt(4)=2\nU_2=(1/2)*(2)=(1) (writing as a row vector to emphasize similarity with above)\nU_'33*U_33 = A_33 - U'_2*U_2 = 2-1*1=1\n\nFinally, u_33=sqrt(1)=1 and the algorithm terminates.  So your matrix U is\n\nU=(1 2 3,0 2 1,0 0 1)\n\nIf at any point, your proposed value for u_jj is the square root of a negative number, the matrix you started with is not positive definite.
437960
why not try Danube's Enigma, it is easy and fulfills the requirements of all matlab projects!\n:)
1159782
you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.
21714
Divide and conquer was a successful military strategy long before it became an algorithm design paradigm. Generals observed that it was easier to defeat one army of 50,000 men, followed by another army of 50,000 men than it was to beat a single 100,000 man army. Thus the wise general would attack so as to divide the enemy army into two forces and then mop up one after the other.  \n\nTo use divide and conquer as an algorithm design technique, we must divide the problem into two smaller subproblems, solve each of them recursively, and then meld the two partial solutions into one solution to the full problem. Whenever the merging takes less time than solving the two subproblems, we get an efficient algorithm. Mergesort, discussed in Section gif, is the classic example of a divide-and-conquer algorithm. It takes only linear time to merge two sorted lists of n/2 elements each of which was obtained in tex2html_wrap_inline24820 time.  \n\nDivide and conquer is a design technique with many important algorithms to its credit, including mergesort, the fast Fourier transform, and Strassen's matrix multiplication algorithm. However, with the exception of binary search, I find it to be a difficult technique to apply in practice. Therefore, the examples below will illustrate binary search and its variants, which yield simple algorithms to solve a surprisingly wide range of problems. \nSee link for more info.
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In order to decode the message, we interpret each pair of encoded numbers as a vector. There are 10 vectors:\n\n<8, 21>, <- 15, - 10>, <- 13, - 13>, <5, 10>, <5, 25>, <5, 19>, <- 1, 6>, <20, 40>, <- 18, - 18>, <1, 16>\n\nSince _RON corresponds to the vectors <0, 18>, <15, 14>, we know that:\n\nM<0, 18> = <-18, -18>\nM<15, 14> = <1, 16>\n\nWe know that for any 2×2 matrix M, the first column of M is M<1, 0> and the second is M<0, 1>. Thus, we need to find M<1, 0>. Fortunately, M is a linear transformation, so if there are constants a and b such that a<0, 18> + b<15, 14> = <1, 0>, then:\n\nM<1, 0> = M(a<0, 18> + b<15, 14>) = aM<0, 18> + bM<15, 14> = a<-18, -18> + b<1, 16>\n\nTo find a and b, we equate x and y components of the vectors, giving the system of equations:\n\n15b=1\n18a+14b=0\n\nWhich has the solution:\n\na = -7/135, b = 1/15\n\nSo M<1, 0> = -7/135<-18, -18> + 1/15<1, 16> = <1, 2>\n\nNow to find M<0, 1>. First we solve:\n\n15b=0\n18a+14b=1\n\nThe solution is:\n\na=1/18, d=0\n\nSo <0, 1> = 1/18<0, 18>, and:\n\nM<0, 1> = 1/18<-18, -18> = <-1, -1>\n\nThus M is the matrix:\n\n[1, -1]\n[2, -1]\n\nTo decode the message, we multiply every vector by M^-1, which is:\n\n[-1, 1]\n[-2, 1]\n\nThe decoded vectors are:\n\n<13, 5>, <5, 20>, <0, 13>, <5, 0>, <20, 15>, <14, 9>, <7, 8>, <20, 0>, <0, 18>, <15, 14>\n\nWhich spells out the message:\n\nMEET_ME_TONIGHT__RON
187387
(n)(3n-1)(3n-2).......(n+1)
1062782
you should use matlab.
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I did check the above answerers answers, and they are wrong :P\na,b,c,d,e,f = 1,2,5,5,4,6.\nYou could go and solve this by the usual elimination methods, but for something this large its far better to use a calculator or online solver. Google for 'simultaneous equation solver' and you'll find  heaps.\n(Just to confirm above answers are wrong: for those numbers you get 4a + b + 5d + 2e + 3f = 69, not 57.)
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well, in case I don't answer sufficiently, I provided a link for Wikipedia to define the ranks of matrices.\n\nBasically, ranks are when the matrices are reduced and you want to know the number of independent rows and columns.\n\nSince A^2 = I, this confirms A is nxn, so we're off to a good start.\n\nThe product of each row i x column i = 1, and row i x column j = 0, when i <> j.\n\nI think rank(I+A) + rank(I-A) somewhere is the solution by comparing the number of independent rows left of I+A and noticing there's only so many independent rows left in I-A.\n\nOne might wonder what does rank(I+A) + rank(I-A) suggest about the equation... (I+A)x(I-A)\n\n(I+A)x(I-A) = matrix[I(i,j)+A(i,j)] x matrix[I(k,l)-A(k,l)]\n            = matrix[sigma(row(i)xcolum(l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-I(i,j)*A(j,l)+I(i,j)*A(j,l) -A(i,j)*A(j,l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-A(i,j)*A(j,l))]\n            = I^2-A^2\n            = I - I\n            = 0\n\nSorry I couldn't answer your question completely.