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Amy has a master's degree in secondary education and has taught math at a public charter high school.
This video lesson is about inconsistent and dependent systems or collections of equations. What are they? Inconsistent systems are those systems that have no solution. Dependent systems are those systems that have an infinite number of solutions. Think of inconsistent and dependent as a traffic light. An inconsistent traffic light never works when you get there. You get a red light every time you get there, and you see all the other cars go, but your light is still red. You wait several rounds, and your light is still red. After a while, you avoid this inconsistent traffic light because you know you won't get anywhere. A dependent traffic light, on the other hand, always gives you the green light on time. You rely on this traffic light time and time again because you know whenever you get there, you will get a green light. So, while you will never go to an inconsistent traffic light, as it has no solutions for you, you will always go to the dependent traffic light because it has an infinite number of green lights for you.
Why should you learn about these? You will come across these types of systems as you go along in your math classes. Once you spot that the system you are working with is either inconsistent or dependent, then you can say that the system has no unique solution because it either has no solution or it has an infinite number of solutions. What causes these situations? For the inconsistent system scenario, this happens when at least two of the equations do not intersect when graphed. This means that they never meet or touch. So, for lines, it means that at least two of the lines are parallel. For planes, it means that at least two of the planes are parallel to each other. For a dependent system, it means that all the equations graph out the same line or plane. Because all the equations are the same, there is no unique solution. Instead, we have an infinite number of solutions because all the equations intersect at all the points.
Now that we know what inconsistent and dependent systems are, we can now ask whether or not we can use Gaussian elimination to help us solve them. Gaussian elimination is the process of turning the system of equations into a matrix, then using matrix operations to change the matrix into row echelon form where the lower diagonal is all zeros. At this point, we can use the last equation to solve for the last variable. We can then substitute this answer and substitute it into the second to last equation to find the next variable. We keep working our way back up until we have all our variables. Take a moment to refresh your Gaussian elimination skills if you need to. Of course, Gaussian elimination works if we have a unique solution, but will this work for inconsistent or dependent systems?
The short answer is no, it won't work. Why doesn't it work? What happens when we try to solve these types of systems using Gaussian elimination? Let's look at a couple examples to see what happens.
Let's first look at a system that is inconsistent. Let's see what happens when we apply Gaussian elimination to it.
Applying Gaussian elimination, we create our matrix by writing down the numbers associated with the variables as well as the constant numbers. We get this augmented matrix:
We want to eliminate the beginning 1 in the second row and the beginning 1 in the third row. To eliminate the 1 in the second row, we can go ahead and subtract the first row from the second to create a new second row. We get 0, 0, 0, 3. Whoa, whoa, whoa! Is this even possible? If we translate this row back into equation form, we get 0 = 3. Is this a valid statement? No, it's not. What does that mean? It means that we can't continue because there is no unique solution.
We see that for inconsistent systems, when we try to use Gaussian elimination, we end up with a false statement. That tells us that there is no unique solution, and we can't continue.
Well, what about the case of the dependent system? What happens when we try to use Gaussian elimination for this type of system? Let's take a look.
We first change this into matrix form:
Applying Gaussian elimination, we need to make the beginning 1 in the second row 0, and we need to make the first 4 and then the -8 in the third row 0. We can make the beginning 1 in the second row 0 by multiplying the second row by -2 and then adding it to the first row to create a new second row. Doing this we get 0, 0, 0 and 0. Okay, that's interesting. It's not a false statement, so let's keep going.
For the third row, we can multiply the first row by -2 and add it to the third to get a new third row. Doing this we get 0, 0, 0 and 0 for the new third row. Hmm. That's interesting too. This leaves me with just the one equation on the top since the other two equations are 0 = 0, which means nothing. Well, we need to stop because we see that we can't go any further to get a unique solution.
If you take these same systems and you try other methods of solving them, you will come across other obstacles. It all means that there is no method to solving inconsistent or dependent systems because there is no unique solution that can be found.
Let's review what we've learned now. We learned that inconsistent systems are those systems that have no solution, and dependent systems are those systems that have an infinite number of solutions. Gaussian elimination, one method of solving systems of equations, cannot be used to solve inconsistent and dependent systems. Because neither type of system has a unique solution, no method of solving them can be used. They will all yield results that don't mean anything or that don't make sense.
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Back To CourseAlgebra II Textbook
26 chapters | 256 lessons