How do you solve a matrix with 2 unknowns and 3 equations?
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.
292019
If the unit vector has components "a" and "b" (the unknowns for which you want to solve) then your first equation is\n\n1.  a^2 + b^2 = 1\n\nIf you have f and the directional derivative then you can take the partial derivatives of f with respect to x and y (call these f_x and f_y) and get your second equation\n\n2.  a f_x + b f_y = directional derivative\n\nWith two equations in the two unknowns a and b you then try to solve for a and b.
994889
Start with a general cubic equation:\nx^3 +ax^2 +bx+c=0.\nFirst, we want to get rid of the x^2 term, so let \nx=y-(a/3).\nIf you plug this in and expand, the y^2 terms will drop out and you will get an equation of the form\ny^3 +Cy+D=0.\nNow, if you let y=u+v (to get an extra bit of freedom), and expand, you get\nu^3 +v^3 +(3uv+C)(u+v)+D=0.\nChoose v so that 3uv+C=0, i.e. v=-C/(3u). Plug this into\nu^3 +v^3 +D=0\nand multiply through by u^3 to get a quadratic in u^3. Solve\nthis by the quadratic formula and take a cube root to find\nu. There will be three cube roots in the complex plane, so find all of them. For each u, find v=-C/(3u), then y=u+v, and finally x=y-(a/3).\n\nThis is often a rather cumbersome proceedure, even if you do some of it ahead of time. Also, the roots may not 'look' like what you expect. If you apply this to an equation that has a root of, say, 4, you will obtain 4 as a sum of two cube roots of quadratic expressions. To recognize the answer as 4 is a challenge in and of itself.
187542
Answer:\ny=-1.25x + 3\n\nProof:\nWe are given two points to work with:\nPoint 1: (x1=0, y1=3)\nPoint 2: (x2=4, y2=-2)\n\nSo, use the "two point equation" to solve this problem.\ny-y1 = (y2-y1)/(x2-x1) * (x-x1)\n\ny-3 = (-2-3)/(4-0) * (x-0)\ny-3 = -5/4 * x\ny = -5/4 * x + 3\n\ny=-1.25x + 3
589722
Step 1: Solve one equation for one variable in terms of the other.\nStep 2: Plug that into the other equation and solve for the second variable.\nStep 3: Plug the second variable into the first equation and solve for the first variable.
685851
x=-1/2.. firstly form a matrix with the following rows\n\nR1 (x -1/2 -1/2)\nR2 (-1/2 x -1/2)\nR3 (-1/2 -1/2 x)\n\nthink of R1, R2, R3 as ur A.\n\nNow Multiply A  with x.\n\nin this case, x will be the vector (a, b, c)transpose.\n\nthe result must be (0,0,0).\n\nso now u have a set of equations in matrix form. all u do is solve for x. apply gaussian elimination and voila, vous avez votre reponse.. shudnt be tough..
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
752603
The inversion of a 3x3 matrix is very easy to solve, however the equation is fairly long. You can find it on wikipedia at http://en.wikipedia.org/wiki/Inverse_matrix#Inversion_of_3_x_3_matrices
565039
For this you are asked to take the the equation y = 2n + 3 and substitute the rest of the values in for n and solve the equation. The implied operation between 2 and n is multiplication.\n\nSo you take the n out of the equation and put in the values that you have listed.\n\nFirst Value:\ny = 2(2) + 3  Do the multiplication first then add the 3\ny = 4 + 3\ny = 7\n\nSecond Value:\ny = 2(1/2) + 3\ny = 1 + 3\ny = 4\n\nThird Value:\ny = 2(0) + 3\ny = 3
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.
978722
You can't, algebraically.  You have two equations and three variables.  Sorry.  You need one more equation, such as a = b.
1159782
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.
240066
First of all, let me know at what standard u need the answer.  Accordingly I will answer.  Essentially, SVD is used in Matrix algebra.  If you are below graduation, here is the simplest answer.\n\nSuppose you have two linear equations, with two unknowns, we will try to solve them as simultaneous equations.  similarly, if there are three quations with three unknowns, still we can attempt for solution.  Suppose there are mor equations than the number of unknowns, as in the case of randomised variables, that is four equations but three unknowns then the problem becomes serious.  we need to find the unknowns but satisfying all the equations.  Here we use this technique.  We try to find the solution of such cases using matrix inversion using Marquardt's algorithm or the solution of the characteristic equation leads to finding the eigen values.  So and so the theory continues.  That is why I asked at what level u need the answer?
429670
In matlab, type\n>> syms x\n>> solve('x^3+4*x^2-5=0')\n \nans =\n \n[                1]\n[ -5/2+1/2*5^(1/2)]\n[ -5/2-1/2*5^(1/2)]
756126
you need to do the following:\nT( (1,-1))= (0, 2) = -6 (1,-1) - 2 (-3,2)  so the first column of your matrix is:\n-6\n-2\n\nnow\nT ( (-3,2) ) = ( -1, -5) = 17 (1,-1) + 6 (-3,2), so the second column of your matrix is:\n17\n6\n\nwhen you put them together you get:\n-6 17\n-2 6\n'
225505
First manipulate a little to get everything lined up\n\n3x - 2y = -8\n9x - 3y = -21\n\nNow to cancel out one of the variables.  It's easiest to cancel out x, since 9 is a multiple of 3.  Multiply the 1st equation by -3\n\n-9x + 6y = 24\n9x -3y = -21\n\nNow add the 2 equations together.  The -9x and 9x cancel eachother out and you are left with\n\n6y - 3y = 24 - 21\n3y = 3\n\nFinish solving for y, and then plug that number in for y in either one of the equations above and solve for x. Good luck!
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)