Solving Linear Systems: Incompatibility And Summation

Hey guys! Let's dive into a cool math problem. We're going to explore the concept of incompatible systems of linear equations, figure out when they pop up, and then do some cool calculations. This is a great way to brush up on your algebra skills and see how different mathematical ideas connect. So, buckle up and get ready to learn!

Understanding the System and Setting the Stage

Okay, so we've got a system of linear equations. This is what it looks like:

2x + y + mz = 1
x - y + m²z = m
2x + (m + 1)z = m²

Here, x, y, and z are the variables we want to solve for, and m is a parameter – think of it as a variable that can take on different values. Our goal is to find the values of m that make this system incompatible. Basically, an incompatible system is one that has no solutions. There are no values for x, y, and z that can satisfy all three equations simultaneously. This can happen for various reasons, such as when the equations contradict each other. The set of all these m values will be M, and we need to figure out what they are. After determining M, our final task will be to compute the sum of all elements in M, which we denote as S.

To get started, we need a plan. We can use methods like Gaussian elimination or Cramer's rule to analyze the system and see how the parameter m affects the solution. Let's start by trying to eliminate variables using the first two equations. We can multiply the first equation by some number and add it to the second equation to eliminate either x or y. This process will help us reduce the system and make it easier to identify conditions for incompatibility. Also we can use the concept of the determinant to find when the system can not be solved. This usually requires setting up the coefficient matrix and checking when the determinant is zero. When the determinant is zero, the system is either inconsistent (no solution) or has infinitely many solutions. We need to distinguish between these two cases.

So, the key is to figure out under what conditions on m does this system fall apart, or more precisely, does not have a solution. This can be an interesting and useful exercise in understanding linear algebra.

Diving into the Calculations: Finding M

Alright, let's get our hands dirty with some calculations! We're going to use a systematic approach, and I'll try to break it down into easy-to-follow steps. First, we can use Gaussian elimination to simplify the system. This involves manipulating the equations (multiplying them by constants, adding or subtracting them) to eliminate variables and hopefully reveal some patterns. Let's start by eliminating x from the second and third equations. We could multiply the second equation by -2 and add it to the first equation. Then, we can multiply the second equation by -2 and add it to the third equation.

Let's begin with the first two equations, let us multiply the second equation by -2 and add to the first one. We will get:

(2x + y + mz) - 2(x - y + m²z) = 1 - 2m
=> 3y + (m - 2m²)z = 1 - 2m

Now let's move to the third equation. We can subtract the first one to cancel the x. This is actually the same action we performed on the first equation. Let's multiply the second equation by -2 and add to the third equation:

[2x + (m + 1)z] - 2(x - y + m²z) = m² - 2m
=> 2y + (m + 1 - 2m²)z = m² - 2m

Now we have a reduced system of equations with just y and z:

3y + (m - 2m²)z = 1 - 2m
2y + (m + 1 - 2m²)z = m² - 2m

At this point, the equations have become simpler and we are ready to analyze the system. We can multiply the first equation by 2 and the second equation by 3 to eliminate the y variable. The resulting system helps us analyze conditions for incompatibility. Let's multiply the first equation by 2 and the second by 3 and then subtract the second from the first.

2[3y + (m - 2m²)z] = 2(1 - 2m)
3[2y + (m + 1 - 2m²)z] = 3(m² - 2m)

6y + (2m - 4m²)z = 2 - 4m
6y + (3m + 3 - 6m²)z = 3m² - 6m

Subtracting the second equation from the first will result in:

(2m - 4m² - 3m - 3 + 6m²)z = 2 - 4m - 3m² + 6m
=> (2m² - m - 3)z = -3m² + 2m + 2

From this point, we can easily get z and substitute it into the other equations. This step helps us isolate z and potentially find values for m that make the system inconsistent.

z = (-3m² + 2m + 2) / (2m² - m - 3)

Now, let's factorize the denominator and simplify:

z = (-3m² + 2m + 2) / [(2m - 3)(m + 1)]

The denominator is the critical part. If the denominator is zero, z is undefined, and there is a high chance that the system might not have a solution. Thus we can say that when m = 3/2 and m = -1, the system is very likely to be incompatible, but we must analyze the numerator too. Let's test these values. If m = -1, the equation becomes: (2(-1)² - (-1) - 3)z = -3(-1)² + 2(-1) + 2, which is 0z = -3. If m = 3/2, the equation becomes 0z = -3(9/4) + 2(3/2) + 2. So, when m is -1, the system is incompatible, and when m is 3/2, the system is also incompatible, so M = {-1, 3/2}.

Finding the Sum S: The Grand Finale

Alright, we've done the hard work, guys! We've identified the values of m that make the system incompatible. Now, the last step is a piece of cake. We need to calculate the sum of all the elements in the set M. In our case, M = {-1, 3/2}. Therefore, the sum S is:

S = -1 + (3/2) = 1/2

So, the sum S of the values of m that make the system incompatible is 1/2. And that's it!

Conclusion: Wrapping it Up

Well done, everyone! We've successfully navigated through a system of linear equations, identified the conditions for incompatibility, and calculated the sum of specific values. This journey involved algebraic manipulation, careful analysis, and a bit of critical thinking. Remember, the key takeaways here are:

• Understanding Incompatibility: Knowing what it means for a system to have no solutions. Be aware of the different types of equations. Learn about when the system of linear equations is inconsistent. Be aware of the determinant and linear independence of the vectors. Learn about the homogeneous equations.
• Strategic Manipulation: Using techniques like Gaussian elimination to simplify equations and reveal hidden relationships. The best and most efficient way to solve a linear equation is to find the parameters. There are many methods, such as the substitution method, elimination method, Cramer's rule, and matrix methods. Try to get the solution by using the minimum number of steps.
• Careful Analysis: Examining the results to identify conditions that lead to no solutions. Take each step and see whether each manipulation is logical and correct. Double-check each step to be sure of your results. Avoid unnecessary complexity, choose the simplest and most elegant method.

I hope this was a helpful and fun journey. Keep practicing, and you'll become a pro in no time! Until next time, keep those math muscles flexing!