Matrices (Transition Matrices and limits)?
407787
You should get steady-state probabilities of 45%, 35%, and 20% for states 1, 2, and 3 respectively.\n\nYou need to solve the system:\n0.8x + 0.2y + 0.1z = x\n0.1x + 0.7y + 0.3z = y\n0.1x + 0.1y + 0.6z = z\n\nOr equivalently in standard form:\n\n-0.2x + 0.2y + 0.1z = 0\n 0.1x - 0.3y + 0.3z = 0\n 0.1x + 0.1y - 0.4z = 0\n\nYou can see that one of the equations is redundant (the first equation is the opposite of the sum of the other two equations, for example) so you can get rid of one. It doesn't matter which. This will always be the case. You can also pick an arbitrary value for any of the variables if you think it will help you out. In this case I'll let z = 10 and drop the first equation. This leaves us with:\n\n0.1x - 0.3y = -3\n0.1x + 0.1y = 4\n\nTake the difference to get 0.4y = 7, y = 17.5, x = 22.5.\n\nBut now x + y + z = 50 and it's supposed to equal 1. This is no problem, as we can just divide everything by 50 to get: x = 0.45, y = 0.35, and z = 0.2.\n\nIf it seems like I pulled a lot of things out of nowhere here, take a look at some theory regarding the homogeneous system Ax = 0 when A is singular, particularly the part where it mentions that if you've got a solution x to the homogenous system, then any scalar multiple of x is also a solution (this explains why we can just divide everything by 50 at the end and still have a solution to the original system). Consult a linear algebra book for more details.
930026
wellz why not just help me with that matrix problem? :oP
596773
Matrix means rows and columns. There lots of problems based on Matrix in mathematics like Determinants. \n\nThe movie matrix is sci-fi movie. Its not real.
816270
is this for a algebra class? calculs class? matrix methods?\n\nTry this.\n\nWe want to simplify the complicated list of 3 constraints to two simple constraints: a range for y and a range for x.\n\nThe lower bound is simple: constraint III says all x and y are greater than or equal to zero.\n\nFind a range for y. Algebra can be done on these eqations the same as with an equal sign as long as the alligator is in the correct direction (it is). So eliminate x from the equations by using, "Constraint II minus Constraint I"; or;\n\n  x+3y<_6\n-(x+ y<_2)\n\n=2y<_4....... or, y<_2. Therefore, 0<_y<_2.\n\nSimliarly for range x, eliminate y by "Three times Constraint I plus Constraint II.", or:\n\n3*(x-y<_2) =  3x- 3y<_6\n                    +( x+3y<_6)\n                    = 4x<_12.... or, x<_3. Therefore, 0<_x<_3.\n\nNow we simply have an equation of three variables with two constraints.\n\nz=3x+2y;  0<_y<_2\n                 0<_x<_3\n\nI'm not sure what you want from here, but I think I understand.\n\nThe graph will be a plane in the x,y,z coordinate system.\n\nThe maximum z is achieved by using maximum values of x and y.\n\nThe corners will be defined by the extreme values of x and y with the correseponding z.\n\nFor instance, with (x,y,z)...\n\n(0,2,z1)\n(3,0,z2)\n(0,0,z3)\n(3,2,z4)\n\nThis will create a "polygon...?" in the x,y,z plane with the corners at the locations shown above.\n\nThis might be able to be done more simply by using matrices but I can't see an obvious way.
584514
Use 'Gaussian elimination' to obtain a basis for the span of matrix columns or rows.\n\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
113822
A 100 by 100 matrix has 100 eigenvectors. I would use some software to do this. This is too hard to do by hand. Buy the student version of MATLAB for example. You might be able to find some free software on the internet as well.
114568
I don't know if this will be specific enough for you...but\nif x is within range of P (a matrix) ...when multiplying P by a constant, it gets distributed....hence multiplying x term and its corresponding solution.  Therefore, if c>=x, then cP-xx>=0  [If c=x, it will be =0, if c>x, it will be positive.\n\nMatrix P: {1x + 0y = a      cP [ cx    + c(0)y = ca\n           0x + 1y = b]          c(0)x + cy    = cb(note that I have already worked matrix P down to solution form of\n  [1x + 0y = a\n   0x + 1y = b]\nIf x>=0, and c>=0, then cx>=0 and xx>=0.  Hence, if c>=x, then cx>=xx, so cP-xx>=0\n\nIf you didn't mean x as a variable, but instead another matrix...disregard this solution
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.
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.
1087743
Matrix - the no. of rows and columns are not necessarily equal.\nDeterminants - they r equal.\nMatrix - elements are enclosed in (), { }, or [ ]\nDeterminants - elements r shown between two vertical lines. I I\nMatrix - it cant hav a value. its just an arrangement.\nDeterminant - has a value.\nMatrix - if all elements r multiplied by a constant the matrix is multiplied to the constant.\nDeterminant - if each of any row or column is multiplied by a constant, the determinant is mulitplied to the constant.\nPhew * Need some lemonade...
554827
Well, I suppose you want a program that computes the trace of a matrix. The trace is simply the sum of the elements in the diagonol in a matrix. I won't show it to you with OOP since there's no reason to.\n\nint trace(int matrix[][],int numrows,int numcols) {\n\nint tracetotal = 0;\n\nif (numcols != numrows) {\nstd::cout<<"Error: Can't compute trace. Not square matrix.&#92;n";\nexit(1);\n}\n\nfor (int i = 0; i < numrows; i++) {\ntracetotal+=matrix[i][i];\n}\n\nreturn tracetotal;\n}
350177
Sometimes adding some indents will help figure out what's going on.  In your algorithm, you have a lot of nested if and while statements, so space them out like this first to see how each is related to each other (I'm using periods because Yahoo won't show spaces):\n\ni <- 0\nwhile (i < n) do\n...j <- i+1\n...m <- a[i]\n...while (j < n) do\n......if (m > a[j]) then\n.........m <- a[j]\n......end if\n......j <- j + 1\n...end while\n...b[i] <- m\n...i <- i + 1\nend while\n\nYour i is a counter that starts with 0, and your j is a counter that starts with 1.  j will always be more than i.\n\nThe "while (i < n) do" line limits this algorithm to 3 loops, since n is 3.\n\nDirectly after this statement, j is established as a counter, and m is defined to be the i-th element of your matrix.  The next while statement then compares m (the i-th element in the matrix) to the element after this (using j to read the next element).\n\nIf m is larger, then the if statement makes m take on the smaller value.  If m is smaller, nothing happens and j is indexed to the next value.  Then the process repeats itself.  If we get to the end of the matrix and m isn't replaced anymore, it becomes the i-th value in matrix B.  After this, i is indexed to the next element in matrix A and the entire process is repeated once again, until all of matrix A has been checked.\n\nThe result then is that through m, matrix A is sorted least to greatest and is written into matrix B, which is answer (d).
458991
No, but if you go to List mode you can enter your matrix as a list and manipulate it.  Check out the owner's manual for help on how to do it.  I've included a link to it.  Hope this helps.
910755
In mathematics, a matrix (plural matrices) is a rectangular table of numbers or, more generally, a table consisting of abstract quantities that can be added and multiplied. Matrices are used to describe linear equations, keep track of the coefficients of linear transformations and to record data that depend on two parameters. Matrices can be added, multiplied, and decomposed in various ways, making them a key concept in linear algebra and matrix theory.
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
1084623
| . x . x |\n| -x . -x |\n\nwhere x is any real number and x ≠ 0 (The last part is only needed so that A will be non-zero). The transpose also works.\n\nMultiply it out yourself. The result has nothing to do with triangular matrices, so I also suggest that you do not merely cite special characteristics of such matrices in your answer.
696922
Let's say it's like this:\n[1 2 3]\n[4 5 6]   } Pretend the whole thing is surrounded by two big\n[7 8 9]     brackets\n\n[9 8 7]\n[6 5 4] And that the top is multiplied by this\n[3 2 1]\n\n[1 2 3]------[9 8 7]   (I'm using---as spaces)\n[4 5 6]--X--[6 5 4] \n[7 8 9]------[3 2 1]\n Multiply the first number in row one of the 1st matrix by the first column of the 2nd matrix ( 9 6 3 ). Then multiply the second number in row one by the second column ( 8 5 2 ) and then 3rd number by the 3rd column ( 7 4 1 ). When you're done, you should have the first row of a newer giant matrix set up...\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n\nNow, take the 2nd row of of the first matrix and multiply it the same way...(now you have the first and second rows I showed you)\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n[4(9) + 5(6) + 6(3)---4(8) + 5(5) + 6(2)---4(7) + 5(4) + 6(1)]\n\nDo you notice a pattern? The column numbers that are multiplied are lined up right on top of each other... this will save you time in the future. The row numbers change, but you know they are always in the same order.\nNow, multiply the last row of the first matrix by the columns of the second matrix...(here is how the final matrix looks)\n\n[1(9) + 2(6) + 3(3)---1(8) + 2(5) + 3(2)---1(7) + 2(4) + 3(1)]\n[4(9) + 5(6) + 6(3)---4(8) + 5(5) + 6(2)---4(7) + 5(4) + 6(1)]\n[7(9) + 8(6) + 9(3)---7(8) + 8(5) + 9(2)---7(7) + 8(4) + 9(1)]\n\nNow, multiply the numbers...\n[9 + 12 + 9---8 + 10 + 6---7 + 8 + 3]\n[36 + 30 + 18---24 + 25 + 12---28 + 20 + 6]\n[63 + 48 + 27---56 + 40 + 18---49 + 32 + 9]\n\nAdd up the numbers to get the final resulting matrix...\n[30  24  18]\n[74  61  54]\n[111 114  90] \nP.S. Sorry I didn't use your numbers... this is the answer to the matrix I made up, but at least it shows you how to do it right? Hope this helps!!!