Question

1. {5 points) The solution to the following system of linear equations is (2.0). Use a method of your choice to show how this answer could be arrived at. 3x + y = 6 2x + 5y = 4 2. {5 points) The following system of equations has no solution. Use the echelon method to show how this conclusion was arrived at. 2x - 3y = 2 4x - y = 5 3. {5 points) The solution to the system below is (-2,0). Use Gauss-Jordan method to show how this solution could be obtained. 3x + 2y = -6 5x - 2y = -10

Answer 1

1. We have, 3x + y = 6 2x + 5y = 4 Using the elimination method, We multiply the first equation by -5 and add it to the second equation -15x - 5y = -30 2x + 5y = 4 By doing this we transformed our original system into an equivalent system as 3x + y = 6 -13x = -26 We divide the second equation by -13 to get the next equivalent system 3x + y = 6 x = 2 Now we multiply the second equation by -3 and add it to the fist equation to arrive at y = 0 x = 2 Hence the solution is (2,0) 2. We have, 2x – 3y = 2 4x - y = 5 Eliminating x term by multiplying equation 1 by 2 and adding it to the equation 2 -4x + y = -6 4x - y = 5 Adding these two gives 0=-1 As the LHS is not equal to RHS above, we can conclude that the system of equations does not have a unique solution as the two lines are parallel to each other.

3. We have 3x + 2y = -6 5x - 2y = -10 The augmented matrix for the given system is as follows: 13 2 - 6 3x + 2y = -6 15 -2 1-10 = We now add row 2 to row 1 58 01 - 16 5-2 1-10 ] 8x = -16 5x - 2y = -10 We now divide row 1 by 8 0 1 - 2 15 -2 1-10 [ ] [5x -+27=2-10] We now multiply row 1 by -5 and it to row 2 0 1 - 2 LO -21 0 [6 ] = -21 1-2y = 0 We now divide row 2 by -2 [ 0 -2 10 ] [3 = 6]