1. Systems of Linear Equations Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

3. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

7. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

8. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Next Step: Replace R2 with 2R1-3R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

9. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

10. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Next Step: Divide R2 by -7 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

11. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

12. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Next Step: Replace R1 with R1-R2 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

13. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Replace R1 with R1-R2 Next Step: Divide R1 by 3 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix You already know how to solve this: Eliminate x by subtracting 2(E1)-3(E2) Solve for y Substitute this value into (E1) Solve for x a) Here is the Gaussian elimination version: Replace R2 with 2R1-3R2 Divide R2 by -7 Replace R1 with R1-R2 Divide R1 by 3 The solution is right there in the matrix: x=-3, y=4 So the solution is x=-3, y=-4 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

16. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

21. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

22. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

23. Solve the following system by a) Gaussian elimination b) Finding the inverse of the coefficient matrix b) Here is the inverse method: The system of equations is in the form We can find the inverse of A by the following method: Put A next to the identity matrix, then perform Gaussian elimination until the identity is on the left, and the inverse matrix will be on the right. This is the inverse matrix Now that we have the inverse, we can multiply on both sides of the original equation: where This is the same answer we got before: X=-3, y=4 *Note: You might be thinking that the inverse matrix way was a whole lot of work to get an answer that we just got faster using the other method. And you would be correct. However, it is a much more general method that will apply to more complicated problems in the future (that would not be so easily solved by the other methods). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24. Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25. Solve for x and y: The solution is x= -2, y= 3 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26. Solve for x and y: The solution is x= -2, y= 3 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB The solution is x=1, y=2

27. Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

28. Solve for x and y: The solution is x=-4, y=4 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

29. Solve for x and y: The solution is x=-4, y=4 Solve for x and y: The solution is x=1, y=-10 Solve for x and y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

30. Solve for x and y: The solution is x=-4, y=4 Solve for x and y: The solution is x=1, y=-10 Solve for x and y: The solution is x=6, y=5 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

31. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

32. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

33. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

34. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

35. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

36. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

37. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

38. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

39. Solve for x, y and z: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

40. Solve for x, y and z: The solution is x=0, y=-1, z=1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

41. Solve this system: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

42. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

43. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

44. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

45. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

46. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

47. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

48. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

49. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. We can write out the solution by choosing z as our parameter, call it t: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

50. Solve this system: Notice that there are only 2 equations, but 3 unknowns. This means that we will NOT be able to find a unique solution. Instead we will get a family of infinitely many solutions (if there are any solutions at all). Here is the augmented matrix for this system. We can interpret this final form of our matrix as follows: x and y are basic variables, and z is a free variable here, because the first 2 columns are pivot columns. We can write out the solution by choosing z as our parameter, call it t: If we separate into homogeneous and particular: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB