Matrices: Find x,y,z, and w by the GAUSE- JORDAN method.?
926227
Well Aq = b. where A is the matrix of coefficients, q is [x; y; z; w], and b is the answer [-2; 2; -11; -3] So we are looking for the Rx = d, that corresponds to the original equation. The aumented matrix is \n     5     3     1     5    -2\n     3     2     2     4     2\n     2     4     3    -3   -11\n     4    -3    -2     2    -3\nThe first step is to create a 1 in 1,1 so we divide Row1 by 5.\nThen subtract 3xRow1 from row 2, 2xRow1 from Row3, 4xRow1 from Row4\nThis is the result\n         1    0.6     0.2    1       -0.4\n         0    0.2     1.4    1        3.2\n         0    2.8     2.6   -5    -10\n         0   -5.4    -2.8   -2      -1.4\nContinue down and to the right with row operations in the same vein until the matrix is in row echelon form then go up and to the left in order to create 0s in all the non-step positions and get the matrix in reduced row echelon form.\n\nThe reduced row echlon form of  the augmented matrix [A b] is \n     1     0     0     0    -2\n     0     1     0     0    -1\n     0     0     1     0     1\n     0     0     0     1     2\n(Heh look at the nice numbers. You got this out of a textbook didn't ya?)\nThis was of course gotten by the Gauss-Jordan Method.\nSo now it's just a matter of reading off the coefficients\n1q1 + 0q2 + 0q3 + 0q4 = -2\n0q1 + 1q2 + 0q3 + 0q4 = -1\n0q1 + 0q2 + 1q3 + 0q4 =  1\n0q1 + 0q2 + 0q3 + 1q4 =  2\n\nSo q = [-2; -1; 1; 2] and x = -2, y = -1, z = 1, w =2.\n\nIf you want all the row operations, well there are an awful lot of them and they're all simple and straightforward. It's simply a matter of carefully working through the algorithm, so you are better off working them out for yourself. You'll probably get them before I can type them all in.
943382
If l,w and y are separate variables then\n(lw - lwy)/lw=lw/lw - lwy/lwy = 1 - y\n\nIf lw and lwy are each a vaiable then\n(lw - lwy)/lw= lw/lw - lwy/lw = 1-lwy/lw\n\nIf lw and y are variables then\n(lw - lwy)/lw = lw/lw - lwy/lw = 1 -y
387063
X=[-(30y-433)]/1232\nY=[-(1232x-433)]/30
1126894
This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
913768
y = ax² + bx + c\n\nWhen x = 2 y = 30 so 4a + 2b + c = 30 ...... (1)\nWhen x = 3 y = 15 so 9a + 3b + c = 15 ...... (2)\nWhen x = 4 y = 6 so 16a + 4b + c = 6 ........ (3)\n\n(2) - (1) 5a + b = -15 ....... (4)\n(3) - (2) 7a + b = -9 ......... (5)\n(5) - (4) 2a = 6\na = 3 so b = - 30 and c = 78\n\nie y = 3x² - 30x + 78
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
677039
(1) { -2x-y-z=0\n(2) { -x-2y-z=0\n(3) { -x-y-2z=0\n\nLet's look at (2):\n-x-2y-z=0 => x=-2y-z\nNow at (3):\n-x-y-2z=0 => x=-y-2z\nNow we can see that\n-2y-z=-y-2z => *** y=z ***\n\nNow let's put these values into (1):\n-2x-y-z=0 => -2x-y-y=0 => -2x-2y=0 => 2x=-2y => *** x=-y ***\n\nFor now, we know that x=-y=-z\nLet's put all of what we know into (2), and express everything using 'x':\n-x-2y-z=0 => -x-(-2x)-(-x)=0 => -x+2x+x=0 => 2x=0 => *** x=0 ***\n\nThus, x=0 => y=-0=0 => z=0\nSo the only solution is (0,0,0).\n\nGood luck!
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.
481388
The only way you'll be able to get a solution to 3 linear equations with 2 unknowns is if there's a dependence relation on the rows or equivalently the columns.  Otherwise, there will be no solution.\n\nI haven't checked the determinant of your matrix but if there are no dependence relations on the rows or columns of the matrix that you gave, Gaussian Elimination will give you a single solution.  Alternatively, for small matrices like the one you have you can use Cramer's Rule more efficiently.\n\nEither way, your coefficient matrix seemes to represent 3 equations and 3 unknowns so I would proceed with Gaussian Elimination.
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1
496301
b. there are infantly many solutions.  \n\nx=3t + 6; y=-2t + 7; z=t\n\nfor all t that are in the Real Numbers.\n\nthis assumes your matrix is really \n\n1  0  -3 | 6\n0  1   2 | 7\n0  0   0 | 0.
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.
793917
1)x+y-z=0.........(eqn. 1)\n2x-y+z=3...........(eqn. 2)\n4x+2y-2z=-2       (eqn.3)\nMultiplying eqn.1 by by 2 and subtracting it from eqn.2,we get\n-3y+3z=3...........(eqn. 4)\nMultiplying eqn.2 by 2 and subtracting it from eqn.3,we get\n4y-3z=-4...........(eqn 5)\nAdding eqn, 4 and 5 we get,y=7\nPutting the value of y in eqn 4 .we get\n-21+3z=3 or,3z=24  or z=8\nputting the value of y and z in eqn.1,we get\nx+7-8=0 \n=>x-1=0 or x=1\nTherefore,x=1.y=7 and z=8  ans\n2)x+2y=9....(eqn 1)\n 2x-3y=4     (eqn.2)\nmutiplying eqn.1 by 2 and subtracting it from eqn 2,we get\n-7y=-14 or y=2\nPutting the value of y in eqn 1,we get.\nx+4=9 or x=5\nTherefore x=5 and y=2 ans.
323643
kate1976 has solved the problem using the traditional way, just using the concept of coordinate Geometry.  Usin the cross product of two vectors, the method is different.\n\nDot and Cross Products:\nSuppose [x1,y1] = [ r1 cos(q1), r1 sin(q1)] and \n[x2,y2] = [ r2 cos(q2), r2 sin(q2)] are points in a plane. Then their dot product\n\n[x1,y1].[x2,y2] = x1x2+y1y2  (dot product definition)  ..(1)\n\nand their cross product\n\n[x1,y1].[x2,y2] = x1y2 - y1x2 (cross product definition).(2)\n\nmay be expressed in terms of  trigonometric functions and the angles between the two points, or more precisely their position vectors.\nTo obtain the geometric interpretation,  observe the polar and rectangular ways to multiply the first point by the complex complex conjugate of the second, when both are viewed as complex numbers: That is,\n\n [x1,y1][x2,-y2] = (r1,q1)(r2,-q2)\n\nFrom the equality of two different ways to multiply points in the plane,  observe\n\n[x1x2+y1y2, x1y2 - y1x2]  = (r1r2,q1-q2)\n\nbut\n\n(r1r2,q1- q2) = [ r1r2  cos( q1- q2), r1r2  sin( q1- q2)]\n\nTherefore comparison (equality) of real and imaginary parts yields:\n\nx1x2 + y1y2 = r1r2  cos(q)\n\nand\n\n x1y2 - y1x2 = r1r2  sin( q)\n\n where\n\n q = q1- q2\n\nis the angle between the two points  [x1,y1] = (r1 , q1) and [x2,y2] = ( r2 , q2)\n\n!!!!!!!!!!! Please read q as theta !!!!!!!!!!!\n\nNow proceed using the equations (1) and (2)
978893
The last link was the most informative.\nThe function is as follows:\n(x' y')T = [(1 1)T (1 -1)T] (x y)T\n\nIf you multiply the basis vectors by this matrix you get:\n(0 2)T for the (1 -1)T vector and\n(-1 -5)T for the (-3 2)T vector.\n\n(0, 2) = c1 * (1, -1) + c2 * (-3, 2)\nso:\n0 = c1 - 3 * c2\n2 = - c1 + 2 * c2\nc1 = -6\nc2 = -2\n\nlikewise:\n(-1, -5) = c3 * (1, -1) + c4 * (-3, 2)\nc3 = 17 and c4 = 6\n\nM = [(-6 -2)T (17 6)T]
194771
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. :-(