A=-ΤΟ 1000 0 2000030 00(3)Consider the following linear system: Axb, where A is given by (3), as in Problem 4, and vectorb = [0, 0, 0, 2023]T. Since matrix A is singular, there are infinitely many least squares solutions. Find theleast squares solution x* with the minimal Euclidean norm.

The given matrix is and .We need to find the least square solution of the system with minimal…

Answered: A = 0 1 00 0020 00 03 000 0 = (3)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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7. (a) Find the equation of the straight line y = ao + a₁t which best fits the following points: 12 13 4 t y 1 3 4 2 2 3 (b) Find the projection matrix onto C(A), the column space of A, where A = 1 1 1 2 1
Suppose that Ax=b has no solution and the columns of A are linearly independent. Let t(A) denote the transpose of matrix A. Select all true statements. O If the A matrix and b vector represent data values, then the least-squares solution gives coefficients for a function that make this function the line that best fits the data values. O The system has a least-squares solution. The solution to t(A)Ax=t(A)b is the least-squares solution. O In the system Ax-b=e, it is possible for e to be the zero vector. O The system has a least squares solution but it may not be unique. O The inverse of t(A)A exists. O The least-squares solution (if it exists) yields a vector in the column space of A that is the "closest" to the vector b.
A. SOLUTION OF NON-HOMOGENEOUS SYSTEM GIVEN: 7 -3 4 0 -7 A = 2 1 -1 4 6 010-3 -3 -5 1.2 TRANSFORM TO LOWER TRIANGULAR MATRIX.
2. (a) Use the linearity of matrix-vector multiplication to prove the following: if the vectors x1, X2 are solutions to Ах = 0 (3.1) then the linear combination x = cịx1 + c2x2 is also a solution, for any constants c1, c2. (You may use the results you showed in the previous problem) (b) Show that the results is not true for nonhomogeneous systems: if the vectors x1, X2 are solutions to Ax = b (3.2) with b + 0, then the linear combination x = c1x1 + c2x2 is not a solution. (c) Show that if, however, x1 and x2 are solutions to the homogenous problem (3.1) and is a particular solution to the nonhomogeneous problem (3.2), then Хр х— С1X1 + C2X2 + Хр is a solution for the nonhomogeneous problem (3.2), for any c1, C2.
1. In each part of this question, you have been given a system of equations and the corresponding matrix taken to RREF. Write the general solution as a vector equation. • List the basic solutions. a. b. (x₁ - x₂ + 2x4 = 0 [1 -X₁ + x₂ + 3x3 - 11x4 0 2x₁ - 2x₂ + 4x4 0 0 -4x₁ + 4x₂ + 3x3 - 17x4 = 0 Lo - = 0 " -1 0 2 01 0 1 -3 0 0 0 0 0 0 00 (3w+9y15z = 0 1 0 -14w + 33x - 75y - 62z = 0 0 1 -5w+ 14x - 29y - 31z = 0 0 0 (52w 164x + 320y + 396z = 0 Lo 0 3 -1 -4 0 0
Consider the following linear system: 1 2 X1 3 (6) X2 Assume that, due to numerical error, the linear system (6) is solved with the same matrix, but with a slightly different right hand side: 1 2 X1 3.01 (7) 3 4 X2 7.02 Without calculating the solutions of the linear systems (6) and (7), mate the relative error in the 1-norm, | : 1. esti-
6. LADW #2.2.1bf, page 46 (you only need to do (b) and (f)) - Note that you need to solve the system and write your solution in vector form. You can use technology to help (ie, you do not need to do row reductions by hand), but you shuld show that main steps (ie, what the reduced echelon form looks like and the corresponding system of equations). 2.2.1(b) 2.2.1(f) x12x2x3 = 1 2x13x2 + x3 = 6 3x1 5x2 = 7 x1 + 5x3 = 9 2x1 2x2x3 + 6x4 - 2x5 = 1 x1x2 + x3 + 2x4 x5 = 2 4x1 - 4x2 + 5x3 +7x4 x5 = 6
By using the method of least squares, find the best line through the points: (−2,−1), (2,1), (0,–3). Step 1. The general equation of a line is c + ₁ = y. Plugging the data points into this formula gives a matrix equation Ac = y. Step 2. The matrix equation Ac = y has no solution, so instead we use the normal equation ATA ĉ = A¹y ATA= ATy = Step 3. Solving the normal equation gives the answer Ĉ= which corresponds to the formula y = Analysis. Compute the predicted y values: y ŷ = Compute the error vector: e=y-ŷ. = Aĉ. Compute the total error: SSE = e² + e² + ez. SSE =
q1
2. The augmented matrix for the system of equations 211 + 2r2 + I3 = 5, 2i + 3r2 + a3 = 2, 3r1 + 4r2 + 5z3 = 1, is given by O (2 2 1 1 4 5 2x 2x2 3r2 3z, 4r2 5z3 1 2 2 1 3. 1 2 4 5 1 '5 2 2 1 2. 3 4 5 1 1 1 213 213

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A = 0 1 00 0020 00 03 000 0 = (3) Consider the following linear system: Ax b, where A is given by (3), as in Problem 4, and vector 6 = [0, 0, 0, 2023] T. Since matrix A is singular, there are infinitely many least squares solutions. Find the east squares solution x* with the minimal Euclidean norm.