Solving Linear Equations: Finding Values For Solutions

Hey guys! Today, we're diving deep into the fascinating world of linear equations. Specifically, we're going to tackle a problem where we need to find the number of values for a parameter, let's call it 'm,' that allow a system of linear equations to have at least one solution in a particular form. Sounds like a mouthful, right? But trust me, we'll break it down step by step, making it super easy to understand. So, grab your favorite beverage, settle in, and let's get started!

Understanding the Problem

Before we jump into the math, let's make sure we're all on the same page. We're given a system of three linear equations with three variables: x, y, and z. These equations also involve a parameter 'm,' which is essentially a variable that can take on different values. Our goal is to find out how many different values of 'm' will allow the system to have at least one solution where x is equal to 1, and y and z can be any values (which we'll call α and β, respectively).

In simpler terms, we're looking for values of 'm' that make the equations work when we plug in 1 for x. This is a classic problem in linear algebra, and it often involves concepts like determinants and matrix operations. But don't worry if those terms sound intimidating – we'll explain everything as we go along.

Now, let's write down the system of equations so we can see it clearly:

1. x + (m - 1)y + (2m - 3)z = 1
2. mx + (2m - 2)y + 2z = 0
3. (m + 1)x + (3m - 3)y + (m^2 - 1)z = 0

Our mission, should we choose to accept it, is to find the number of 'm' values that make this system have solutions of the form (1, α, β).

Setting up the Equations with x = 1

Okay, so the first step is to substitute x = 1 into our system of equations. This simplifies things quite a bit, as it reduces the number of variables we need to worry about directly. When we plug in x = 1, our equations become:

1. 1 + (m - 1)y + (2m - 3)z = 1
2. m + (2m - 2)y + 2z = 0
3. (m + 1) + (3m - 3)y + (m^2 - 1)z = 0

Now, let's clean these up a little bit to make them even easier to work with. We can subtract 1 from both sides of the first equation and rearrange the others to get:

1. (m - 1)y + (2m - 3)z = 0
2. (2m - 2)y + 2z = -m
3. (3m - 3)y + (m^2 - 1)z = -(m + 1)

See? Already, things are looking a bit more manageable. We've got a system of three equations with two unknowns, y and z, and our parameter 'm.' Now we need to figure out how the values of 'm' affect the solutions to this system.

Using Determinants to Find Solutions

Here's where things get a little more interesting. To determine whether this system has solutions, we can use the concept of determinants. For a system of linear equations to have a non-trivial solution (i.e., a solution other than y = 0 and z = 0), the determinant of the coefficient matrix must be equal to zero. This is a fundamental concept in linear algebra, and it's super useful for problems like this.

So, what's the coefficient matrix? It's the matrix formed by the coefficients of y and z in our equations. In this case, it looks like this:

To find the determinant, we'll focus on the first two equations initially and then check if the solution satisfies the third equation. The determinant of the 2x2 matrix formed by the first two equations is:

Det = (m - 1)(2) - (2m - 3)(2m - 2)

Let's expand and simplify this expression:

Det = 2m - 2 - (4m^2 - 4m - 6m + 6) Det = 2m - 2 - 4m^2 + 10m - 6 Det = -4m^2 + 12m - 8

For the system to have a solution, this determinant must be zero. So, let's set it equal to zero and solve for 'm':

-4m^2 + 12m - 8 = 0

We can divide the entire equation by -4 to simplify it:

m^2 - 3m + 2 = 0

Solving the Quadratic Equation for 'm'

Now we have a quadratic equation that we can solve for 'm.' This should be pretty straightforward. We can factor the quadratic as follows:

(m - 1)(m - 2) = 0

This gives us two possible values for 'm':

m = 1 or m = 2

So, it looks like we have two potential values for 'm' that could make the system have a solution. But we're not done yet! We need to check if these values actually work in the original system of equations.

Checking the Solutions for 'm'

It's crucial to plug these values of 'm' back into the original equations to make sure they actually lead to a consistent solution. Sometimes, a value might make the determinant zero, but it could still lead to a system with no solutions or infinitely many solutions that don't fit our (1, α, β) form.

Let's start with m = 1. If we substitute m = 1 into our simplified equations (after plugging in x = 1), we get:

1. (1 - 1)y + (2(1) - 3)z = 0 => -z = 0
2. (2(1) - 2)y + 2z = -1 => 2z = -1
3. (3(1) - 3)y + (1^2 - 1)z = -(1 + 1) => 0 = -2

From the first equation, we get z = 0. From the second equation, we get 2z = -1, which means z = -1/2. These two equations give us conflicting values for z, which means that m = 1 does not lead to a consistent solution. Also, the third equation 0 = -2 is a contradiction, confirming that m = 1 is not a valid solution.

Now, let's try m = 2. Substituting m = 2 into our simplified equations, we get:

1. (2 - 1)y + (2(2) - 3)z = 0 => y + z = 0
2. (2(2) - 2)y + 2z = -2 => 2y + 2z = -2
3. (3(2) - 3)y + (2^2 - 1)z = -(2 + 1) => 3y + 3z = -3

We can see that the second equation is just the first equation multiplied by 2, and the third equation is the first equation multiplied by 3. This means we essentially have only one independent equation: y + z = 0. This equation has infinitely many solutions, but we need to make sure at least one of them works with our original condition.

From y + z = 0, we have y = -z. We can choose any value for z, and we'll get a corresponding value for y. So, there are indeed solutions of the form (1, α, β) when m = 2.

Final Answer

After all that work, we've found that only one value of 'm' allows the system of equations to have at least one solution of the form (1, α, β). That value is m = 2.

So, the final answer is that there is only one value of m for which the system possesses at least one solution of the type (1, α, β).

Key Takeaways

• Determinants are your friends: When dealing with systems of linear equations, determinants can help you determine if solutions exist. If the determinant of the coefficient matrix is zero, there's a good chance you have a solution (or infinitely many!).
• Always check your solutions: Just because you find a potential value for a parameter doesn't mean it's a valid solution. Plug it back into the original equations to make sure it works.
• Don't be afraid to simplify: Linear equations can look scary, but breaking them down step by step makes the problem much more manageable. Simplify equations whenever you can to make your life easier.

Alright, guys, that's it for today's deep dive into linear equations! I hope you found this explanation helpful and that you're feeling more confident about tackling similar problems in the future. Remember, practice makes perfect, so keep working at it, and you'll become a linear equation whiz in no time! Until next time, keep those brains buzzing! 😉