in linear algebra solving matrices: i thing that there is no effect to switching between columns, is it right?
153486
No , it will change the order of variales.\nThat is you will get the vlue of xp insteat of xq. If you change p-th column q-th column.
48764
I found a website where you can download a program which uses this method.\n\nHere is how it is presented:\n\nSolving a System of Linear Equations v 1.0\nThis program solves a system of linear equations by Montante's method which have the peculiarity that once the system is solve in the main diagonal is the value of the determinant of the system. It works only with integers.  \nSize: 1 Kb  Downloads: 1112 Updated: 2003-7-23
1117062
if ur just solving that equation...u dont need to switch the sign.....\nwatch (im not gonna have the equal sign on the bottom takes too long). \n\nx-1<-3\n(add one to both sides now)\nx<-2\n\nthers ur answer
963940
Surely,you can do it.It reduces your head-ache of doing big multiplications and any proess which reduces your work-load but gives the correct answer is vaid.To tell you frankly,in higher classes,that is the procedure we follow to solve a problem in a quicker way.
159422
Is this a trick question? For a given system of linear equations, there is one solution that will yield a particular answer. There are infinite possible systems that will yield that answer (x=-2). There are also multiple ways to arrive at the same  correct answer for a given system. I think we need more info to be of any real help...
919233
Yes, when a list is already sorted, bubblesort will pass through the list once, and find that it does not need to swap any elements. \n\nIt will also use considerably less time if the elements in the list are not too far from their sorted places.
601270
I can do it... but let me think a little on it.  Can I take the full 4 months?  :-)\n\nOkay, I think I have a solution... took me about 5 minutes I think:\n\nIf I move the dot from B1 -> B2 and C1 -> C2, then B and C will still have 4 dots, and row four will still have 4 dots, but now row two will have 4 dots.\n\nYou didn't say they couldn't share the same square!!!!\n------------\nI'm back\n\nYou didn't exactly state what a "row" was.  I would have called that a "line" but maybe that's what you meant.\n\nOkay, I have a couple revised solutions.  A "row" is now defined as 4 dots that are colinear (on the same line) but each point must be distinct.  This invalidates my previous answer.\n\nOne solution matches with what has already been stated, which is:\nA4 --> B5  and  D4 --> C5.  This makes two pairs of overlapping lines each with 4 dots.\n\nHowever, my new solution is to take C1 and place it halfway between columns B and C and halfway between rows 2 and 3.\n\nThen you have a line on row 4, a line on column B, and two diagonal lines!!!!\n\nOne dot moved, 4 lines created!!!!\n\nAnd if you *must* move two dots, then just move B1 --> B5 and you still have a valid answer.
216212
Yes, but you have to be more clear what you want in general.\n\nSuppose we are working in n-dimensional Euclidean space. Suppose also that you have a hyperplane going through the origin. If you take n-1 independent vectors in that hyperplane and do a determinant construction similar to that for the cross product, you will get a vector normal to your hyperplane. But you will need n-1 *independent* vectors and you will have to do a nxn determinant (with the top row consisting of the unit tangent vectors along the axes).\n\nIf you don't have a full hyperplane, you will be wanting a basis for the normal subspace. Other than the previously mentioned Graham-Schmidt process, I don't know a good way to do that. Hmmm.....
145647
In general it used in negetive sense.\nFor example you are supposed to implement Matrix multiplication( or built a matrix object).\nThere is two way to show that your implementation works,\n1. create some matrix in your code and show with that. to test for another value you have to change the code.\n2. Take matrix as input from user.\n\nFirst practice is known as hardcore progrmming.
252498
block matrices are those with equal number of rows and columns.\n\nex:\n\n1 1 1\n1 2 2 \n3 4 5\n\nthat's a 3x3 matrix.  same number of rows and columns.\n\nthey're the only matrices that can be multiplied. \n\nmatrices can never be divided by another matrix.
841292
When properly balanced, the coefficients read from left to right, 1,3,2,3. The balancing used by the other two people doesnt even balance correctly!!
741630
if the matrix is squared, and it has an inverse, then the answer is YES
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
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?
599681
It depends on exactly what you mean by a matrix. If you're talking about multi-dimensional or nested arrays in ActionScript, then here is a link to get you started...
241815
If this matrix is A, we want to solve an equation AX=B, to solve this equation we have to cancel A, so multiply both sides by an inverse, A^(-1) and obtain X=BA^(-1).\nIf the matrix in non invertible, we cannot cancel it as easily. At least I assume that was the cancellation you requested.
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.