AMS 361 Lecture 12: AMS 361 Undetermined Coefficients

In Exercises 1-6, determine the size of the matrix. [1 30 2 -4 1 -4 6 2] [1 2 -1 -2] [2 -6 -1 2 -1 0 1 1] [-1] [3 2 1 1 6 1 1 -1 4 -7 -1 2 1 4 2 0 3 1 1 0] [1 2 3 4 -10] In Exercises 7-10, identify the elementary row operation(s) being performed to obtain the new equivalent matrix. [-2 3 5 -1 1 -8] [13 3 0 -1 -39 -8] [3 -4 -1 3 -4 7] [3 5 -1 0 -4 -5] [0 -1 4 -1 3 -5 -5 -7 1 5 6 3] [-1 0 0 3 -1 7 -7 -5 -27 6 5 -27] [-1 2 5 -2 -5 4 3 1 -7 -2 -7 6] [-1 0 0 -2 -9 -6 3 7 8 -2 -11 -4] In Exercises 11-18, find the solution of the system of linear equation represented by the augmented matrix. [1 0 0 1 0 2] [1 0 0 1 2 3] [1 0 0 -1 1 0 0 -2 1 3 1 -1] [1 0 0 2 0 0 1 1 0 0 -1 0] [2 1 0 1 -1 1 -1 1 2 3 0 1] [2 1 1 1 -2 0 1 1 1 0 -2 0] [1 0 0 0 2 1 0 0 0 2 1 0 1 1 2 1 3 3 1 4] [1 0 0 0 2 1 0 0 0 3 1 0 1 0 2 0 3 1 0 2] In Exercises 15-24, determine whether the matrix is in row- from, if it is the, determine whether it is also in reduced row- forms. [1 0 0 0 1 0 0 1 0 0 2 0] [0 1 1 0 0 2 0 1] [2 0 0 0 -1 0 1 1 0 3 4 1] [1 0 0 0 1 0 2 3 1 1 4 0] [0 0 0 0 0 0 1 0 0 0 1 2 0 0 0] [1 0 0 0 0 0 0 0 0 0 1 0] In Exercises 25-38, solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. x + 2y = 7 2x + y = 8 -x + 2y = 1.5 2x - 4y = 3 2x - y = -0.1 3x + 2y = 1.6 -3x + 5y = -22 3x + 4y = 4 4x - 8y = 32 x + 2y = 0 x + y = 6 3x - 2y = 8 2x + 6y = 16 -2x - 6y = -16

Use the row reduction algorithm to transform the matrix into echelon form or reduced echelon form as indicated. 3) Find the echelon form of the given matrix. [1 -3 2 4 -11 2 -2 9 5 3 -5 -1] A) [1 0 0 4 1 0 -2 3 27 3 4 17] B) [1 0 0 4 1 -6 -2 3 9 3 4 -7] C) [1 0 0 4 1 0 -2 3 27 3 4 0] D) [1 0 0 4 1 0 -2 3 15 3 4 -1] Solve the problem. 4) Let a_1 = [3 4 -4], a_2 = [-4 1 1], and b = [2 -10 6]. Determine whether b can be written as a linear combination of a_1 and a_2. In other words, determine whether we x_1 and x_2 exist, such that x_1 a_1 + x_2 a_2 = b. Determine the weights x_1 and x_2 if possible. A) x_1 = -1, x_2 = -3 B) x_1 = -2, x_2 = -2 C) x_1 = -2, x_2 = -1 D) No solution

Solve the following equations. Remember that equations may have no solution or many solutions. 2x + 3x + 5(2x - 3) = 30 4x^2 + 7x - 2x (2x - 5) = 17 (x - 2) (x + 4) = 0 (x - 2) (x + 4) = 2x. (x - 2) (x + 4) = -8 x(x - 2) (x + 4) = 0 4x(x - 2) (x - 2) = 0 (y + 2) - 2y (x + 2) = 0 (x + 2) (y + 2) = 0 (x + 2) (y + 21) + (x - 3) (y + 2) = 0 (x - 2) (x + 4) - 2 (x - 4) = 0 (x + 2)^2 + (x - 2)^2 = 0 (x + 2)^2 - (x - 2)^2 = 0 x(x^2 + 2) = 0 x/3 - x/2 = 2/3 x/3 = 2x 2/x - 3/2x = 1 4x(x -4) (x + 3.8)/x^4 - 4x^+ 7x^2 - 5x + 102 = 0 (-12P/P) (3P/2 + P/2) = 480 2/Q = 0 (25/x)/x = 1 (5Q/P + 1)/(1/P + 2) = 20 10 - 1/10Q = 0 4x + 8(x - 2) = 12 2/x = 4 2 - 0.75x = x - 1.5 3x(x - 2) = 0 (x - 2)^2 = 4 100(1 - x) (1 + x) = 0 100(1 - x) (1 + x) = 100 1 - 3/5 + 2/x - = 3/10 7/2x - x/9 = x Refer in exchange rates given in Table 1.1 to answer question 34 to 40. Given answer correct to two decimal places. Am at 400 Canadian Calculate the price of the How many U