Matrices and Determinants of Order 3

Lessons

Senior 5 · Mathematics

Matrices and Determinants of Order 3

5 lessons

Lesson 3 of 5

Determinants of order 3 and their properties

40 min

One number that measures a transformation

The shear of lesson 2 turned the unit cube into a slanted box, yet the box has the same base and the same height, so its volume is still 1. A reflection also keeps volume but turns space "inside out", like a left shoe in a mirror becoming a right shoe, and the projection onto the $x y$ -plane squashes every solid flat, to volume 0. One number attached to a $3 \times 3$ matrix records all of this: its determinant. In S4 you met the $2 \times 2$ case, $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$ , which measures how a transformation of the plane scales area. Order 3 is built from it.

Definition

For $A = (a_{ij})$ of order 3:

the minor $M_{ij}$ is the $2 \times 2$ determinant left when row $i$ and column $j$ are deleted;
the cofactor is $C_{ij} = (-1)^{i+j}M_{ij}$ : the minor with the sign from the pattern $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$ ;
the determinant of $A$ , written $\det A$ or $∣ A ∣$ , is the expansion along row 1:

\det A = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 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} .

Expanding along any row or column, each entry times its own cofactor, gives the same number (Laplace's theorem). Choose the row or column with the most zeros.

Sarrus' rule

Multiplying out the expansion gives six products of three entries, one from each row and each column. Sarrus' rule is a picture of them: repeat the first two columns, add the three products along the diagonals going down to the right, and subtract the three going up to the right.

Worked example 1: the same determinant two ways

Evaluate $\det A$ for $A = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 4 & -2 \\ 5 & 0 & 1 \end{pmatrix}$ .

By cofactors along row 1:

\det A = 2\begin{vmatrix} 4 & -2 \\ 0 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 1 & -2 \\ 5 & 1 \end{vmatrix} + 3\begin{vmatrix} 1 & 4 \\ 5 & 0 \end{vmatrix} = 2(4) + 1(11) + 3(-20) = -41 .

By Sarrus: down-right products $2 \cdot 4 \cdot 1 + (-1)(-2)(5) + 3 \cdot 1 \cdot 0 = 8 + 10 + 0 = 18$ ; up-right products $3 \cdot 4 \cdot 5 + 2(-2)(0) + (-1)(1)(1) = 60 + 0 - 1 = 59$ . So $\det A = 18 - 59 = -41$ . ✓

Check by column 2, which has a zero: $-(-1)\begin{vmatrix} 1 & -2 \\ 5 & 1 \end{vmatrix} + 4\begin{vmatrix} 2 & 3 \\ 5 & 1 \end{vmatrix} - 0 = 11 + 4(-13) = -41$ . ✓

Important

Properties of determinants ( $A, B$ of order 3, $k$ real). Each statement about rows is also true for columns, because of property 1.

$\det A^T = \det A$ .
Swapping two rows changes the sign of the determinant.
If two rows are equal or proportional, $\det A = 0$ . A row of zeros also gives 0.
Multiplying one row by $k$ multiplies the determinant by $k$ . Hence a common factor of a row can be taken outside, and $\det(kA) = k^3\det A$ .
Adding a multiple of one row to another row leaves the determinant unchanged: $R_i \to R_i + kR_j$ is "free".
The determinant of a triangular (or diagonal) matrix is the product of its diagonal entries; $\det I_3 = 1$ .
$\det(AB) = \det A \cdot \det B$ .

Reasons. (2) Swapping two rows swaps the factors in each $2 \times 2$ minor or moves an entry to a place of opposite sign. (3) Swapping two equal rows changes nothing, yet by (2) changes the sign, so $\det A = -\det A$ and $\det A = 0$ . (5) Expanding along row $i$ , the new determinant is $\det A$ plus $k$ times a determinant with two equal rows, which is 0. Property 7 is proved in later courses; check it with the matrices of lesson 2: $\det P = -4$ , $\det Q = -22$ and $\det(PQ) = 88$ .

Worked example 2: simplify first, then expand

Evaluate $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 11 \end{vmatrix}$ using property 5.

Solution. $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$ do not change the value:

\begin{vmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ 3 & 7 & 11 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 1 & 2 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = 1(2 - 1) = 1 .

Worked example 3: factorising a determinant

Show that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (b - a)(c - a)(c - b)$ .

Solution. $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ , then expand along row 1:

\begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^2 & b^2 - a^2 & c^2 - a^2 \end{vmatrix} = (b - a)(c^2 - a^2) - (c - a)(b^2 - a^2) .

Take out $(b - a) (c - a)$ : the bracket left is $(c + a) - (b + a) = c - b$ . So the determinant is $(b - a) (c - a) (c - b)$ . It is zero exactly when two of $a, b, c$ are equal, which agrees with property 3: two equal columns.

Worked example 4: an equation with a determinant

Solve $\begin{vmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 1 & 1 & 1 \end{vmatrix} = 0$ .

Solution. Expand along row 1:

x (x - 1) - 2 (2 - 1) + 1 (2 - x) = x^{2} - x - 2 + 2 - x = x^{2} - 2 x = x (x - 2) .

So $x = 0$ or $x = 2$ . Check $x = 2$ : rows 1 and 2 become equal, so the determinant is 0 by property 3. ✓

What the determinant means

If $M$ is the matrix of a linear transformation, then $|\det M|$ is the volume of the image of the unit cube, and every volume is multiplied by $|\det M|$ . The sign tells whether orientation is kept ( $+$ ) or reversed ( $-$ , as in a mirror).

Transformation (lesson 2)	$\det$	Effect on a solid
Shear $(x, y + z, z)$	$1$	same volume, same orientation
Rotation about the $z$ -axis	$\cos^2\theta + \sin^2\theta = 1$	same volume, same orientation
Reflection in the $x y$ -plane	$- 1$	same volume, orientation reversed
Enlargement, scale factor $k$	$k^{3}$	volume multiplied by $k^{3}$
Projection onto the $x y$ -plane	$0$	flattened: volume 0

A determinant of 0 means space is flattened onto a plane, a line or a point, so different points land on the same image and the transformation cannot be undone. Lesson 4 turns this into the test for an inverse.

Warning

Keep the sign pattern. The middle term of a row-1 expansion has a minus sign: $-a_{12}M_{12}$ . Most wrong answers lose it.
$\det(kA) = k^3\det A$ for order 3, not $k\det A$ : every one of the three rows is multiplied by $k$ .
$\det(A + B) \ne \det A + \det B$ in general.
Sarrus' rule works only for order 3. It has no version for $4 \times 4$ determinants.

Tip

Before expanding, look for a zero, a common factor, or two rows that are almost equal. One row operation that creates two zeros in a row or column turns the determinant into a single $2 \times 2$ one.

3Blue1Brown: The determinant, Chapter 6 of Essence of Linear Algebra (about 10 min). In the 3-D part, watch the unit cube become a slanted box whose volume is the determinant, and notice what a negative determinant does to orientation.

✦

Practice

Questions 1 to 3 are routine; 4 to 6 need more care; 7 asks for a proof.

Evaluate $\begin{vmatrix} 3 & 0 & -2 \\ 1 & 4 & 5 \\ -1 & 2 & 1 \end{vmatrix}$ by cofactors, then check with Sarrus' rule.
Write down $\begin{vmatrix} 2 & 1 & 4 \\ 0 & -3 & 7 \\ 0 & 0 & 5 \end{vmatrix}$ without expanding.
Explain without calculating why $\begin{vmatrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 7 & -1 & 4 \end{vmatrix} = 0$ .
A $3 \times 3$ matrix $A$ has $\det A = 4$ . Find $\det(2A)$ , $\det A^T$ , $\det(A^2)$ and $\det(-A)$ .
Evaluate $\begin{vmatrix} 2 & 4 & 6 \\ 1 & 3 & 5 \\ 4 & 1 & 7 \end{vmatrix}$ , first taking a factor out of row 1.
Solve $\begin{vmatrix} x - 1 & 2 & 0 \\ 2 & x - 1 & 0 \\ 1 & 3 & x + 2 \end{vmatrix} = 0$ .
Prove that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ b + c & c + a & a + b \end{vmatrix} = 0$ for all real $a, b, c$ .

Show answersHide answers

$3 (4 - 10) - 0 + (- 2) (2 + 4) = - 18 - 12 = - 30$ .
$2 \times (-3) \times 5 = -30$ (triangular).
Row 2 is 2 × row 1: two proportional rows give 0.
$2^3 \times 4 = 32$ ; $4$ ; $4^{2} = 16$ ; $(-1)^3 \times 4 = -4$ .
$2\begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 4 & 1 & 7 \end{vmatrix} = 2 \times 9 = 18$ .
Expand along column 3: $(x + 2)\left[(x - 1)^2 - 4\right] = (x + 2)(x - 3)(x + 1)$ , so $x = - 2$ , $- 1$ or $3$ .
$R_3 \to R_3 + R_2$ gives row 3 $= (a + b + c) (1, 1, 1)$ , a multiple of row 1, so the determinant is 0.

Sources and credits

Written by Ganzaa: the explanations, the Sarrus diagram, the worked examples, the practice items and the quiz questions. Every determinant and root was checked by computer.
Video: The determinant | Chapter 6, Essence of linear algebra, 3Blue1Brown (Grant Sanderson), embedded from YouTube.

1One number that measures a transformation

2Sarrus' rule

3What the determinant means

One number that measures a transformation

Sarrus' rule

What the determinant means