If you have this transition matrix ............. what would  the initial-state matrix be?
953148
For found an answer watch here:\nhttp://en.wikipedia.org/wiki/Transition_matrix
897524
Kinetic energy is .5 * mass * velocity ^ 2\nSo the change in kinetic energy, or delta K.E, would be:\n\n.5 * mass * initial-velocity ^ 2 - .5 * mass * final-velocity ^ 2\n\nSince Kinetic Energy is lost, the initial velocity goes first.\n\nGood luck on your test!
333863
Diagonalisation is the answer here.  Not sure how this didn't turn up in a web search, as I found plenty of resources for it, but the idea is that you find a matrix that diagonalises your initial matrix, and this allows powers to be done trivially.\n\nAssuming you're aiming to get a matrix A to the power n:\n\nFind P such that:\nP^-1 A P = D, where D is diagonal.  Then:\nA = (P D P^-1) and:\nA^n = (P D P^-1)(P D P^-1)(P D P^-1)....\n\nThe P^-1 P terms cancel and you're left with:\nA^n = P D^n P^-1\n\n... and this is much easier to calculate as D^n is trivial :)\n\nA good reference is in the sources.  Hope that helps!
414316
Hi,\nThe answer is C.\n\nYou could make a ray-diagram and see for yourself...
136546
here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
268402
Consider the matrix of the real parts of a Hermitian matrix as A and the matrix of the imaginary parts as B. Use the definition of Hermitian.\nAnother hint: If a_ij = a+ib, what would a_ji have to be if the matrix A is Hermitian ? What is then true about the diagonal of a Hermitian matrix?
1047731
Assume dc/dz is constant, n(t=T)=-DA(dc/dz) T\nBut in reality dc/dz often changes.  Normally dc/dz should decrease with time.  If it is a close system, it will reach the equilibrium state.
414270
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
251771
Suppose P(k) is the probability that k of the N pots are occupied. Considering the process to begin on the second throw, there are N possible states (k=1,2,3...N). The initial probability distribution (after the first rock is thrown) is:\n\nP(1)=1, P(2)=P(3)=...=P(N)=0\n\nEach time a rock is thrown at random, the probability distribution undergoes a constant linear transformation (this is a Markov process):\n\nP --> P'(k) = (k/N)*P(k) + ((k-1)/N)*P(k-1)\n\nTo answer the question it's necessary to calculate P(N) after r transitions, equate it to .95 and solve for r. That can be done in principle for various specific values of N. For example for N=2, it's not to hard to see that:\n\nP(1) = (1/2)^r\nP(2) = 1 - (1/2)^r\n\nSolving (1/2)^r = 1-.95 = .05 gives about r = 4 (i.e. 5 actual rocks).\n\nI can also solve the case N=3 exactly. In this case, the transition P -> P' caused by each throw is: \n\nP'(1) = (1/3)P(1)\nP'(2) = (2/3)P(1) + (2/3)P(2)\nP'(3) = P(3) + (1/3)P(2)\n\nThis has solution (P(k) after r transitions):\n\nP(1) = (1/3)^r\nP(2) = (2^(r+1) - 2)/3^r\nP(3) = 1-P(1)-P(2)\n\nSo the exact answer in this case is the solution of:\n\nP(1) + P(2) = (2^(r+1)-1)/3^r = (2^(r+1)-1)/3^r = .05\n\n==> r is between 7 and 8 (i.e. between 8 and 9 actual throws).\n\nIf there's an expression for P(N) for general values of N, it's probably relatively complicated. \n\nHowever for large N it might be a pretty good approximation to treat the occupation states of each pot as independent (even though they're not really). Taking r as the actual number of rocks thrown, the probability that any given pot is occupied is exactly:\n\nP = 1 - ((N-1)/N)^r\n\nIf these events were independent, the probability that all pots are occupied would be:\n\nP^N = [1-((N-1)/N)^r]^N = .95\n\n==>\n\nr = ln(1- .95^(1/N))/(ln(N-1)-ln(N))\n\nFor N=3 this gives r = 10 which is a bit larger than the exact value, but it's probably more accurate for larger N.
346932
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.
438919
I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
330850
S A D + M A D \n A D = A D\n     = MS
201316
This website may help you.  Good luck.\n\nhttp://www.sosmath.com/matrix/matrix.html
906577
Hi!\n(A) \nBy definition di=dm * X^2 , where di is the amount added to I when the mass dm is added to the system. I~Inertia . In this case for the particle X=d.\n\nEach mass m=0.85 kg add to the system (delta I)1 = m * d^2.\n\nEach rod has an Inertia of (delta I)2 = (1/3) M * d^2 . This is because for a rod you have to calculate:\n             S { dm * x^2 = (M/d)*dx * x^2 } ==> (M/d) * (1/3)(X^3) with X=d. => (1/3)M * d^2  \n\nso the total inertia of the system is I= 2*[ m*d^2+ (1/3) M*d^2] because we have 2 rods and tho particles.:\n    I= 7.84 * 10^(-3) kg / m^2.    Where ^ is used for power!\n\n(B) kinetic energy= (1/2) Inertia  * (angularspeed)^2 =3.528 /10^4\n\nIf you want to know more try the link below!
1058187
for a rigid rotor, the transition lines are equally spaced in the wavenumber space. However, this is not always the case, except for the rigid rotor model. \n\nfor non-rigid rotor model, we need to consider changes in the moment of inertia of the molecule due to centrifugal distortion and the effects of vibration on the rotation.\n\nthe formulas are hard to write here, better to look at the references in \nhttp://en.wikipedia.org/wiki/Rotational_spectroscopy\nor elsewhere.
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