Determinant Calculation: A Step-by-Step Guide

Calculating the determinant of a 3x3 matrix might seem daunting at first, but with a systematic approach, it becomes a manageable task. In this article, we'll walk you through the process of calculating the determinant of the following matrix:

$$\begin{bmatrix} 1 & -3 & 3 \\ 5 & 0 & 4 \\ 2 & -1 & 6 \end{bmatrix}$$

By the end of this guide, you'll not only know the answer but also understand the method behind it.

Understanding Determinants

Before diving into the calculation, let's briefly discuss what a determinant is. A determinant is a scalar value that can be computed from the elements of a square matrix. It provides important information about the matrix, such as whether the matrix is invertible (i.e., whether it has an inverse). A non-zero determinant indicates that the matrix is invertible, while a zero determinant means it is not.

For a 2x2 matrix, the determinant is calculated as follows:

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$$

However, for a 3x3 matrix, the calculation is a bit more involved, but don't worry, we'll break it down step by step.

Methods for Calculating the Determinant of a 3x3 Matrix

There are a couple of common methods for calculating the determinant of a 3x3 matrix. Here, we'll focus on the cofactor expansion method, which is widely used and relatively straightforward.

Cofactor Expansion Method

The cofactor expansion method involves expanding the determinant along a row or a column. Let's expand along the first row for our example matrix. The formula for the determinant of a 3x3 matrix using cofactor expansion along the first row is:

$$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = a \begin{vmatrix} e & f \\ h & i \end{vmatrix} - b \begin{vmatrix} d & f \\ g & i \end{vmatrix} + c \begin{vmatrix} d & e \\ g & h \end{vmatrix}$$

Notice the alternating signs (+, -, +) and how each element in the first row is multiplied by the determinant of the 2x2 matrix that remains after removing the row and column containing that element. These 2x2 determinants are called minors, and the minors multiplied by the appropriate sign are called cofactors.

Step-by-Step Calculation

Now, let's apply the cofactor expansion method to our matrix:

$$\begin{bmatrix} 1 & -3 & 3 \\ 5 & 0 & 4 \\ 2 & -1 & 6 \end{bmatrix}$$

1. Expand along the first row:

   $$$$

det = 1 \begin{vmatrix} 0 & 4 \ -1 & 6 \end{vmatrix} - (-3) \begin{vmatrix} 5 & 4 \ 2 & 6 \end{vmatrix} + 3 \begin{vmatrix} 5 & 0 \ 2 & -1 \end{vmatrix} $

2. Calculate the 2x2 determinants:

   • $$\begin{vmatrix} 0 & 4 \\ -1 & 6 \end{vmatrix} = (0 \cdot 6) - (4 \cdot -1) = 0 + 4 = 4$$

   • $$\begin{vmatrix} 5 & 4 \\ 2 & 6 \end{vmatrix} = (5 \cdot 6) - (4 \cdot 2) = 30 - 8 = 22$$

   • $$\begin{vmatrix} 5 & 0 \\ 2 & -1 \end{vmatrix} = (5 \cdot -1) - (0 \cdot 2) = -5 - 0 = -5$$

3. Substitute the 2x2 determinants back into the equation:

   $$$$

det = 1(4) - (-3)(22) + 3(-5) $

4. Simplify:

   $$$$

det = 4 + 66 - 15 $

5. Calculate the final result:

   $$$$

det = 55$

Therefore, the determinant of the given matrix is 55.

Alternative Method: Diagonal Method (Sarrus' Rule)

Another method, specifically for 3x3 matrices, is the diagonal method, sometimes called Sarrus' Rule. This method provides a more visual way to calculate the determinant. You rewrite the first two columns of the matrix to the right of the matrix, then sum the products of the diagonals going from upper-left to lower-right, and subtract the products of the diagonals going from upper-right to lower-left.

For our matrix:

$$\begin{bmatrix} 1 & -3 & 3 \\ 5 & 0 & 4 \\ 2 & -1 & 6 \end{bmatrix}$$

Rewrite the first two columns:

$$\begin{bmatrix} 1 & -3 & 3 | & 1 & -3 \\ 5 & 0 & 4 | & 5 & 0 \\ 2 & -1 & 6 | & 2 & -1 \end{bmatrix}$$

Now, calculate the sum of the products of the diagonals from upper-left to lower-right:

$$(1 \cdot 0 \cdot 6) + (-3 \cdot 4 \cdot 2) + (3 \cdot 5 \cdot -1) = 0 - 24 - 15 = -39$$

Next, calculate the sum of the products of the diagonals from upper-right to lower-left:

$$(3 \cdot 0 \cdot 2) + (1 \cdot 4 \cdot -1) + (-3 \cdot 5 \cdot 6) = 0 - 4 - 90 = -94$$

Finally, subtract the second sum from the first sum:

$$det = -39 - (-94) = -39 + 94 = 55$$

As you can see, both methods yield the same result: the determinant of the matrix is 55. The diagonal method is often quicker for 3x3 matrices, but it only works for 3x3 matrices, while cofactor expansion can be generalized to larger matrices.

Conclusion

Calculating the determinant of a 3x3 matrix is a fundamental skill in linear algebra. We've demonstrated the cofactor expansion method and Sarrus' Rule, providing you with the tools to tackle similar problems. Remember to practice these methods to solidify your understanding. With practice, calculating determinants will become second nature!

For further learning on determinants and matrices, you can explore resources like Khan Academy's Linear Algebra section.