The 3×3 Determinant

Introduction

The determinant is a scalar value that can be computed from a square matrix. For a 3×3 matrix, the determinant provides important geometric and algebraic information, including whether the matrix is invertible and the volume of a parallelepiped formed by its column (or row) vectors. In physics, the 3×3 determinant is particularly useful for calculating cross products in a compact and elegant form.

Definition of a 3×3 Determinant:

For a 3×3 matrix:

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}

the determinant, denoted $\det(A)$ or $|A|$, is a scalar value defined by:

\det(A) = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

Method 1: Expansion by Minors (Laplace Expansion)

The most general method for calculating a 3×3 determinant is expansion by minors (also called Laplace expansion). This method works by expanding along any row or column.

Expansion Along the First Row:

The determinant can be calculated by expanding along the first row:

\det(A) = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}

where each 2×2 determinant is calculated as:

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

Expanding this gives:

\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})

Pattern for Expansion:

When expanding along a row or column, use the following sign pattern:

\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}

The sign alternates starting with $+$ in the top-left corner. The $(i,j)$ entry uses sign $(-1)^{i+j}$.

Method 2: Sarrus' Rule

Sarrus' rule is a special method that works only for 3×3 determinants. It provides a visual mnemonic for calculating the determinant.

Sarrus' Rule:

To apply Sarrus' rule:

Write the first two columns of the matrix to the right of the matrix
Sum the products along the three main diagonals (top-left to bottom-right)
Subtract the sum of products along the three anti-diagonals (top-right to bottom-left)

\begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}

The determinant is:

\det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32} - a_{12}a_{21}a_{33}

Properties of Determinants

Key Properties:

Row/Column operations: Swapping two rows (or columns) changes the sign of the determinant
Scaling: Multiplying a row (or column) by a scalar $k$ multiplies the determinant by $k$
Adding rows: Adding a multiple of one row to another doesn't change the determinant
Transpose: $\det(A) = \det(A^T)$ (determinant of transpose equals determinant of original)
Product: $\det(AB) = \det(A)\det(B)$
Invertibility: A matrix is invertible if and only if its determinant is non-zero
Zero determinant: If any row or column is a linear combination of others, the determinant is zero

Examples

Example 1: Calculating a Determinant Using Expansion by Minors

Calculate the determinant of:

A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 1 & 2 \\ 3 & 2 & 1 \end{pmatrix}

Solution:

Expanding along the first row:

\det(A) = 2 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 3 \begin{vmatrix} 4 & 2 \\ 3 & 1 \end{vmatrix} + 1 \begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix}

Calculating each 2×2 determinant:

\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3

\begin{vmatrix} 4 & 2 \\ 3 & 1 \end{vmatrix} = (4)(1) - (2)(3) = 4 - 6 = -2

\begin{vmatrix} 4 & 1 \\ 3 & 2 \end{vmatrix} = (4)(2) - (1)(3) = 8 - 3 = 5

Therefore:

\det(A) = 2(-3) - 3(-2) + 1(5) = -6 + 6 + 5 = 5

Example 2: Using Sarrus' Rule

Calculate the determinant of the same matrix using Sarrus' rule:

A = \begin{pmatrix} 2 & 3 & 1 \\ 4 & 1 & 2 \\ 3 & 2 & 1 \end{pmatrix}

Solution:

Main diagonals (sum):

$$(2)(1)(1) + (3)(2)(3) + (1)(4)(2) = 2 + 18 + 8 = 28$$

Anti-diagonals (subtract):

$$(1)(1)(3) + (2)(2)(2) + (1)(4)(3) = 3 + 8 + 12 = 23$$

Therefore:

\det(A) = 28 - 23 = 5

This matches the result from Example 1.

Example 3: Determinant with Zero Row

Calculate the determinant of:

B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 4 & 5 & 6 \end{pmatrix}

Solution:

Since the second row is all zeros, the determinant is zero. We can verify by expansion along the second row:

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

Application to Cross Products

The cross product of two vectors can be elegantly expressed using a determinant. This provides a compact and systematic way to calculate cross products.

Cross Product as a Determinant:

For two vectors $\mathbf{a} = (a_x, a_y, a_z)$ and $\mathbf{b} = (b_x, b_y, b_z)$, the cross product is given by:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}

Expanding this determinant along the first row gives:

\mathbf{a} \times \mathbf{b} = \hat{x} \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} - \hat{y} \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} + \hat{z} \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix}

Evaluating the 2×2 determinants:

\mathbf{a} \times \mathbf{b} = \hat{x}(a_y b_z - a_z b_y) - \hat{y}(a_x b_z - a_z b_x) + \hat{z}(a_x b_y - a_y b_x)

Rearranging the signs:

\mathbf{a} \times \mathbf{b} = (a_y b_z - a_z b_y)\hat{x} + (a_z b_x - a_x b_z)\hat{y} + (a_x b_y - a_y b_x)\hat{z}

This matches the standard component form of the cross product!

Examples Using Determinants for Cross Products

Example 4: Cross Product Using Determinant

Calculate $\mathbf{a} \times \mathbf{b}$ where $\mathbf{a} = (3, 4, 0)$ and $\mathbf{b} = (2, -1, 5)$ using the determinant method.

Solution:

Set up the determinant:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 3 & 4 & 0 \\ 2 & -1 & 5 \end{vmatrix}

Expanding along the first row:

\mathbf{a} \times \mathbf{b} = \hat{x} \begin{vmatrix} 4 & 0 \\ -1 & 5 \end{vmatrix} - \hat{y} \begin{vmatrix} 3 & 0 \\ 2 & 5 \end{vmatrix} + \hat{z} \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix}

Calculating each 2×2 determinant:

\begin{vmatrix} 4 & 0 \\ -1 & 5 \end{vmatrix} = (4)(5) - (0)(-1) = 20

\begin{vmatrix} 3 & 0 \\ 2 & 5 \end{vmatrix} = (3)(5) - (0)(2) = 15

\begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (4)(2) = -3 - 8 = -11

Therefore:

\mathbf{a} \times \mathbf{b} = 20\hat{x} - 15\hat{y} - 11\hat{z} = (20, -15, -11)

Example 5: Cross Product of Unit Vectors

Use the determinant method to show that $\hat{x} \times \hat{y} = \hat{z}$.

Solution:

With $\hat{x} = (1, 0, 0)$ and $\hat{y} = (0, 1, 0)$:

\hat{x} \times \hat{y} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{vmatrix}

Expanding along the first row:

\hat{x} \times \hat{y} = \hat{x} \begin{vmatrix} 0 & 0 \\ 1 & 0 \end{vmatrix} - \hat{y} \begin{vmatrix} 1 & 0 \\ 0 & 0 \end{vmatrix} + \hat{z} \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}

The first two 2×2 determinants are zero, and the third is:

\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1

Therefore:

\hat{x} \times \hat{y} = \hat{z}

Example 6: Parallel Vectors

Show that if $\mathbf{a} = (2, 4, 6)$ and $\mathbf{b} = (1, 2, 3)$ are parallel, then $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ using the determinant method.

Solution:

Set up the determinant:

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 2 & 4 & 6 \\ 1 & 2 & 3 \end{vmatrix}

Notice that the second row is exactly twice the third row: $(2, 4, 6) = 2(1, 2, 3)$. When one row is a scalar multiple of another, the determinant is zero. Expanding to verify:

\mathbf{a} \times \mathbf{b} = \hat{x}(4 \cdot 3 - 6 \cdot 2) - \hat{y}(2 \cdot 3 - 6 \cdot 1) + \hat{z}(2 \cdot 2 - 4 \cdot 1)

= \hat{x}(12 - 12) - \hat{y}(6 - 6) + \hat{z}(4 - 4) = \mathbf{0}

Geometric Interpretation

The determinant has important geometric meanings:

Volume of parallelepiped: The absolute value of the determinant of a 3×3 matrix equals the volume of the parallelepiped formed by its three column (or row) vectors.
Orientation: The sign of the determinant indicates the orientation (right-handed vs. left-handed) of the coordinate system formed by the vectors.
Linear independence: If the determinant is zero, the vectors are linearly dependent (they lie in a plane or are collinear).

Advantages of the Determinant Method

Why Use Determinants for Cross Products?

Systematic approach: The determinant provides a systematic, algorithmic way to compute cross products.
Easy to remember: The mnemonic of writing unit vectors in the first row makes the formula easy to recall.
Visual structure: The matrix structure makes it clear which components are being combined.
Connection to linear algebra: Links vector operations to matrix theory and provides deeper mathematical insight.
Generalization: The determinant approach generalizes to higher dimensions and other vector operations.