Matriks: Determinan, Kofaktor, Adjoin, & Invers

Hey guys! Let's dive into the fascinating world of matrices today. Specifically, we're going to tackle a problem that involves finding the determinant, cofactors, the adjoint, and the inverse of a given matrix. These are fundamental concepts in linear algebra and super useful in various fields, from computer graphics to engineering. So, grab your calculators, sharpen your pencils, and let's get this done!

Our mission, should we choose to accept it, is to work with the following matrix, let's call it matrix A:

$A = \begin{bmatrix} 3 & 6 & 4 \\ 1 & 5 & 8 \\ 2 & 9 & 7 \end{bmatrix}$

We need to find four key things:

1. Determinant of A (Det(A))
2. Cofactors of A
3. Adjoint of A (Adj(A))
4. Inverse of A ($A^{-1}$)

Ready to break it down step-by-step?

1. Det (A): Unlocking the Determinant's Secrets

The determinant is a scalar value that can be computed from the elements of a square matrix. It tells us a lot about the matrix, like whether it's invertible or not. For a 3x3 matrix like our matrix A, there are a few ways to calculate the determinant. We'll use the cofactor expansion method here, which also sets us up nicely for finding the cofactors later!

To find the determinant of A, denoted as Det(A) or |A|, we can expand along any row or any column. Let's choose to expand along the first row, as it's often the most straightforward. The formula involves multiplying each element of the first row by its corresponding cofactor and summing them up. The formula looks like this:

Det(A) = $a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

Where $a_{ij}$ is the element in the i-th row and j-th column, and $C_{ij}$ is its cofactor. But wait, we haven't calculated the cofactors yet! Don't worry, the determinant calculation itself uses a simpler form involving minors. A minor, $M_{ij}$, is the determinant of the submatrix formed by deleting the i-th row and j-th column. The cofactor $C_{ij}$ is related to the minor by $C_{ij} = (-1)^{i+j}M_{ij}$.

So, let's first find the minors for the first row:

• Minor $M_{11}$: Delete row 1 and column 1. The remaining submatrix is $\begin{bmatrix} 5 & 8 \\ 9 & 7 \end{bmatrix}$. Its determinant is $(5 \times 7) - (8 \times 9) = 35 - 72 = -37$.
• Minor $M_{12}$: Delete row 1 and column 2. The remaining submatrix is $\begin{bmatrix} 1 & 8 \\ 2 & 7 \end{bmatrix}$. Its determinant is $(1 \times 7) - (8 \times 2) = 7 - 16 = -9$.
• Minor $M_{13}$: Delete row 1 and column 3. The remaining submatrix is $\begin{bmatrix} 1 & 5 \\ 2 & 9 \end{bmatrix}$. Its determinant is $(1 \times 9) - (5 \times 2) = 9 - 10 = -1$.

Now, let's calculate the cofactors for the first row using the formula $C_{ij} = (-1)^{i+j}M_{ij}$:

• Cofactor $C_{11}$: $(-1)^{1+1}M_{11} = (+1) \times (-37) = -37$.
• Cofactor $C_{12}$: $(-1)^{1+2}M_{12} = (-1) \times (-9) = 9$.
• Cofactor $C_{13}$: $(-1)^{1+3}M_{13} = (+1) \times (-1) = -1$.

Finally, we can calculate the determinant of A using the elements of the first row and their cofactors:

Det(A) = $a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$

Det(A) = $(3 \times -37) + (6 \times 9) + (4 \times -1)$

Det(A) = $-111 + 54 - 4$

Det(A) = $-61$

So, the determinant of matrix A is -61. Since the determinant is non-zero, we know that matrix A is invertible, and we can proceed to find its inverse!

Pro-tip: You can calculate the determinant by expanding along any row or column. Try expanding along a different row or column to double-check your answer. It should always yield the same result!

2. Kofaktor (A): Building the Cofactor Matrix

Alright guys, now that we've dipped our toes into calculating cofactors for the first row, let's get the cofactors of A for the entire matrix. Remember, the cofactor $C_{ij}$ is calculated as $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$. We've already found the minors and cofactors for the first row. Let's find the rest!

We need to calculate 9 cofactors in total for a 3x3 matrix.

• Cofactors for Row 1 (Already calculated):

  • $C_{11} = -37$
  • $C_{12} = 9$
  • $C_{13} = -1$

• Cofactors for Row 2:

  • Minor $M_{21}$: Delete row 2, column 1. Submatrix: $\begin{bmatrix} 6 & 4 \\ 9 & 7 \end{bmatrix}$. Determinant: $(6 \times 7) - (4 \times 9) = 42 - 36 = 6$.
    • $C_{21} = (-1)^{2+1}M_{21} = (-1) \times 6 = -6$.
  • Minor $M_{22}$: Delete row 2, column 2. Submatrix: $\begin{bmatrix} 3 & 4 \\ 2 & 7 \end{bmatrix}$. Determinant: $(3 \times 7) - (4 \times 2) = 21 - 8 = 13$.
    • $C_{22} = (-1)^{2+2}M_{22} = (+1) \times 13 = 13$.
  • Minor $M_{23}$: Delete row 2, column 3. Submatrix: $\begin{bmatrix} 3 & 6 \\ 2 & 9 \end{bmatrix}$. Determinant: $(3 \times 9) - (6 \times 2) = 27 - 12 = 15$.
    • $C_{23} = (-1)^{2+3}M_{23} = (-1) \times 15 = -15$.

• Cofactors for Row 3:

  • Minor $M_{31}$: Delete row 3, column 1. Submatrix: $\begin{bmatrix} 6 & 4 \\ 5 & 8 \end{bmatrix}$. Determinant: $(6 \times 8) - (4 \times 5) = 48 - 20 = 28$.
    • $C_{31} = (-1)^{3+1}M_{31} = (+1) \times 28 = 28$.
  • Minor $M_{32}$: Delete row 3, column 2. Submatrix: $\begin{bmatrix} 3 & 4 \\ 1 & 8 \end{bmatrix}$. Determinant: $(3 \times 8) - (4 \times 1) = 24 - 4 = 20$.
    • $C_{32} = (-1)^{3+2}M_{32} = (-1) \times 20 = -20$.
  • Minor $M_{33}$: Delete row 3, column 3. Submatrix: $\begin{bmatrix} 3 & 6 \\ 1 & 5 \end{bmatrix}$. Determinant: $(3 \times 5) - (6 \times 1) = 15 - 6 = 9$.
    • $C_{33} = (-1)^{3+3}M_{33} = (+1) \times 9 = 9$.

Now, we can assemble these cofactors into a matrix, which we'll call the Cofactor Matrix (C):

$C = \begin{bmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix} = \begin{bmatrix} -37 & 9 & -1 \\ -6 & 13 & -15 \\ 28 & -20 & 9 \end{bmatrix}$

This Cofactor Matrix is a crucial stepping stone to finding the adjoint matrix. Pretty cool, right?

3. Adj (A): The Adjoint Matrix Revealed

Next up, we're going to find the Adjoint of A (Adj(A)). The adjoint matrix is simply the transpose of the cofactor matrix. Transposing a matrix means swapping its rows and columns. So, the first row of the cofactor matrix becomes the first column of the adjoint matrix, the second row becomes the second column, and so on.

Let's take our Cofactor Matrix C that we just calculated:

$C = \begin{bmatrix} -37 & 9 & -1 \\ -6 & 13 & -15 \\ 28 & -20 & 9 \end{bmatrix}$

To find the Adjoint matrix, Adj(A), we transpose C:

Adj(A) = $C^T = \begin{bmatrix} -37 & -6 & 28 \\ 9 & 13 & -20 \\ -1 & -15 & 9 \end{bmatrix}$

And there you have it! The adjoint matrix of A. This is the final piece we need to calculate the inverse of matrix A. Making progress, team!

Key takeaway: Don't forget that the adjoint is the transpose of the cofactor matrix. It's an easy step to miss if you're rushing!

4. $A^{-1}$ (Invers): The Grand Finale - Calculating the Inverse

Finally, guys, we've reached the grand finale: calculating the Inverse of A ($A^{-1}$)! The formula for the inverse of a matrix A is elegantly simple, provided the determinant is not zero (which we've already confirmed for our matrix A):

$A^{-1} = \frac{1}{Det(A)} \times Adj(A)$

We have everything we need: the determinant of A (Det(A) = -61) and the adjoint of A (Adj(A)).

$Adj(A) = \begin{bmatrix} -37 & -6 & 28 \\ 9 & 13 & -20 \\ -1 & -15 & 9 \end{bmatrix}$

Now, let's plug these values into the formula:

$A^{-1} = \frac{1}{-61} \times \begin{bmatrix} -37 & -6 & 28 \\ 9 & 13 & -20 \\ -1 & -15 & 9 \end{bmatrix}$

To get the final inverse matrix, we multiply each element of the adjoint matrix by the scalar $\frac{1}{-61}$:

$A^{-1} = \begin{bmatrix} \frac{-37}{-61} & \frac{-6}{-61} & \frac{28}{-61} \\ \frac{9}{-61} & \frac{13}{-61} & \frac{-20}{-61} \\ \frac{-1}{-61} & \frac{-15}{-61} & \frac{9}{-61} \end{bmatrix}$

Simplifying the fractions (and noting that a negative divided by a negative is positive):

$A^{-1} = \begin{bmatrix} \frac{37}{61} & \frac{6}{61} & -\frac{28}{61} \\ -\frac{9}{61} & -\frac{13}{61} & \frac{20}{61} \\ \frac{1}{61} & \frac{15}{61} & -\frac{9}{61} \end{bmatrix}$

And there you have it – the inverse of matrix A! This means that if you multiply matrix A by this inverse matrix ($A \times A^{-1}$ or $A^{-1} \times A$), you should get the identity matrix (a matrix with 1s on the diagonal and 0s elsewhere). You can perform this multiplication as a final check to ensure your calculations are correct.

Double-checking is key: Always multiply $A \times A^{-1}$ to verify you get the identity matrix. This catches any arithmetic errors you might have made along the way.

Conclusion: Mastering Matrix Operations

So, we've successfully navigated through finding the determinant, cofactors, adjoint, and inverse of matrix A. These steps, while sometimes tedious, are fundamental to understanding how matrices behave and are applied in higher mathematics and various scientific disciplines. Remember the process: calculate minors, use them to find cofactors, transpose the cofactor matrix to get the adjoint, and finally, multiply the adjoint by $1/Det(A)$ to find the inverse. Keep practicing these calculations, and soon you'll be a matrix manipulation pro!

Keep exploring, keep calculating, and never hesitate to ask questions. Happy matrixing, everyone!