1. Systems of Linear Equations
Examples
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2. Solve the following system by a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
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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)
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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:
a) Here is the Gaussian elimination version:
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

7. Solve the following system by a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
You already know how to solve this:
a) Here is the Gaussian elimination version:
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

8. Solve the following system by a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
You already know how to solve this:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Next Step:
Replace R2 with 2R1-3R2
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

9. Solve the following system by a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
You already know how to solve this:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
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

10. Solve the following system by a) Gaussian elimination
b) Finding the inverse of the coefficient matrix
You already know how to solve this:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
Substitute this value into (E1)
Next Step:
Divide R2 by -7
Solve for x
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:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
Substitute this value into (E1)
Divide R2 by -7
Solve for x
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:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
Substitute this value into (E1)
Divide R2 by -7
Solve for x
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:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
Substitute this value into (E1)
Divide R2 by -7
Solve for x
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:
a) Here is the Gaussian elimination version:
Eliminate x by subtracting 2(E1)-3(E2)
Solve for y
Replace R2 with 2R1-3R2
Substitute this value into (E1)
Divide R2 by -7
Solve for x
Replace R1 with R1-R2
Divide R1 by 3
So the solution is x = -3, y = 4
The solution is right there in the matrix: 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
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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
where
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.
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
where
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.
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
where
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.
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
where
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
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
where
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
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
Now that we have the inverse, we can multiply on both sides of the original equation:
b) Here is the inverse method:
The system of equations is in the form
where
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
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
Now that we have the inverse, we can multiply on both sides of the original equation:
b) Here is the inverse method:
The system of equations is in the form
where
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
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
Now that we have the inverse, we can multiply on both sides of the original equation:
b) Here is the inverse method:
The system of equations is in the form
where
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 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 be very useful.
This is the inverse matrix
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For Campus Learning Assistance Services at UCSB

24. Solve for x and y: Solve for x and y: Prepared by Vince Zaccone
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25. Solve for x and y: The solution is x= -2, y= 3 Solve for x and y:
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26. Solve for x and y: The solution is x= -2, y= 3 Solve for x and y:
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The solution is x=1, y=2

27. Solve for x and y: Solve for x and y: Solve for x and y:
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28. Solve for x and y: Solve for x and y: The solution is x= -4, y=4
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29. Solve for x and y: Solve for x and y: The solution is x=-4, y=4
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30. Solve for x and y: Solve for x and y: The solution is x=-4, y=4
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31. Solve for x, y and z: Prepared by Vince Zaccone
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32. Solve for x, y and z: Prepared by Vince Zaccone
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33. Solve for x, y and z: Prepared by Vince Zaccone
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34. Solve for x, y and z: Prepared by Vince Zaccone
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35. Solve for x, y and z: Prepared by Vince Zaccone
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36. Solve for x, y and z: Prepared by Vince Zaccone
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37. Solve for x, y and z: Prepared by Vince Zaccone
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38. Solve for x, y and z: Prepared by Vince Zaccone
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39. Solve for x, y and z: Prepared by Vince Zaccone
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40. The solution is x=0, y= -1, z=1
Solve for x, y and z:
The solution is x=0, y= -1, z=1
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41. Solve this system: Prepared by Vince Zaccone
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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).
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For Campus Learning Assistance Services at UCSB

43. Here is the augmented matrix for this system.
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.
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For Campus Learning Assistance Services at UCSB

44. Here is the augmented matrix for this system.
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. Here is the augmented matrix for this system.
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. Here is the augmented matrix for this system.
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. Here is the augmented matrix for this system.
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. Here is the augmented matrix for this system.
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. Here is the augmented matrix for this system.
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 write out the solution by choosing z as our parameter, call it t:
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

50. Here is the augmented matrix for this system.
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 write out the solution by choosing z as our parameter, call it t:
If we separate into homogeneous and particular:
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

51. Here is the augmented matrix for this system.
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 write out the solution by choosing z as our parameter, call it t:
If we separate into homogeneous and particular:
A geometrical interpretation is that this is a straight line in 3-dimensions, with slopes given by the homogeneous part and intercept given by the particular part.
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

52. Here is the augmented matrix for this system.
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 write out the solution by choosing z as our parameter, call it t:
If we separate into homogeneous and particular:
A geometrical interpretation is that this is a straight line in 3-dimensions, with slopes given by the homogeneous part and intercept given by the particular part.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB