Problem 5
For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. (a) x1 2x2 = 3 2x1 x2 = 9 (b) 2x1 3x2 = 5 4x1 + 6x2 = 8 (c) x1 + x2 = 0 2x1 + 3x2 = 0 3x1 2x2 = 0 (d) 3x1 + 2x2 x3 = 4 x1 2x2 + 2x3 = 1 11x1 + 2x2 + x3 = 14 (e) 2x1 + 3x2 + x3 = 1 x1 + x2 + x3 = 3 3x1 + 4x2 + 2x3 = 4 (f) x1 x2 + 2x3 = 4 2x1 + 3x2 x3 = 1 7x1 + 3x2 + 4x3 = 7 (g) x1 + x2 + x3 + x4 = 0 2x1 + 3x2 x3 x4 = 2 3x1 + 2x2 + x3 + x4 = 5 3x1 + 6x2 x3 x4 = 4 (h) x1 2x2 = 3 2x1 + x2 = 1 5x1 + 8x2 = 4 (i) x1 + 2x2 x3 = 2 2x1 + 2x2 + x3 = 4 3x1 + 2x2 + 2x3 = 5 3x1 + 8x2 + 5x3 = 17 (j) x1 + 2x2 3x3 + x4 = 1 x1 x2 + 4x3 x4 = 6 2x1 4x2 + 7x3 x4 = 1 (k) x1 + 3x2 + x3 + x4 = 3 2x1 2x2 + x3 + 2x4 = 8 x1 5x2 + x4 = 5 (l) x1 3x2 + x3 = 1 2x1 + x2 x3 = 2 x1 + 4x2 2x3 = 1 5x1 8x2 + 2x3 = 5

Step-by-Step Solution:
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The full step-by-step solution to problem: 5 from chapter: 1.2 was answered by , our top Math solution expert on 03/15/18, 05:24PM. The answer to “For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. (a) x1 2x2 = 3 2x1 x2 = 9 (b) 2x1 3x2 = 5 4x1 + 6x2 = 8 (c) x1 + x2 = 0 2x1 + 3x2 = 0 3x1 2x2 = 0 (d) 3x1 + 2x2 x3 = 4 x1 2x2 + 2x3 = 1 11x1 + 2x2 + x3 = 14 (e) 2x1 + 3x2 + x3 = 1 x1 + x2 + x3 = 3 3x1 + 4x2 + 2x3 = 4 (f) x1 x2 + 2x3 = 4 2x1 + 3x2 x3 = 1 7x1 + 3x2 + 4x3 = 7 (g) x1 + x2 + x3 + x4 = 0 2x1 + 3x2 x3 x4 = 2 3x1 + 2x2 + x3 + x4 = 5 3x1 + 6x2 x3 x4 = 4 (h) x1 2x2 = 3 2x1 + x2 = 1 5x1 + 8x2 = 4 (i) x1 + 2x2 x3 = 2 2x1 + 2x2 + x3 = 4 3x1 + 2x2 + 2x3 = 5 3x1 + 8x2 + 5x3 = 17 (j) x1 + 2x2 3x3 + x4 = 1 x1 x2 + 4x3 x4 = 6 2x1 4x2 + 7x3 x4 = 1 (k) x1 + 3x2 + x3 + x4 = 3 2x1 2x2 + x3 + 2x4 = 8 x1 5x2 + x4 = 5 (l) x1 3x2 + x3 = 1 2x1 + x2 x3 = 2 x1 + 4x2 2x3 = 1 5x1 8x2 + 2x3 = 5” is broken down into a number of easy to follow steps, and 314 words. Linear Algebra with Applications was written by and is associated to the ISBN: 9780136009290. Since the solution to 5 from 1.2 chapter was answered, more than 215 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Linear Algebra with Applications, edition: 8. This full solution covers the following key subjects: . This expansive textbook survival guide covers 47 chapters, and 921 solutions.