I'm using the Markov Chain to solve a 3x3 equation..\nafter solving the transition matrix Ax=x how would I apply the normalizing contraint x+y+z=1?\nsay vector x=(x,y,z) would I solve for v/||v||=1? where ||v||=sqrt(x^2+y^2+z^2)
134028
What did you get for vector x=(x,y,z)?\nI would would add them all together\nx + y + z\nYou would have some positive number n which most likely is not 1.  \nDivide x, y, and z by this number\nExample:\nIf vector x = (2,3,5), then\n2 + 3 + 5 = 8\nso the new vector would be\n(2/8, 3/8, 5/8)\n(1/4, 3/8, 5/8)\nThe numbers in the vector have to add to one.
469279
i2=i1R1/R2\ni1=R2i2/R1\n\nR1/R2=i2/i1\n\nR1/Rx = R2/R3\n(R1/R2)*R3=Rx\nRx=(i2/i1)*R3
500579
The trick here is to find three equations where three of the variables are expressed functions of another variable. \nSo for example we could express x, y and w functions of z\nIf you add the first equation to the third and substract the second, you'll find: y= -3-z. We'll call this new equation A\nIf you add the second and the fourth equation and replace y by its value found in (A), you'll have: w= 7/4 + 1/4 z, which we'll call (B)\nNow take the first equation and replace w and y by their value from (A) and (B): \nx= -17/2 -7/2 z, this is the equation (C).\nNow that you have your three equations A,B and C you can substitute them in one of the four initial equations, for example the third one. You'll find:\nz=1\ny= -4\nw=2\nx= -12\nThis is not the entire solution (you have to detail how to get to A, B and C) but you should be able to do it (I don't want to be the only one working here!)
577956
It does work.\n(3x+1)ln2=(x-2)ln3 \nDistribute ln2 on the left side and ln3 on the right side.\n3x*ln2 + ln2 = x*ln3 - 2ln3\nSubtract x*ln3 from each side\n3x*ln2 - x*ln3 + ln2 = -2ln3\nSubtract ln2 from each side\n3x*ln2 - x*ln3 = -2ln3 - ln2\nFactor an x from the left side\nx(3ln2 - ln3) = -2ln3 - ln2\nDivide both sides by 3ln2 - ln3\nx = (-2ln3 -ln2)/(3ln2 -ln3)\nx = -2.9469
660465
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.
285206
At first I thought you were wanting the orthogonal matrices, but your example shows you are not. Orthogonal matrices are those whose inverse is their transpose. Yours are ones where the elements of the transpose are the inverse of the elements of the original.\n\nCertainly such a matrix has to have +-1 on the diagonal. Your matrix is singluar and any 2x2 matrix with your property and all 1 on the diagonal will be also. This means that one eigenvalue is 0 and the other is 2.\n\nI don't see anything special for 3x3 offhand, but I will look further.
433746
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
29309
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.
554392
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
294281
I assume these are vectors.\nv-w = <3-(-5),4-12> = <8,-8>\nu*(v-w) = <-2,1>*<8,-8> = -2(8) + 1(-8)\n= -16 - 8\n= -24
444344
The pre-image of w will be the set of vectors v so that T(v)=w. In other words, you are solving \n\nA*[x;y;z]=[6 4].\n\nThis gives the system of equations\nx+3y+z=6\n2x=4.\n\nClearly, x=2 and 3y+z=4, so the pre-image of w is\n{ (2, y, 4-3y): y is real}
340598
When an eigenvalue is repeated there are a couple of scenarios. Suppose the eigenvalue is r and is repreated k times. In the first scenario, there are k independent eigenvectors and things proceed as usual. In the second, your solutions will not simply be of the form\nA*e^(rt). You will have to consider solutions of the form A*te^(rt), B*t^2 e^(rt), etc. Until you get k independent solutions. I suspect that you are in the second scenario and will have to look at solutions of the form Ate^(6t) where A is a vertical matrix.
340077
The eqaution of the line is\ny-y1 =m(x-x1)\ny-0 = -3[x-(-4)]\ny-0 = -3(x+4)\ny= -3x-12
78169
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. :-(
382205
The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
333049
No solution.\n\nDo (equation 3) minus 2*(equation 1)...\nyou get\ny + z = 13\nwhich contradicts equation 2.  Either you have a mistake, or the answer is no solution.\n\nHope this helps,\nGod bless,\nShane
534029
in order to solve the system of equations you can do one of two things.  The easiest thing to do is to set the two equations to be equal. i.e. \n\nsince y=2x and y = -3x+5, then 2x=-3x+5.  In solving for x you will find that 5x=5 or x=1.  Then you can plug your resulting x back into the equation to get y=2(1) =2.  The main thing you are trying to do is get one of your variables to be equal to some combination of your other variables.  For example, if you are trying to solve 3y+2x-1=5x+3 and 6y=2x+7.  The first thing you want to do is to get a single variable by itself on one side of an equation.\nLet's start with the first equation, 3y+2x-1=5x+3. You would try to group the y terms by themselves on one side of the equation, an group the x terms and constants on the other side of the equation.  In this example, in order to group the terms correctly, you would have 3y=3x+4.  Then solve for y by itself by dividing both sides by 3.  The result is y=x+4/3.  You take this result and substitute for y in the second equation, 6y=2x+7.   You will get 6(x+4/3)=2x+7.  Then you solve for x by itself.  6x+8=2x+7\n4x=-1;  x=-1/4.  Then you can substitute this result back into either of your original equations to produce your solution for this system.  6y=2(-1/4)+7 => 6y=13/2 =>y=13/12.  In order to check your answer, you can graph the two linear equations.  Where the two lines intersect is the solution to your system.\n\n\nIf you are looking for a more advanced way to solve a system of equations, then you may want to consider using an augmented matrix.  Rearrange your equations so that you have only variables on one side of your equations, and only constants on the other side of your equations, then sole the system of Ax=b.  The matrix A is the coefficients of your variables, and b is a vector of your constants.  x is a vector of your variables.  Augment the matrix of your coefficients with your constant vector and do Gaussian Elimination until the matrix is in reduced row echelon form.  This will provide the exact solution to the system of equations.