Solving Permutations And Linear Equations A Step By Step Guide
Hey guys! Today, we're diving deep into two fascinating areas of mathematics permutations and systems of linear equations. We'll tackle a permutation problem that looks intimidating but is actually quite manageable, and then we'll conquer a system of equations using the inverse matrix method. So, grab your thinking caps, and let's get started!
Unraveling Permutations Finding the Value of 'n'
Let's kick things off by deciphering this permutation equation: 4 * nP3 = 5 * (n-1)P3. Permutations, as you know, deal with arranging items in a specific order. The notation nP3 represents the number of ways to arrange 3 items out of a set of n items. Our mission is to find the value of n that satisfies this equation. This section will guide you step-by-step through solving this permutation problem, ensuring you understand every twist and turn.
Breaking Down the Permutation Formula
Before we jump into solving the equation, let's refresh our understanding of the permutation formula. The number of permutations of r items chosen from a set of n items (where order matters) is given by: nP_r = n! / (n-r)!. Here, ! denotes the factorial, meaning the product of all positive integers up to that number (e.g., 5! = 5 * 4 * 3 * 2 * 1). Understanding this formula is crucial for tackling permutation problems effectively. It allows us to quantify the different ways we can arrange a subset of items from a larger set, considering the order of arrangement as significant.
Applying this formula to our equation, we get:
4 * [n! / (n-3)!] = 5 * [(n-1)! / (n-1-3)!]
Which simplifies to:
4 * [n! / (n-3)!] = 5 * [(n-1)! / (n-4)!]
Simplifying the Equation the Key to Unlocking 'n'
Now comes the fun part simplifying! We need to manipulate the factorials to make the equation easier to solve. Remember, n! = n * (n-1) * (n-2) * ... * 2 * 1. We can use this property to expand the factorials in our equation. The goal here is to simplify the factorials and reduce the complexity of the equation. By expanding the factorials, we can identify common terms that can be canceled out, leading to a more manageable expression. This step is essential for isolating 'n' and finding its value.
Let's expand the factorials:
4 * [n * (n-1) * (n-2) * (n-3)! / (n-3)!] = 5 * [(n-1) * (n-2) * (n-3) * (n-4)! / (n-4)!]
Notice that (n-3)! and (n-4)! appear in both the numerator and denominator, allowing us to cancel them out:
4 * n * (n-1) * (n-2) = 5 * (n-1) * (n-2) * (n-3)
Solving the Quadratic Equation Unveiling the Solution
We're getting closer! Now, we can further simplify by dividing both sides by (n-1) * (n-2), but we need to be cautious. We must consider the cases where n = 1 or n = 2 separately, as dividing by zero is a big no-no in mathematics. This step highlights the importance of paying attention to potential singularities or undefined cases when manipulating equations. Overlooking these cases can lead to incorrect solutions.
Assuming n ≠1 and n ≠2, we divide both sides by (n-1) * (n-2):
4n = 5(n-3)
Expanding and rearranging, we get:
4n = 5n - 15
n = 15
Now, let's check if n = 1 or n = 2 are solutions. If we substitute n = 1 or n = 2 into the original equation, we'll find that they don't satisfy the equation. Therefore, the only valid solution is n = 15.
Verifying the Solution Ensuring Accuracy
It's always a good practice to verify our solution. Let's plug n = 15 back into the original equation:
4 * 15P3 = 5 * (15-1)P3
4 * (15! / 12!) = 5 * (14! / 11!)
4 * (15 * 14 * 13) = 5 * (14 * 13 * 12)
4 * 2730 = 5 * 2184
10920 = 10920
The equation holds true! So, we've successfully found the value of n.
Cracking Systems of Linear Equations The Inverse Matrix Method
Now, let's switch gears and tackle a system of linear equations using the inverse matrix method. This method is a powerful tool for solving systems of equations, especially when dealing with larger systems. The inverse matrix method provides a systematic approach to finding the solutions, leveraging matrix algebra to simplify the process. It's a valuable technique for various applications in mathematics, science, and engineering.
We have the following system of equations:
x + 2y - z = 9
2x - y + 3z = -2
3x + 2y + 3z = 9
Transforming the System into Matrix Form A Foundation for Solution
The first step is to represent this system of equations in matrix form. We can write it as AX = B, where:
Ais the coefficient matrix:
| 1 2 -1 |
| 2 -1 3 |
| 3 2 3 |
Xis the variable matrix:
| x |
| y |
| z |
Bis the constant matrix:
| 9 |
| -2 |
| 9 |
Finding the Inverse of the Coefficient Matrix The Core of the Method
To solve for X, we need to find the inverse of matrix A, denoted as A^-1. The inverse matrix, when multiplied by the original matrix, results in the identity matrix. Finding the inverse matrix is a crucial step in this method, as it allows us to isolate the variable matrix and solve for the unknowns. There are several methods to find the inverse, including the adjugate method and Gaussian elimination. We'll use the adjugate method here.
First, we find the determinant of A, denoted as |A|:
|A| = 1 * (-1 * 3 - 3 * 2) - 2 * (2 * 3 - 3 * 3) + (-1) * (2 * 2 - (-1) * 3)
|A| = 1 * (-9) - 2 * (-3) + (-1) * (7)
|A| = -9 + 6 - 7
|A| = -10
Since the determinant is non-zero, the inverse exists. This is a critical check because a matrix is only invertible if its determinant is not zero. A zero determinant indicates that the matrix is singular, and the system of equations may have either no solution or infinitely many solutions.
Next, we find the adjugate of A, which is the transpose of the cofactor matrix. The cofactor of an element is found by multiplying the minor (the determinant of the submatrix formed by deleting the row and column of the element) by (-1)^(row+column). The adjugate matrix is a crucial component in calculating the inverse, as it encapsulates the information about the cofactors of the original matrix.
The cofactor matrix is:
| -9 3 7 |
| -8 6 4 |
| 5 -5 -5 |
The adjugate of A is the transpose of the cofactor matrix:
| -9 -8 5 |
| 3 6 -5 |
| 7 4 -5 |
Now, we can find the inverse of A:
A^-1 = adj(A) / |A|
A^-1 = (1 / -10) * adj(A)
A^-1 =
| 9/10 8/10 -5/10 |
| -3/10 -6/10 5/10 |
| -7/10 -4/10 5/10 |
Solving for the Variables The Final Step
Now that we have A^-1, we can solve for X:
X = A^-1 * B
| x |
| y |
| z |
=
| 9/10 8/10 -5/10 |
| -3/10 -6/10 5/10 |
| -7/10 -4/10 5/10 |
*
| 9 |
| -2 |
| 9 |
Performing the matrix multiplication, we get:
| x |
| y |
| z |
=
| (9/10)*9 + (8/10)*(-2) + (-5/10)*9 |
| (-3/10)*9 + (-6/10)*(-2) + (5/10)*9 |
| (-7/10)*9 + (-4/10)*(-2) + (5/10)*9 |
| x |
| y |
| z |
=
| 81/10 - 16/10 - 45/10 |
| -27/10 + 12/10 + 45/10 |
| -63/10 + 8/10 + 45/10 |
| x |
| y |
| z |
=
| 20/10 |
| 30/10 |
| -10/10 |
| x |
| y |
| z |
=
| 2 |
| 3 |
| -1 |
Therefore, the solution is x = 2, y = 3, and z = -1.
Verifying the Solution Ensuring Accuracy
Let's substitute these values back into the original equations to verify our solution:
2 + 2(3) - (-1) = 9
2 + 6 + 1 = 9
9 = 9
2(2) - 3 + 3(-1) = -2
4 - 3 - 3 = -2
-2 = -2
3(2) + 2(3) + 3(-1) = 9
6 + 6 - 3 = 9
9 = 9
The solution satisfies all three equations! We've successfully solved the system of linear equations using the inverse matrix method.
Conclusion Mastering Mathematical Techniques
So, there you have it! We've conquered a permutation problem and solved a system of linear equations using the inverse matrix method. Remember, practice is key to mastering these techniques. Keep exploring different problems and challenging yourself. You've got this!