Systems and Matrices Final Exam Review Key

University:
University of North Georgia
Course:
MATH 1113 | Precalculus
Academic year:

2024
Views:

247

Pages:

2
Author:

5CelestialChronicle

Systems and Matrices Solve each system by hand. State which method you used: Substitution or Elimination. y = 40x + 20 y = -10x - 30 Solution: (-1, -20) Method: Substitution 6x + 3y = 6 3x + 5y = 12 Solution: (-1, 4) Method: Elimination 2x - 3y = -2 2x + y = 24 Solution: (8.75, 6.5) Method: Elimination 4x - 10 = y 2x - 12 = 3(4 - 10) Solution: (3, 2) Method: Substitution State the dimensions of each matrix. [2 -5 3; 1 -4 6] Dimensions: 2 x 3 [1 -4; 5 -6] Dimensions: 2 x 2 [3] Dimensions: 3 x 1 [4 5 -6] Dimensions: 1 x 3 For #9-#10, use the matrices below. R = [3 -5; -2 -2] S = [1 -5; -2 2] R + S = [[3+1 -5+(-5); -2+(-2) -2+2]] = [[4 -10; -4 0]] 2R - 3S = [[6 -10; -4 -4]] - [[3 -15; -6 6]] = [[3 -5; 2 -10]] For #11-#12, use the matrices below. C = [1 -5; -2 2] D = [0 4 -3; 6 1 5] CD Not possible. The number of columns in C (2) does not equal the number of rows in D (2). DC = [[0(1) + 4(-2) 0(-5) + 4(2) 0(-3) + 4(5); 6(1) + 1(-2) 6(-5) + 1(2) 6(-3) + 1(5)]] = [[-8 8 20; 4 -28 -13]] Use the following matrices for the multiplication problems: A = [1 -2] [12 10] B = [-13 6] [15 -1] G = [0 11 -12] [6 1 5] [-3 4 5] H = [-2 -5 10] [-4 0 1] [22 -33 11] 13. AB [1 -2] * [-13 6] [12 10] [15 -1] = [1*(-13) + (-2)*15 1*6 + (-2)*(-1)] [12*(-13) + 10*15 12*6 + 10*(-1)] = [-43 8] [-6 62] 14. BA [-13 6] * [1 -2] [15 -1] [12 10] = [(-13)*1 + 6*12 (-13)*(-2) + 6*10] [15*1 + (-1)*12 15*(-2) + (-1)*10] = [59 86] [3 -40] 15. GH [0 11 -12] * [-2 -5 10] [6 1 5] [-4 0 1] [-3 4 5] [22 -33 11] = [0*(-2) + 11*(-4) + (-12)*22 0*(-5) + 11*0 + (-12)*(-33) 0*10 + 11*1 + (-12)*11] [6*(-2) + 1*(-4) + 5*22 6*(-5) + 1*0 + 5*(-33) 6*10 + 1*1 + 5*11] [-3*(-2) + 4*(-4) + 5*22 -3*(-5) + 4*0 + 5*(-33) -3*10 + 4*1 + 5*11] = [-308 396 -121] [94 -195 116] [100 -150 29] 16. What are the dimensions of the product matrix when the following two matrices are multiplied? 5x2 * 2x3 Answer: The product matrix has dimensions 5x3 17. Find the inverse of matrix F = [1 2; 4 5] using your calculator. F^-1 = [5/3 -2/3] [-4/3 1/3] 18. Find the inverse of matrix K = [-2 3; 5 -1] using your calculator. K^-1 = [1/13 3/13] [5/13 -2/13] 19. Use the inverse matrix and matrix multiplication to solve the following system of equations. Start by filling in the Coefficient, Variable, and Constant matrices. Remember to write your answer as an ordered pair. 3x - 2y = 7 5x + 3y = -1 [3 -2] [x] = [7] [5 3] [y] [-1] A^-1 * B = [1 -2] * [7] [5 3] [-1] = [1*7 + (-2)*(-1)] [5*7 + 3*(-1)] = [9] [32] So, the solution is (9, 32). 20. Use the inverse matrix and matrix multiplication to solve the following system of equations. Start by filling in the Coefficient, Variable, and Constant matrices. Remember to write your answer as an ordered triple. 2x - y + 2z = 15 -x + y + z = 3 3x - y + 2z = 16 [2 -1 2] [x] = [15] [-1 1 1] [y] [3] [3 -1 2] [z] [16] A^-1 * B = [3 1 5] [1 -2 3] [-2 3 -1] = [3*15 + 1*3 + 5*16] [1*15 + (-2)*3 + 3*16] [(-2)*15 + 3*3 + (-1)*16] = [120] [48] [-23] So, the solution is (120, 48, -23).

Systems and Matrices Final Exam Review Key

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