Solving The System Of Equations X+ky + Kz = K For Different Values Of K

Hey guys! Let's dive into this interesting math problem where we need to figure out for which values of k the following system of equations has solutions:

• x + ky + kz = k
• x + y + z = k
• ky + 2z = k

This is a classic problem in linear algebra, and it involves understanding how the coefficients of the variables and the constants on the right-hand side interact to determine the existence and uniqueness of solutions. So, grab your thinking caps, and let’s get started!

Understanding the Problem

Before we jump into solving, let's break down what the question is really asking. We have three equations with three unknowns (x, y, and z), and a parameter k. The values of k will affect whether the system has:

• A unique solution (one set of x, y, z that satisfies all equations).
• Infinitely many solutions.
• No solution.

Our goal is to find the specific values of k that lead to these different scenarios. To do this, we’ll use techniques from linear algebra, such as Gaussian elimination and determinants.

Setting up the Augmented Matrix

The first step in solving this system is to represent it in matrix form. This makes it easier to manipulate the equations systematically. We'll create an augmented matrix, which includes the coefficients of x, y, and z, as well as the constants on the right side.

The augmented matrix for our system is:

|	1	  k	  k	|	 k	|
|	1	  1	  1	|	 k	|
|	0	  k	  2	|	 k	|

Each row represents one of our equations, and the vertical line separates the coefficient matrix from the constants. Now, we’ll use row operations to simplify this matrix.

Performing Gaussian Elimination

Gaussian elimination is a method for transforming a matrix into row-echelon form, which makes it easier to solve the system. We'll use elementary row operations, which include:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding a multiple of one row to another row.

Let's start by subtracting the first row from the second row (R2 = R2 - R1) to eliminate the x term in the second equation:

|	1	  k	  k	|	 k	|
|	0	 1-k	 1-k	|	 0	|
|	0	  k	  2	|	 k	|

Now, let's analyze the second row. We have (1-k)y + (1-k)z = 0. This gives us a crucial insight: if k ≠ 1, we can potentially simplify this further. But, let's consider the case when k = 1 separately, as it might lead to a special situation.

Case 1: k = 1

If k = 1, our augmented matrix becomes:

|	1	  1	  1	|	 1	|
|	0	  0	  0	|	 0	|
|	0	  1	  2	|	 1	|

Notice that the second row is all zeros, which means we have a redundant equation. We effectively have two equations with three unknowns:

• x + y + z = 1
• y + 2z = 1

This system has infinitely many solutions. We can express x and y in terms of z, or vice versa. So, k = 1 is one value we're looking for.

Case 2: k ≠ 1

Now, let's go back to our matrix after the first row operation, but this time, assume k ≠ 1:

|	1	  k	  k	|	 k	|
|	0	 1-k	 1-k	|	 0	|
|	0	  k	  2	|	 k	|

Since k ≠ 1, we can divide the second row by (1-k):

|	1	  k	  k	|	 k	|
|	0	  1	  1	|	 0	|
|	0	  k	  2	|	 k	|

Next, we'll eliminate the k in the third row by subtracting k times the second row from the third row (R3 = R3 - kR2):

|	1	  k	  k	|	 k	|
|	0	  1	  1	|	 0	|
|	0	  0	 2-k	|	 k	|

Now we have an upper triangular matrix, which is great! The last row gives us the equation (2-k)z = k. We need to consider two sub-cases here:

Sub-case 2.1: k = 2

If k = 2, the last row becomes 0z = 2, which is impossible. This means there is no solution when k = 2.

Sub-case 2.2: k ≠ 2

If k ≠ 2, we can solve for z: z = k / (2-k). Then we can back-substitute to find y and x. In this case, we have a unique solution.

Summarizing the Results

Alright, guys, we've done some serious matrix maneuvering! Let's recap what we've found:

• k = 1: The system has infinitely many solutions.
• k = 2: The system has no solution.
• k ≠ 1 and k ≠ 2: The system has a unique solution.

So, the different values of k that change the nature of the solutions are k = 1 and k = 2. For any other value of k, we get a unique solution.

Why This Matters

Understanding how systems of equations behave for different parameter values is super important in many areas, such as:

• Engineering: Designing structures or systems that need to meet certain conditions.
• Economics: Modeling markets and predicting economic behavior.
• Computer Graphics: Transforming objects in 3D space.

By understanding the conditions for unique solutions, infinite solutions, or no solutions, we can make informed decisions and avoid potential problems.

Final Thoughts

Solving systems of equations with parameters can seem daunting at first, but by breaking it down step by step and using techniques like Gaussian elimination, we can tackle even the trickiest problems. Remember, the key is to be systematic and consider all possible cases.

I hope this explanation helped you understand how to solve this type of problem. Keep practicing, and you'll become a master of linear equations in no time! Keep up the great work, guys! You've got this! Figuring out the values of k that make this happen involves some linear algebra magic, so let's get to it.

Step 1: Rewrite the System

First, let’s rewrite our system of equations to make it look a bit cleaner:

1. x + ky + kz = k
2. x + y + z = k
3. ky + 2z = k

Step 2: The Matrix Form

To really dive into this, we're going to turn these equations into matrix form. It’s like giving our problem a superhero outfit – suddenly, it’s ready to take on anything!

We create a coefficient matrix (A), a variable matrix (X), and a constant matrix (B):

A = | 1 k k | | 1 1 1 | | 0 k 2 |

X = | x | | y | | z |

B = | k | | k | | k |

So, our system is now AX = B.

Step 3: Determinant Time

The determinant of matrix A (det(A)) is super important. It tells us whether our system has a unique solution. If det(A) is not zero, we're in the clear for a unique solution. If it's zero, things get interesting – we might have infinitely many solutions or none at all.

Let’s calculate det(A):

det(A) = 1*(12 - 1k) - k*(12 - 01) + k*(1k - 01) = 2 - k - 2k + k^2 = k^2 - 3k + 2

Step 4: Finding Critical Values of k

We need to find when det(A) = 0 because that’s where our solution landscape changes. So, let's solve:

k^2 - 3k + 2 = 0 (k - 1)(k - 2) = 0

So, k = 1 and k = 2 are our critical values. These are the points where we need to investigate further.

Step 5: Diving into k = 1

Let's plug k = 1 back into our original equations:

1. x + y + z = 1
2. x + y + z = 1
3. y + 2z = 1

Notice that equations 1 and 2 are the same. We really have two equations with three unknowns. This usually means we’ve got infinitely many solutions. We can express x in terms of y and z (or vice versa), so there are a bunch of solutions that fit.

Step 6: Let's Check Out k = 2

Now, let’s see what happens when k = 2:

1. x + 2y + 2z = 2
2. x + y + z = 2
3. 2y + 2z = 2

This looks a bit tricky. Let's subtract equation 2 from equation 1:

(x + 2y + 2z) - (x + y + z) = 2 - 2 y + z = 0

But equation 3 simplifies to:

y + z = 1

Wait a minute! We’ve got y + z = 0 and y + z = 1. That’s a contradiction! This means when k = 2, we have no solution. The system is inconsistent.

Step 7: The Grand Finale

Alright, let's bring it all together:

• k = 1: We have infinitely many solutions.
• k = 2: We have no solution.
• k ≠ 1 and k ≠ 2: We have a unique solution. This is because det(A) is not zero, so we can solve for x, y, and z.

Wrapping It Up

So, there you have it! The system of equations behaves differently based on the value of k. We found the critical values by looking at the determinant and then checked those values in the original equations to see what happens.

Understanding these kinds of problems is super useful in all sorts of fields, from engineering to economics, because real-world systems often have parameters that change, and we need to know how those changes affect the solutions.

Keep up the great work, guys, and remember: math is like a puzzle, and every problem is a chance to find another piece!