Find the least-squares solution x* of the system?
Find the least-squares solution x* of the system: (the following is in the form of Ax=b) [2; -5; 2] * x = [-3; 21; 6] [ 2 ] [ -3 ] l -5 l x = l 21l [ 2 ] [ 6 ] x* = [ ???????? ]
Find the least-squares solution of the system?
( 1 -2 ( 0 -1 2 x = 6 4 5 ) 14 ) Find x b.) Determine the projection p = Ax c.) complete the residual b - Ax
Find the least-squares solution x∗ of the system?
[ 1 -1 -1 ] [ 1 -1. 1 ] x [ 1. 1. 1 ] [ 1. 1 -1 ] = [.. 1 ] [.. 5 ] [. -5 ] [ 11 ] A = [ 1 -1 -1 ] [ 1 -1. 1 ] [ 1. 1. 1 ] [ 1. 1 -1 ] c = [.. 1 ] [.. 5 ] [. -5 ] [ 11 ] A^t • A = [. 1. 1. 1. 1 ]. . [1 -1 -1 ] [ -1 -1. 1. 1 ] • [ 1 -1. 1 ] [ -1. 1. 1 -1 ]. .[ 1. 1. 1 ] . . . . . . . . . . . [ 1. 1 -1 ] = [ 4 0 0 ] [ 0 4 0 ] [ 0 0 4 ] A^t • c = [. 1. 1. 1. 1 ]. . [. 1 ] [ -1 -1. 1. 1 ] • [. 5 ] [ -1. 1. 1 -1 ]. .[ -5 ] . . . . . . . . . . . [ 11 ] = [ 12 ] [.. 0 ] [ -12 ] [ 4 0 0 ]. . . . .[. 12 ] [ 0 4 0 ] • x = [. . 0 ] [ 0 0 4 ]. . . . .[ -12 ] 4x1 = 12 … x1 means x sub 1 x1 = 3 4x2 = 0 x2 = 0 4x3 = -12 x3 = -3 . . . .[. 3 ] x* = [. 0 ] . . . .[ -3 ]
Least-squares solution to a system?
Since you didn't specify a method, I'll use the normal equation: AT*Ax = AT*b Where A is the coefficient matrix and b is the constant matrix, and AT is the transpose of A. X is the least squares vector. To solve for x, we'll multiply on the left of both sides (AT*A)^-1, which gives x = (AT*A)^(-1)*AT*b For this problem specifically: A = [1 2 1] [1 3 2] [2 5 3] [2 0 1] [3 1 1] and b = [-1] [2] [0] [1] [-2] Using a calculator to save time, we find that x = (AT*A)^(-1)*AT*b = [-23/15] [-28/15] [64/15] = [a] [b] [c] I hope this was helpful.
Find the pseudoinverse of the matrix A (1 1, 2 3, 2 1) (A is a 3x2 matrix). Find the least squares solution?
Pseudoinverse: The pseudoinverse of A gives the least squares solution for any system A v "=" z ( indeed (A v - z)² min (eq 1)) => v = A⁺ z where A⁺ is the pseudoinverse of A v= [a] [b] and z is a 3x1 matrix. Writing (eq 1) in component we have (x₁ a + y₁ b - z₁)² + (x₂ a + y₂ b - z₂)² + (x₃ a + y₃ b - z₃)² min where x₁ = A₁₁ = 1 y₁ = A₁₂ = 1 x₂ = A₂₁ = 2 y₂ = A₂₂ = 3 x₃ = A₁₁ = 2 y₃ = A₁₂ = 1 The (3) columns of A⁺ are the solutions of the above system for z = [1] [0] [0] (first column) z= [0] [1] [0] (second column) and z= [0] [0] [1] (third column). The values of a and b that minimize the sum of squares are those which null the partial derivatives: x₁(x₁ a + y₁ b - z₁) + x₂(x₂ a + y₂ b - z₂) + x₃(x₃ a + y₃ b - z₃) = 0 y₁(x₁ a + y₁ b - z₁) + y₂(x₂ a + y₂ b - z₂) + y₃(x₃ a + y₃ b - z₃) = 0 Then D = (Sx2)(Sy2) - (Sxy)² a = [(Sxz)(Sy2) - (Syz)(Sxy)]/D b = [(Syz)(Sx2) - (Sxz)(Sxy)]/D Performing the calculations for z = [1 0 0]⁺ (transposed, a 3x1 matrix) we obtain [a b]⁺ = [1/9, 0]⁺ If z = [0 1 0]⁺ then [a, b]⁺ = [-5/18, 1/2]⁺ and for z = [0 0 1]⁺ [a, b]⁺ = [13/18, -1/2]⁺ => A⁺ = [(1/9) (-5/18) (13/18)] [ (0) (1/2) (-1/2)] Least squares solution for z=[1,2,-1]⁺ [-7/6] [3/2]
Linear algebra least-squares solution?
Describe all least-squares solutions of the following system. explain what the answer means and draw a picture. x+y = 2 x+y = 4 Already this looks like trouble, however I found a least squares solution where x-hat = (6, -12) and I tried to draw the whole thing out but I can't find the meaning to all of it. How to I construct a subspace out of the same vector is it just a line from ) to (1,1) and A(x-hat) is so far from b it doesn't make sense
How do I solve system of linear equations?
There are many ways to solve a system of linear equations. If the system is small (e.g., 2 Equations in 2 Unknowns), it could even be done with pen, paper, and a hand calculator.As the systems get larger, it becomes much more tedious to compute solutions. If you just want to solve the system, the simplest method is probably Gaussian elimination. You would describe the system in terms of the coefficient matrix [A], the solution vector (x), and the constant vector (b):[A](x) = (b)The task now becomes to solve for (x).Gaussian elimination reduces the [A] matrix to the point it is lower triangular or upper triangular. At that point, one solution has been found. That solution is then back-substituted into the next row to solve for the next solution. These two solutions are back-substituted into the next row to solve for the next solution. And so on until all the solutions are found. This method is probably the quickest and easiest to compute the solution vector (x).Another easy method is Gauss-Jordan elimination. This method takes a bit more work. Instead of finding one solution and then back-substituting to find the other solutions, it reduces the [A] matrix all the way to an explicit solution for all the solution values.Or, the inverse of {A] could be found. Then the result is stated as[math](x) = [A] ^{-1} (b)[/math]However, inverting [A] usually involves a lot more work than the first two methods mentioned.In addition, these methods can be modified somewhat to make them more effective. For example, pivoting may be done, or other techniques added to make them more sophisticated.On the other hand, if you are happy with numerical results, many software products are available that can do this task for you, for example, MATLAB, MathCAD, Maple, Mathematica, etc. And there are many free products available too, that can be downloaded as apps, executable or used online. For example, Simultaneous Equation Solver solves a system of up to 12 Equations in 12 Unknowns.There are many other resources available.
[Linear Algebra] Find the least-squares solution x* of the system?
[A|B]= [1 -1 1 | -9 ] [1 1 -1 | 1 ] [1 1 1 | -3 ] [1 -1 -1 | -9 ] by R2-->R2- R1 R3-->R3- R1 R4-->R4- R1 we get [1 -1 1 | -9 ] [0 2 -2 | 10 ] [0 2 0 | 6 ] [0 0 -2 | 0 ] by R3-->R3- R2 [1 -1 1 | -9 ] [0 2 -2 | 10 ] [0 0 2 | -4 ] [0 0 -2 | 0 ] [1 -1 1 | -9 ] [0 2 -2 | 10 ] [0 0 2 | -4 ] [0 0 0 | -4 ] rank([A|B]) ≠ rank (A) therefore no solution
How can I solve a 20-variable system of equations?
If you have 23 equations in 20 variables, the system is overdetermined, and there may not be a solution. Your best bet is to write the system in matrix form as [math]Ax = b[/math] and then find the least squares solution by minimizing [math]||Ax - b||_2^2[/math]. If a solution exists, you'll get it, and if not, you'll get the best approximation in a reasonable sense. There's all kinds of least squares software out there, so pick one that looks easy to use.
Help!!!Least-squares solution and least-squares error???
Let's let MATLAB do a little work here. >> A = [1 3 ; -2 -6 ; 3 9]; >> B = [1 ; 0 ; 1]; >> x = A\B Warning: Rank deficient, rank = 1, tol = 7.4773e-015. x = 0 2/21 >> (A*x - B).^2 ans = 25/49 16/49 1/49 >> x = pinv(A)*B x = 1/35 3/35 >> (A*x - B).^2 ans = 25/49 16/49 1/49