Show that the solution vectors of a consistent nonhomogenous system of m linear equations in n unknowns do not form a subspace of R^n
137873
I know what you are talking about.\n\nIts in Linear Algebra.\n\nThis system can be expressed quite compactly by writing Ax=b, where A is an m x n  matrix whose elements are the coefficients, aij, in the linear system. We use A to denote both the linear map and the matrix; note that A is linear map from R^n to R^m . The column vector of unknowns is x which belongs to R^n and the right-side column vector is b which belongs to R^n . When b=0, the system is said to be homogeneous.\n\nFirst consider the homogeneous system Ax=0. The solution set of this system is the null space of A, denoted null A. The question is whether or not the solution set includes more than the zero vector. We have seen that if A is not injective (one-to-one), then the null space contains more than {0} . Furthermore, we have seen that if m<n, then A cannot be injective. Putting these observations together, we see that a homogeneous system with more unknowns than equations (n>m) must have nonzero solutions. This makes sense: there are not enough equations (constraints) to determine a solution uniquely.\n\nOops you asked about "nonhomogenus solutions". We will get there now. :)\n\nNow consider the nonhomogeneous system Ax=b. In order for a solution to exist, b must be a linear combination of the columns of A (with weights given by the components of x). Using familiar terminology, this means that b must lie in the range of A = R^m. Introducing new terminology that is used with matrices, this means that b must lie in the column space of A, which is the span of the columns of A. If , then there is at least one solution for all choices of b belong to R^m. This means that if A is surjective (onto), then there are solutions for all . Conversely, if n<m, then A is not surjective and no solution exists for some choices of b. We conclude that for a nonhomogeneous system with more equations than unknowns (m>n), no solution exists for some choices of b. This also makes sense: it is possible that there are too many equations (constraints) to determine a solution. \n\nWhat about m=n?\nA special case arises when m=n. In this case, if the columns of A are linearly independent, they span R^m and all b belong to R^m  lie in the column space of A (including b=0). Thus, all vectors b admit a solution, x, whose components give the weights in expressing b as a linear combination of the columns of A. Because this representation is unique, the system Ax=b has a unique solution for all b belong to R^m , in the case that m=n and the columns of A are linearly independent. \n\nThus, \n\nAx=b has solutions for ALL b. It does NOT depend on the subspace of R^n.\n\nRegards
244205
First of all, the characteristic equation is m^3-3m^2+3m-1=0, which factors as (m-1)^3=0, for m=1 a triply repreated root.\n\nThis gives e^x, xe^x, and x^2 e^x as three independent solutions for the *homogeneous* equation (the one with a 0 on the right hand side). Now you have to find a *particular* solution of the non-homogeneous equation. The problem is that e^x is already a solution of the homogeneous version. So now you go up in the heierarchy to try a particular solution of the form\ny_p=Ax^3 e^x.\nPlug this into the original equation to solve for A. Then add this particular solution to the general solution of the homogeneous part to get your general solution. Check to be sure there are three arbitrary constants!
259743
I LOVE U\n\nFirst line:\n= I [ (L+N)² - N(2L+N) ] O / L\n= I [ L² + 2LN + N² ) - ( 2LN + N² ) ] O / L\n= I [ L² ] O / L\n= I L O\n\nSecond line:\n= [(v+r)(v-r)][E/V(U³/U²)]  (Note: that is U^3/U^2)\n= [v² - r²] [E/V(U)]\n\nNow we know that we "r" (are) nothing without each other, so r = 0\n\n= [V²] [E/V(U)]\n= V E U\n\nAll together our equation is I LOVE U
29345
these are data from end fixing of the structure you should introduce to solve the system of equations, otherwise you would get infinite solutions or (rigide body motion)-Good Luck
96897
http://kr.cs.ait.ac.th/~radok/math/mat7/step15.htm go here and you will find your Answer\nYou also can read -\n*Numerical Methods of Engineers\n -Steven C. Chapra\n -Raymond P. Canale
37342
A quotation from\nhttp://www.ams.org/featurecolumn/archive/sylvester3.html\n\n"In retrospect, it is clear that one could easily see that there had to be at least three ordinary lines for any point-line configuration. This follows from Melchior's work. Later it was shown that the number of ordinary points had to grow larger as the number of points in a plane finite configuration grew. Both G. Dirac and T. Motzkin were among those who explicitly conjectured that there would be at least n/2 ordinary points for a set of n plane points. However, it should again be noted that for n =7 and 13, there are configurations for which this bound does not hold. The additional requirement that n be large enough has to be added. It was announced by Hansen that he had a proof that there were always at least n/2 (n > 13), but the proof, unfortunately, had an error. Despite this, there were many correct new results in Hansen's investigations related to Sylvester-Gallai results. The number of ordinary points is known to be at least 3n/7 (n > 2), as was shown by L. Kelly and W. Moser, and the best known result (due to J. Csima and E. Sawyer) is that the number of ordinary lines is at least 6n/13 (n > 6). "
458120
false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
76377
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
45758
#include <iostream>\n#include <vector>\nusing namespace std;\n#define rep(i, n) for((i) = 0; (i) < (n); (i)++)\n#define rep2(i, k, n) for((i) = k; (i) < (n); (i)++)\nvoid f21();\nint n;\ndouble mat[100][100];\nint main()\n{\nwhile(true)\n{\ncin >> n;\nif(!n) break;\nmemset(mat, 0, sizeof(mat));\nf21();\n}\nreturn 0;\n}\n\nvoid f21()\n{\ncout << "****************************************" << endl;\ncout << "*          INVERSE OF MATRIX           *" << endl;\ncout << "****************************************" << endl;\n\ncout << "ORDER OF THE MATRIX: " << n << endl;\ncout << "ENTER ELEMENT ROW-WISE" << endl;\nint m = 2*n;\nint i, j, k;\nrep(i, n) {\nrep(j, n) {\ncin >> mat[i][j];\ncout <<"&#92;t";\nprintf("%5.3lf", mat[i][j]);\n}\nrep2(j, n, m) {\nif(i == (j-n)){\nmat[i][j] = 1;\ncontinue;\n}\nmat[i][j] = 0;\n}\ncout << endl;\n}\n\ncout << endl << "INVERSE MATRIX: " << endl;\ndouble pivot, comm;\nrep(i, n) {\npivot = mat[i][i];\nif(mat[i][i] == 0){\ncout << "DIVISION BY ZERO." <<endl;\nreturn;\n}\nrep(j, m) mat[i][j] = mat[i][j]/pivot;\nrep(j, n) {\nif(i == j) continue;\ncomm = mat[j][i];\nrep2(k, i, m) {\nmat[j][k] = mat[j][k] - (comm*mat[i][k]);\n}\n}\n}\n\nrep(i, n) {\nrep2(j, n, m) {\ncout <<"&#92;t";\nprintf("%6.4lf", mat[i][j]);\n}\ncout << endl;\n}\ncout << endl;\n}\n\n/*This is writen in C++ this will output the inverse of a Matrix\ninput:\n4\n1 2 3 4\n2 1 4 2\n2 5 7 12\n1 -1 3 11\n\n5\n1.173 -0.634 0 0 0\n-0.539 1.173 -0.634 0 0\n0 -0.539 1.173 -0.634 0\n0 0 -0.539 1.173 -0.634\n0 0 0 -0.539 1.173\n\n0\n*/
444022
it can be the original way you had it (c=mx +y) but it is alot simpler in the long run to rearrange it to (y=mx+c) (y being y, x being x, m being gradient, c being a constant). this way it is easyer to solve.
419963
I have gone over the problem twice.  This system of linear equations has NO SOLUTION!!!  This means that not even a single set of points (x,y,z) exists that can satisfy ALL three equations simultaneously.  Here is my work to prove it:\n\n(1) 4x+2y+3z=6\n(2) 3x+y+2z=8\n(3) x+y+z=5\n\n(1) 4x+2y+3z=6\n(2) 3x+y+2z=8 (multiply by -2, and then add equations to cancel y)\n\n4x+2y+3z=6\n-6x-2y-4z=-16\n----------------------> (4) -2x-z=-10\nrearrange this ---> (4) z=10-2x\n\n(2) 3x+y+2z=8\n(3) x+y+z=5 (multiply by -2, and then add equations to cancel y)\n\n3x+y+2z=8\n-2x-2y-2z= -10\n----------------------> (5) x-y=-2\nrearrange this ---> (5) y = 2+x\n\nPlug (4) and (5) into (1):\n\n4x+2(2+x)+3(10-2x)=6\n\n4x+4+2x+30-6x=6\n\n34 ≠ 6\n\nTherefore, there's NO SOLUTION!!!!!!
196260
why not try Danube's Enigma, it is easy and fulfills the requirements of all matlab projects!\n:)
420791
It is not known to be a fact.  However, it would be a special case of the Fermat-Catalan conjecture about integer solutions of x^p + y^q = z^r, and it has been proved that a similar special case to yours has no solutions.
396924
The same equation, in this case, cannot have two different answers. So there is no solution to this system (to understand it, try graphing it)
668921
both of your example equations contain infinite solutions
67168
The answer is:\n2y^3 - y^2\nOR\n(2y-1)*y^2
196194
Actually, you get X''/X=C and Y''/Y=-C where C does not have to be 0. Also, remember that sums of solutions are solutions, so the solutions you get from separation of variables only give a basis for the vector space of solutions, not all the solutions. The specifics are difficult in this forum, but you have to set up the series and use the boundary conditions to find the coefficients of the series.