Calculating Determinant And Inverse Of Matrix K Using Cofactor Expansion
Hey guys! Today, we're diving into a fun math problem: calculating the determinant and inverse of a matrix using cofactor expansion. It might sound intimidating, but trust me, we'll break it down step by step. We're going to tackle this matrix:
So, grab your calculators (or your brainpower!) and let's get started!
Understanding the Basics
Before we jump into the calculations, let's make sure we're all on the same page with some key concepts. We will understand the importance of determinants and cofactor expansion, and how they are our tools for solving this problem. This section is crucial, so stick with me!
What is a Determinant?
In simple terms, the determinant is a special number that can be calculated from a square matrix. Think of it as a secret code that tells us important things about the matrix. For example, the determinant can tell us if a matrix has an inverse (more on that later!) and helps in solving systems of linear equations.
The determinant of a matrix A is often written as det(A) or |A|. For a 2x2 matrix, it's calculated like this:
If $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, then det(A) = ad - bc
But what about larger matrices like our 4x4 matrix K? That's where cofactor expansion comes in!
What is Cofactor Expansion?
Cofactor expansion is a method for calculating the determinant of a matrix of any size. The basic idea is to break down the larger matrix into smaller matrices (minors) and use their determinants to find the determinant of the original matrix. It might seem complicated, but it's actually a pretty systematic process.
The cofactor expansion involves choosing a row or column, and then for each element in that row or column, we calculate its cofactor. The cofactor is a signed minor, where the sign depends on the position of the element. We then multiply each element by its cofactor and sum the results. The main advantage of cofactor expansion is its ability to handle larger matrices by breaking them down into smaller, more manageable components.
Why is This Important?
Understanding determinants and cofactor expansion is crucial in various fields like engineering, physics, and computer science. They're used in solving linear systems, finding eigenvalues, and even in computer graphics for transformations. So, what we're learning here isn't just abstract math – it has real-world applications!
Now that we have a solid understanding of the basics, let's move on to the fun part: actually calculating the determinant of our matrix K.
Calculating the Determinant of Matrix K
Alright, let's get our hands dirty and calculate the determinant of matrix K. Remember our matrix?
We're going to use cofactor expansion, and a clever strategy to make our lives easier. The strategy for cofactor expansion can significantly reduce the computational effort involved, especially for matrices with many zero elements.
Choosing the Right Row or Column
The key to efficient cofactor expansion is choosing a row or column with as many zeros as possible. Why? Because when we multiply an element by its cofactor, if the element is zero, the whole term becomes zero! This simplifies our calculations significantly.
Looking at matrix K, we can see that the third and fourth columns have two zeros each. Let's choose the fourth column for our expansion. It has the elements 0, 0, 0, and a. This means we only have one non-zero element to worry about!
Applying Cofactor Expansion
The determinant of K can be calculated by expanding along the fourth column as follows:
det(K) = (0)C₁₄ + (0)C₂₄ + (0)C₃₄ + (a)C₄₄
Where Cᵢⱼ represents the cofactor of the element in the i-th row and j-th column.
Notice how the first three terms are zero? That's the power of choosing the right column! We're left with:
det(K) = aC₄₄
So, we only need to calculate C₄₄, which is the cofactor of the element a in the bottom right corner.
Calculating the Cofactor C₄₄
The cofactor C₄₄ is calculated as (-1)^(4+4) times the determinant of the 3x3 matrix formed by removing the fourth row and fourth column from K. This 3x3 matrix is:
So, C₄₄ = (-1)⁸ * det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -1 & 5 & 7 \end{pmatrix}$
Since (-1)⁸ = 1, we just need to calculate the determinant of this 3x3 matrix. We can use cofactor expansion again! This time, let's expand along the third column, which has two zeros:
det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -1 & 5 & 7 \end{pmatrix}$ = (0)C₁₃ + (0)C₂₃ + (7)C₃₃
Again, two terms are zero, leaving us with:
= 7 * C₃₃
C₃₃ is (-1)^(3+3) times the determinant of the 2x2 matrix formed by removing the third row and third column:
So, C₃₃ = (-1)⁶ * det$\begin{pmatrix} 1 & b \ 3 & 6 \end{pmatrix}$ = (1) * (16 - 3b) = 6 - 3b
Putting It All Together
Now we can substitute back to find C₄₄:
C₄₄ = 7 * (6 - 3b) = 42 - 21b
And finally, the determinant of K:
det(K) = aC₄₄ = a(42 - 21b) = 21a(2 - b)
The determinant of matrix K is 21a(2 - b). Isn't it cool how we broke down a 4x4 determinant into smaller, manageable pieces? Next, we'll tackle the inverse, but first, let's take a breather and appreciate our progress!
Finding the Inverse of Matrix K
Okay, now that we've conquered the determinant, let's move on to finding the inverse of matrix K. This is where things get a little more involved, but don't worry, we'll take it one step at a time. Understanding the process of finding the inverse is a crucial skill in linear algebra, and it builds directly on our determinant calculation.
When Does a Matrix Have an Inverse?
First things first, a matrix has an inverse if and only if its determinant is non-zero. Remember, the determinant is like a secret code – if it's zero, it means the matrix is singular and doesn't have an inverse. It's like trying to divide by zero; it just doesn't work!
In our case, det(K) = 21a(2 - b). So, K has an inverse if 21a(2 - b) ≠ 0. This means a ≠ 0 and b ≠ 2.
The Adjugate Matrix
The adjugate (or adjoint) of a matrix is a key component in finding the inverse. The adjugate of a matrix is the transpose of its cofactor matrix. Sounds like a mouthful, right? Let's break it down:
- Cofactor Matrix: We calculate the cofactor for each element in the original matrix, just like we did for the determinant. We arrange these cofactors into a new matrix.
- Transpose: The transpose of a matrix is obtained by swapping its rows and columns. So, the first row becomes the first column, the second row becomes the second column, and so on.
If we denote the cofactor matrix of K as C, then the adjugate of K, written as adj(K), is Cᵀ (the transpose of C).
The Inverse Formula
Now for the grand finale! The inverse of a matrix K, denoted as K⁻¹, is given by the following formula:
K⁻¹ = (1 / det(K)) * adj(K)
In other words, we divide the adjugate of K by the determinant of K. This formula elegantly combines the concepts we've learned so far.
Steps to Find the Inverse
Let's summarize the steps we need to take to find the inverse of matrix K:
- Calculate the determinant of K: We already did this! det(K) = 21a(2 - b)
- Find the cofactor matrix of K: This is the most tedious part, but we'll break it down. It involves calculating the cofactor for each of the 16 elements in K.
- Find the adjugate of K: Transpose the cofactor matrix.
- Apply the inverse formula: Multiply the adjugate by 1 / det(K).
Now, let's roll up our sleeves and start calculating those cofactors!
Calculating the Cofactor Matrix
This is the most computationally intensive part of finding the inverse, but we'll tackle it systematically. We need to find the cofactor for each of the 16 elements in matrix K. Remember, the process for finding each cofactor is consistent, making it manageable even for a 4x4 matrix.
Let's denote the cofactor of the element in the i-th row and j-th column as Cᵢⱼ. We'll go through each cofactor one by one.
Cofactor C₁₁
C₁₁ is (-1)^(1+1) times the determinant of the 3x3 matrix formed by removing the first row and first column:
C₁₁ = (-1)² * det$\begin{pmatrix} 6 & 0 & 0 \ 5 & 7 & 0 \ -2 & 9 & a \end{pmatrix}$
We can expand this determinant along the first row:
= 1 * [6 * det$\begin{pmatrix} 7 & 0 \ 9 & a \end{pmatrix}$] = 6 * (7a - 09) = 42a
So, C₁₁ = 42a
Cofactor C₁₂
C₁₂ is (-1)^(1+2) times the determinant of the 3x3 matrix formed by removing the first row and second column:
C₁₂ = (-1)³ * det$\begin{pmatrix} 3 & 0 & 0 \ -1 & 7 & 0 \ -2 & 9 & a \end{pmatrix}$
= -1 * [3 * det$\begin{pmatrix} 7 & 0 \ 9 & a \end{pmatrix}$] = -3 * (7a - 09) = -21a
So, C₁₂ = -21a
Cofactor C₁₃
C₁₃ is (-1)^(1+3) times the determinant of the 3x3 matrix formed by removing the first row and third column:
C₁₃ = (-1)⁴ * det$\begin{pmatrix} 3 & 6 & 0 \ -1 & 5 & 0 \ -2 & -2 & a \end{pmatrix}$
We can expand this determinant along the third column:
= 1 * [a * det$\begin{pmatrix} 3 & 6 \ -1 & 5 \end{pmatrix}$] = a * (35 - 6(-1)) = 21a
So, C₁₃ = 21a
Cofactor C₁₄
C₁₄ is (-1)^(1+4) times the determinant of the 3x3 matrix formed by removing the first row and fourth column:
C₁₄ = (-1)⁵ * det$\begin{pmatrix} 3 & 6 & 0 \ -1 & 5 & 7 \ -2 & -2 & 9 \end{pmatrix}$
We can expand this determinant along the first row:
= -1 * [3 * det$\begin{pmatrix} 5 & 7 \ -2 & 9 \end{pmatrix}$ - 6 * det$\begin{pmatrix} -1 & 7 \ -2 & 9 \end{pmatrix}$] = -1 * [3 * (59 - 7(-2)) - 6 * (-19 - 7(-2))] = -1 * [3 * 59 - 6 * 5] = -147
So, C₁₄ = -147
We'll continue this process for the remaining cofactors. It's a bit lengthy, but it's crucial for finding the inverse. We're essentially systematically breaking down the problem, making it more manageable.
Cofactor C₂₁
C₂₁ = (-1)^(2+1) * det$\begin{pmatrix} b & 0 & 0 \ 5 & 7 & 0 \ -2 & 9 & a \end{pmatrix}$ = -1 * [b * det$\begin{pmatrix} 7 & 0 \ 9 & a \end{pmatrix}$] = -7ab
Cofactor C₂₂
C₂₂ = (-1)^(2+2) * det$\begin{pmatrix} 1 & 0 & 0 \ -1 & 7 & 0 \ -2 & 9 & a \end{pmatrix}$ = 1 * [1 * det$\begin{pmatrix} 7 & 0 \ 9 & a \end{pmatrix}$] = 7a
Cofactor C₂₃
C₂₃ = (-1)^(2+3) * det$\begin{pmatrix} 1 & b & 0 \ -1 & 5 & 0 \ -2 & -2 & a \end{pmatrix}$ = -1 * [a * det$\begin{pmatrix} 1 & b \ -1 & 5 \end{pmatrix}$] = -a * (5 + b)
Cofactor C₂₄
C₂₄ = (-1)^(2+4) * det$\begin{pmatrix} 1 & b & 0 \ -1 & 5 & 7 \ -2 & -2 & 9 \end{pmatrix}$ = 1 * [1 * det$\begin{pmatrix} 5 & 7 \ -2 & 9 \end{pmatrix}$ - b * det$\begin{pmatrix} -1 & 7 \ -2 & 9 \
Cofactor C₃₁
C₃₁ = (-1)^(3+1) * det$\begin{pmatrix} b & 0 & 0 \ 6 & 0 & 0 \ -2 & 9 & a \end{pmatrix}$ = 1 * [b * det$\begin{pmatrix} 0 & 0 \ 9 & a \end{pmatrix}$- 0 + 0] = 0
Cofactor C₃₂
C₃₂ = (-1)^(3+2) * det$\begin{pmatrix} 1 & 0 & 0 \ 3 & 0 & 0 \ -2 & 9 & a \end{pmatrix}$ = -1 * [1 * det$\begin{pmatrix} 0 & 0 \ 9 & a \end{pmatrix}$- 0 + 0] = 0
Cofactor C₃₃
C₃₃ = (-1)^(3+3) * det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -2 & -2 & a \end{pmatrix}$ = 1 * [a * det$\begin{pmatrix} 1 & b \ 3 & 6 \end{pmatrix}$] = a * (6 - 3b) = 3a(2-b)
Cofactor C₃₄
C₃₄ = (-1)^(3+4) * det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -2 & -2 & 9 \end{pmatrix}$ = -1 * [9 * det$\begin{pmatrix} 1 & b \ 3 & 6 \end{pmatrix}$] = -9 * (6 - 3b) = -27(2-b)
Cofactor C₄₁
C₄₁ = (-1)^(4+1) * det$\begin{pmatrix} b & 0 & 0 \ 6 & 0 & 0 \ 5 & 7 & 0 \end{pmatrix}$ = -1 * [b * det$\begin{pmatrix} 0 & 0 \ 7 & 0 \end{pmatrix}$- 0 + 0] = 0
Cofactor C₄₂
C₄₂ = (-1)^(4+2) * det$\begin{pmatrix} 1 & 0 & 0 \ 3 & 0 & 0 \ -1 & 7 & 0 \end{pmatrix}$ = 1 * [1 * det$\begin{pmatrix} 0 & 0 \ 7 & 0 \end{pmatrix}$- 0 + 0] = 0
Cofactor C₄₃
C₄₃ = (-1)^(4+3) * det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -1 & 5 & 7 \end{pmatrix}$ = -1 * [7 * det$\begin{pmatrix} 1 & b \ 3 & 6 \end{pmatrix}$] = -7 * (6 - 3b) = -21(2-b)
Cofactor C₄₄
C₄₄ = (-1)^(4+4) * det$\begin{pmatrix} 1 & b & 0 \ 3 & 6 & 0 \ -1 & 5 & 7 \end{pmatrix}$ = 1 * [7 * det$\begin{pmatrix} 1 & b \ 3 & 6 \end{pmatrix}$] = 7 * (6 - 3b) = 21(2-b)
Constructing the Cofactor Matrix
Now that we've calculated all the cofactors, we can assemble the cofactor matrix:
C = $\begin{pmatrix} 42a & -21a & 21a & -147 \ -7ab & 7a & -a(5 + b) & -9(5-b) \ 0 & 0 & 3a(2-b) & -27(2-b) \ 0 & 0 & -21(2-b) & 21(2-b) \end{pmatrix}$
Finding the Adjugate and the Inverse
Almost there, guys! We've done the heavy lifting by calculating the cofactor matrix. Now, let's find the adjugate and finally, the inverse of matrix K. This stage is about putting all the pieces together and applying the formulas we discussed earlier.
The Adjugate Matrix
Remember, the adjugate is the transpose of the cofactor matrix. So, we swap the rows and columns of C:
adj(K) = Cᵀ = $\begin{pmatrix} 42a & -7ab & 0 & 0 \ -21a & 7a & 0 & 0 \ 21a & -a(5 + b) & 3a(2-b) & -21(2-b) \ -147 & -9(5-b) & -27(2-b) & 21(2-b) \end{pmatrix}$
The Inverse Matrix
Now we use the formula K⁻¹ = (1 / det(K)) * adj(K). We know det(K) = 21a(2 - b), so:
K⁻¹ = (1 / (21a(2 - b))) * $\begin{pmatrix} 42a & -7ab & 0 & 0 \ -21a & 7a & 0 & 0 \ 21a & -a(5 + b) & 3a(2-b) & -21(2-b) \ -147 & -9(5-b) & -27(2-b) & 21(2-b) \end{pmatrix}$
We can simplify this by multiplying each element of the adjugate by 1 / (21a(2 - b)):
K⁻¹ = $\begin{pmatrix} 2/(2-b) & -b/(3a(2-b)) & 0 & 0 \ -1/(2-b) & 1/(3(2-b)) & 0 & 0 \ 1/(2-b) & -(5 + b)/(21(2-b)) & 1/7 & -1/a \ -7/(a(2-b)) & -3(5-b)/(7a(2-b)) & -9/(7a) & 1/a \end{pmatrix}$
And there you have it! This is the inverse of matrix K. I know it was a long journey, but we did it! We successfully calculated the determinant and the inverse using cofactor expansion. Give yourselves a pat on the back!
Conclusion
So guys, we've reached the end of our mathematical adventure today! We started with a 4x4 matrix, and we conquered it by finding both its determinant and its inverse using cofactor expansion. We've seen how breaking down complex problems into smaller steps can make them much more manageable.
We've covered a lot of ground: understanding determinants, mastering cofactor expansion, and navigating the process of finding an inverse. These are powerful tools in linear algebra and have applications in various fields.
Remember, practice makes perfect! The more you work with matrices and determinants, the more comfortable you'll become with these concepts. So, keep exploring, keep learning, and most importantly, keep having fun with math!
If you have any questions or want to dive deeper into this topic, feel free to ask. Until next time, keep those matrices in order! 😉
