5. Cross Product

a. Determinants

The easiest way to remember the cross product involves determinants. If you are comfortable taking the determinants of \(2\times2\) and \(3\times3\) matrices including the method of expanding on a row or column, skip ahead three pages. If not, use these pages to review.

There are several ways to compute determinants. We will review the method of diagonals, the method of expansion on a row or column, and the use of properties frequently called row and column operations.

1. Method of Diagonals

The simplest way to compute \(2\times2\) and \(3\times3\) determinants is the method of diagonals.

Caution: This method does not work with matrices larger than \(3\times3\).

\(2\times2\) Matrices

The determinant of a \(2\times2\) matrix is defined as: \[ \det\begin{pmatrix} a & c \\ b & d \end{pmatrix} =\begin{vmatrix} a & c \\ b & d \end{vmatrix} =ad-bc \] The entries \(a\) and \(d\) are called the principle diagonal entries, while the entries \(b\) and \(c\) are called the off diagonal entries. So the determinant is the product of the principal diagonal entries minus the product of the off diagonal entries.

Find the determinant of the matrix \(\begin{pmatrix} 2 & 4 \\ 1 & 3 \end{pmatrix}\).

\( \begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix} =2\cdot3-1\cdot4=6-4=2 \)

Find the determinant of the matrix \(\begin{pmatrix} 7 & 6 \\ 3 & 4 \end{pmatrix}\).

The determinant is \( \begin{vmatrix} 7 & 6 \\ 3 & 4 \end{vmatrix}=10\).

The determinant is the product of the principle diagonal entries minus the product of the off diagonal entries: \[ \begin{vmatrix} 7 & 6 \\ 3 & 4 \end{vmatrix} =7\cdot4-6\cdot3=28-18=10 \]

Find the determinant of the matrix \(\begin{pmatrix} 4 & 10 \\ 2 & 5 \end{pmatrix}\).

The determinant is \( \begin{vmatrix} 4 & 10 \\ 2 & 5 \end{vmatrix}=0\).

The determinant is the product of the principle diagonal entries minus the product of the off diagonal entries: \[ \begin{vmatrix} 4 & 10 \\ 2 & 5 \end{vmatrix} =4\cdot5-10\cdot2=20-20=0 \]

\(3\times3\) Matrices

The determinant of a \(3\times3\) matrix is defined as follows: Copy its first two columns to the right, add the products of the entries on each diagonal which is down and to the right and subtract the products of the entries on each diagonal which is down and to the left: \[\begin{aligned} \begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix} &=\begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix}\; \begin{matrix} a & d \\ b & e \\ c & f \end{matrix} \\ &=(aei+dhc+gbf)-(gec+ahf+dbi) \end{aligned}\]

\[\begin{aligned} \begin{vmatrix} 2 & 4 & 1 \\ 3 & 1 & 5 \\ -2 & 1 & 4 \end{vmatrix} &=\begin{vmatrix} 2 & 4 & 1 \\ 3 & 1 & 5 \\ -2 & 1 & 4 \end{vmatrix}\, \begin{matrix} 2 & 4 \\ 3 & 1 \\ -2 & 1 \end{matrix} \\ &=(8+-40+3)-(-2+10+48)=-85 \end{aligned}\]

Find the determinant of the matrix \(\begin{pmatrix} 1 & 4 & 7 \\ 2 & 2 & 1 \\ 3 & 0 & 3 \end{pmatrix}\).

The determinant is \( \begin{vmatrix} 1 & 4 & 7 \\ 2 & 2 & 1 \\ 3 & 0 & 3 \end{vmatrix}=-48\).

\[\begin{aligned} \begin{vmatrix} 1 & 4 & 7 \\ 2 & 2 & 1 \\ 3 & 0 & 3 \end{vmatrix} &=\begin{vmatrix} 1 & 4 & 7 \\ 2 & 2 & 1 \\ 3 & 0 & 3 \end{vmatrix}\; \begin{matrix} 1 & 4 \\ 2 & 2 \\ 3 & 0 \end{matrix} \\ &=(1\cdot2\cdot3)+(4\cdot1\cdot3)+(7\cdot2\cdot0) \\ &\quad-(7\cdot2\cdot3)-(1\cdot1\cdot0)-(4\cdot2\cdot3) \\ &=6+12+0-42-0-24 \\ &=-48 \end{aligned}\]

Find the determinant of the matrix \(\begin{pmatrix} 3 & 1 & 1 \\ 2 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}\).

The determinant is \( \begin{vmatrix} 3 & 1 & 1 \\ 2 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix}=0\).

\[\begin{aligned} \begin{vmatrix} 3 & 1 & 1 \\ 2 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix} &=\begin{vmatrix} 3 & 1 & 1 \\ 2 & 0 & 0 \\ 0 & 1 & 1 \end{vmatrix}\; \begin{matrix} 3 & 1 \\ 2 & 0 \\ 0 & 1 \end{matrix} \\ &=(3\cdot0\cdot1)+(1\cdot0\cdot0)+(1\cdot2\cdot1) \\ &\quad-(1\cdot0\cdot0)-(3\cdot0\cdot1)-(1\cdot2\cdot1) \\ &=0+0+2-0-0-2 \\ &=0 \end{aligned}\]

This is a legitimate way of finding the determinant of a \(3\times3\) matrix, but it does \(not\) generalize to larger matrices. It is a sheer coincidence that the method of diagonals works on \(2\times2\) and \(3\times3\) matrices. However, we can use this method to derive a more robust method for taking the determinant of a \(3\times3\) matrix which does generalize to larger matrices. This will be explored on the next page.

But first we look at

\(4\times4\) Matrices

It is tempting to define the determinant of a \(4\times4\) matrix to be

WRONG!

\[\begin{aligned} \begin{vmatrix} a & e & i & m \\ b & f & j & n \\ c & g & k & o \\ d & h & l & p \end{vmatrix} &=\begin{vmatrix} a & e & i & m \\ b & f & j & n \\ c & g & k & o \\ d & h & l & p \end{vmatrix}\; \begin{matrix} a & e & i \\ b & f & j \\ c & g & k \\ d & h & l \end{matrix} \\ &=(afkp+ejod+inch+mbgl) \\ &\qquad-(mjgd+ankh+ebol+ifcp) \end{aligned}\]