Determinants

Evaluating 2×2 and 3×3 determinants, cofactor expansion along any row or column, Sarrus' rule, and the seven properties that turn a hard determinant into an easy one.

Module contents

A determinant is a single number attached to a square matrix. Written $∣ A ∣$ or $det A$ or $Δ$ . Non-square matrices do not have one.

The number answers a question that runs through the whole module: is this matrix invertible? If $∣ A ∣ \neq = 0$ it is; if $∣ A ∣ = 0$ it is not.

Order 1 and order 2

$∣ a ∣ = a$

$a c b d = a d - b c$

Main diagonal product minus anti-diagonal product.

$3254 = 3 (4) - 5 (2) = 12 - 10 = 2$

Minors and cofactors

For a square matrix $A$ of order $n$ :

Minor $M_{ij}$ — the determinant of the matrix left after deleting row $i$ and column $j$ .

Cofactor $A_{ij}$ — the minor with a sign attached:

$A_{ij} = (- 1)^{i + j} M_{ij}$

The signs form a fixed checkerboard, always starting with $+$ at the top left:

$+ - + - + - + - +$

Order 3 — expansion by cofactors

$∣ A ∣ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$

that is, expansion along the first row. Written out:

$∣ A ∣ = a_{11} a_{22} a_{32} a_{23} a_{33} - a_{12} a_{21} a_{31} a_{23} a_{33} + a_{13} a_{21} a_{31} a_{22} a_{32}$

You may expand along any row or any column and get the same answer. So expand along whichever line has the most zeros — each zero kills a whole $2 \times 2$ determinant. This is the single biggest time-saver in the topic.

Formally, for any fixed $i$ or any fixed $j$ :

$∣ A ∣ = j = 1 \sum n a_{ij} A_{ij} = i = 1 \sum n a_{ij} A_{ij}$

And a useful companion fact: expanding along one row using another row's cofactors always gives zero.

$j \sum a_{ij} A_{k j} = 0 when i \neq = k$

Sarrus' rule (3 × 3 only)

Copy the first two columns to the right, then add the three down-right diagonals and subtract the three down-left ones.

$a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33} a_{11} a_{21} a_{31} a_{12} a_{22} a_{32}$

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

Fast, but it works for order $3$ only. Never attempt it on a $4 \times 4$ .

Properties of determinants

These are the tools for handling anything larger or messier than a plain $3 \times 3$ .

P1 — Transpose. $∣ A^{T} ∣ = ∣ A ∣$ .
Every property stated for rows therefore holds for columns too.

P2 — Row/column interchange. Swapping two rows (or two columns) multiplies the determinant by $- 1$ .

P3 — Identical lines. If two rows (or two columns) are identical, $∣ A ∣ = 0$ .
(Follows from P2: swapping them leaves $A$ unchanged but negates $∣ A ∣$ , so $∣ A ∣ = - ∣ A ∣$ .)

P4 — Common factor. If every element of one row is multiplied by $k$ , the determinant is multiplied by $k$ . Read backwards: a common factor can be taken out of a single row.

$k a c k b d = k a c b d and hence ∣ k A ∣ = k^{n} ∣ A ∣$

P5 — Zero line. If any row or column is entirely zero, $∣ A ∣ = 0$ .

P6 — Proportional lines. If one row is a scalar multiple of another, $∣ A ∣ = 0$ .

P7 — Row operation invariance. Adding a multiple of one row to another row leaves the determinant unchanged:

$R_{i} \to R_{i} + k R_{j} (i \neq = j) \Rightarrow ∣ A ∣ unchanged$

This is the workhorse. Use it to manufacture zeros, then expand along the line you created.

P8 — Sum splitting. If one row is a sum, the determinant splits:

$a + x c b + y d = a c b d + x c y d$

P9 — Triangular matrices. For a triangular (or diagonal) matrix, $∣ A ∣$ is the product of the diagonal elements.

P10 — Product rule. $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ .
Consequences worth memorising: $∣ A^{n} ∣ = ∣ A ∣^{n}$ , and $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣}$ .

Singular versus non-singular

$∣ A ∣ = 0 ⟺ A is singular ⟺ A^{- 1} does not exist$

$∣ A ∣ \neq = 0 ⟺ A is non-singular ⟺ A^{- 1} exists and is unique$

Choosing a method

Situation	Best method
$2 \times 2$	$a d - b c$
$3 \times 3$ , no zeros	Sarrus, or expansion along row 1
$3 \times 3$ with a zero-rich row/column	Expand along that line
$3 \times 3$ with symbols or repeated entries	Row operations (P7) first
$4 \times 4$ and above	Row operations to make a zero-rich line, then expand
Triangular	Multiply the diagonal

Worked examples

4 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Order 3 by expansion along the first row

Question

Evaluate $215 - 1 40 3 - 2 1$ .

Write the expansion with the checkerboard signs
$Δ = 2 40 - 2 1 - (- 1) 15 - 2 1 + 31540$
Evaluate the three 2 × 2 determinants
$40 - 2 1 = 4 (1) - (- 2) (0) = 4$
$15 - 2 1 = 1 (1) - (- 2) (5) = 1 + 10 = 11$
$1540 = 1 (0) - 4 (5) = - 20$
Combine
$Δ = 2 (4) + 1 (11) + 3 (- 20) = 8 + 11 - 60 = - 41$
Cross-check by expanding along column 2
Column 2 is $(- 1, 4, 0)$ with signs $(-, +, -)$ :

$Δ = - (- 1) 15 - 2 1 + 42531 - 0 = 1 (11) + 4 (2 - 15) = 11 - 52 = - 41 ✓$

Answer

$Δ = - 41$

NoteWatch the sign on the middle term: it is −a₁₂ × M₁₂, so a negative a₁₂ produces a positive contribution.

Example 2

Exploiting zeros — expanding along the best line

Question

Evaluate $3271016000204513$ .

Find the emptiest line
Column 3 is $(0, 0, 2, 0)$ — three zeros. Expanding along it leaves a single term.
Expand along column 3
The surviving element is $a_{33} = 2$ , with sign $(- 1)^{3 + 3} = +$ :

$Δ = + 2321010453$
Repeat the trick on the 3 × 3
Its column 2 is $(0, 1, 0)$ — expand along it. The surviving element is at position $(2, 2)$ , sign $+$ :

$321010453 = + 13143 = 9 - 4 = 5$
Combine
$Δ = 2 \times 5 = 10$

Answer

$Δ = 10$

NoteA 4×4 that would take six 3×3 expansions collapsed to one 2×2. Always scan for the emptiest row or column first.

Example 3

Using row operations to manufacture zeros

Question

Evaluate $1472583610$ using properties.

Clear column 1 using row 1
Apply $R_{2} \to R_{2} - 4 R_{1}$ and $R_{3} \to R_{3} - 7 R_{1}$ . By P7 the determinant is unchanged.

$R_{2}$ : $(4 - 4, 5 - 8, 6 - 12) = (0, - 3, - 6)$
$R_{3}$ : $(7 - 7, 8 - 14, 10 - 21) = (0, - 6, - 11)$

$Δ = 100 2 - 3 - 6 3 - 6 - 11$
Expand along the new column 1
Only $a_{11} = 1$ survives, with sign $+$ :

$Δ = 1 - 3 - 6 - 6 - 11$
Evaluate
$= (- 3) (- 11) - (- 6) (- 6) = 33 - 36 = - 3$
Check directly
Expanding the original along row 1:

$1 (50 - 48) - 2 (40 - 42) + 3 (32 - 35) = 2 + 4 - 9 = - 3 ✓$

Answer

$Δ = - 3$

Example 4

A determinant with symbols — factorising by properties

Question

Prove that $1 a a^{2} 1 b b^{2} 1 c c^{2} = (a - b) (b - c) (c - a)$ .

Create zeros with column operations
Apply $C_{1} \to C_{1} - C_{2}$ and $C_{2} \to C_{2} - C_{3}$ (both allowed by P7, applied to columns via P1):

$Δ = 0 a - b a^{2} - b^{2} 0 b - c b^{2} - c^{2} 1 c c^{2}$
Take common factors out of columns 1 and 2
Column 1 has the common factor $(a - b)$ , since $a^{2} - b^{2} = (a - b) (a + b)$ . Column 2 has $(b - c)$ . By P4:

$Δ = (a - b) (b - c) 01 a + b 01 b + c 1 c c^{2}$
Expand along row 1
Only the entry $1$ at position $(1, 3)$ survives, with sign $(- 1)^{1 + 3} = +$ :

$Δ = (a - b) (b - c) 1 a + b 1 b + c$
Evaluate and simplify
$1 a + b 1 b + c = (b + c) - (a + b) = c - a$

$Δ = (a - b) (b - c) (c - a) ■$

Answer

$Δ = (a - b) (b - c) (c - a)$

NoteThis is the 3×3 Vandermonde determinant. The pattern — subtract adjacent columns, then pull out the differences — solves most symbolic determinant questions.

Practice problems

7 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

Evaluate (a) $73 - 2 5$ (b) $cos θ sin θ - sin θ cos θ$

Show solution

(a) $7 (5) - (- 2) (3) = 35 + 6 = 41$

(b) $cos θ cos θ - (- sin θ) (sin θ) = cos^{2} θ + sin^{2} θ = 1$

Result (b) says every rotation matrix has determinant $1$ — rotation preserves area, which is exactly what a determinant of $1$ means geometrically.

Answer(a)

41

(b)

1

Problem 2Basic

Find the minor $M_{23}$ and the cofactor $A_{23}$ of $14 - 1 306 - 2 52$ .

Show solution

Minor. Delete row $2$ and column $3$ :

$M_{23} = 1 - 1 36 = 1 (6) - 3 (- 1) = 6 + 3 = 9$

Cofactor. Attach the sign $(- 1)^{2 + 3} = (- 1)^{5} = - 1$ :

$A_{23} = (- 1)^{2 + 3} M_{23} = - 9$

Answer

M_{23} = 9

A_{23} = - 9

Problem 3Exam level

Evaluate $213 3 - 2 1 - 1 42$ by (a) expansion along row 1 and (b) Sarrus' rule.

Show solution

(a) Expansion along row 1.

$Δ = 2 - 2 1 42 - 31342 + (- 1) 13 - 2 1$

$- 2 1 42 = - 4 - 4 = - 81342 = 2 - 12 = - 10 13 - 2 1 = 1 + 6 = 7$

$Δ = 2 (- 8) - 3 (- 10) - 1 (7) = - 16 + 30 - 7 = 7$

(b) Sarrus' rule.

Down-right diagonals:
$2 (- 2) (2) + 3 (4) (3) + (- 1) (1) (1) = - 8 + 36 - 1 = 27$

Down-left diagonals:
$(- 1) (- 2) (3) + 2 (4) (1) + 3 (1) (2) = 6 + 8 + 6 = 20$

$Δ = 27 - 20 = 7 ✓$

Both methods agree.

Answer

Δ = 7

Problem 4Exam level

Find $x$ if $x 13 2 x 2 31 x = 0$ .

Show solution

Expand along row 1:

$x x 2 1 x - 2 13 1 x + 3 13 x 2 = 0$

$x (x^{2} - 2) - 2 (x - 3) + 3 (2 - 3 x) = 0$

$x^{3} - 2 x - 2 x + 6 + 6 - 9 x = 0$

$x^{3} - 13 x + 12 = 0$

Try $x = 1$ : $1 - 13 + 12 = 0$ ✓, so $(x - 1)$ is a factor.

Dividing: $x^{3} - 13 x + 12 = (x - 1) (x^{2} + x - 12) = (x - 1) (x + 4) (x - 3)$

$x = 1, x = 3, x = - 4$

At each of these values the matrix is singular.

Answer

x = 1, 3, - 4

Problem 5Exam level

Without expanding, show that $111 a b c b + c c + a a + b = 0$ .

Show solution

Apply the column operation $C_{3} \to C_{3} + C_{2}$ , which by P7 leaves the determinant unchanged:

$Δ = 111 a b c a + b + c a + b + c a + b + c$

Column 3 now has the common factor $(a + b + c)$ . Take it out by P4:

$Δ = (a + b + c) 111 a b c 111$

Columns $1$ and $3$ are now identical, so by P3 the determinant is zero:

$Δ = (a + b + c) \times 0 = 0 ■$

No expansion was needed at any point.

Answer

Δ = 0

, since a column operation produces two identical columns.

Problem 6Exam level

If $A$ is a $3 \times 3$ matrix with $∣ A ∣ = 5$ , find (a) $∣3 A ∣$ , (b) $∣ A^{T} ∣$ , (c) $∣ A^{2} ∣$ , (d) $∣ A^{- 1} ∣$ , (e) $∣2 A^{- 1} ∣$ .

Show solution

Here $n = 3$ throughout.

(a) $∣ k A ∣ = k^{n} ∣ A ∣$ , so $∣3 A ∣ = 3^{3} \times 5 = 27 \times 5 = 135$

(b) $∣ A^{T} ∣ = ∣ A ∣ = 5$

(c) $∣ A^{2} ∣ = ∣ A ∣^{2} = 5^{2} = 25$

(d) $∣ A^{- 1} ∣ = \frac{1}{∣ A ∣} = \frac{1}{5}$

(e) $∣2 A^{- 1} ∣ = 2^{3} ∣ A^{- 1} ∣ = 8 \times \frac{1}{5} = \frac{8}{5}$

The trap in (a) and (e) is writing $∣ k A ∣ = k ∣ A ∣$ . The scalar comes out once per row, and there are $n$ rows.

Answer(a)

135

(b)

5

(c)

25

(d)

1/5

(e)

8/5

Problem 7Challenge

Evaluate $a b c b c a c a b$ and state when it is zero.

Show solution

Apply $R_{1} \to R_{1} + R_{2} + R_{3}$ . Every entry of the new row 1 becomes $a + b + c$ :

$Δ = a + b + c b c a + b + c c a a + b + c a b = (a + b + c) 1 b c 1 c a 1 a b$

Now clear row 1 with $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ :

$= (a + b + c) 1 b c 0 c - b a - c 0 a - b b - c$

Expand along row 1:

$= (a + b + c) [(c - b) (b - c) - (a - b) (a - c)]$

$= (a + b + c) [- (b - c)^{2} - (a^{2} - a c - ab + b c)]$

$= (a + b + c) [- (b^{2} - 2 b c + c^{2}) - a^{2} + a c + ab - b c]$

$= (a + b + c) [- a^{2} - b^{2} - c^{2} + ab + b c + c a]$

$Δ = - (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Using the identity $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ :

$Δ = - (a^{3} + b^{3} + c^{3} - 3 ab c) = 3 ab c - a^{3} - b^{3} - c^{3}$

When is it zero? Either $a + b + c = 0$ , or $a^{2} + b^{2} + c^{2} - ab - b c - c a = 0$ . The second factor equals $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ , which for real numbers vanishes only when $a = b = c$ .

So $Δ = 0 ⟺ a + b + c = 0$ or $a = b = c$ .

Answer

Δ = 3 ab c - a^{3} - b^{3} - c^{3}

; zero exactly when

a + b + c = 0

a = b = c