2.2 Properties of Determinants · Linear Algebra Refresher

Cofactor expansion costs roughly

n!

operations — unusable past

4 \times 4

. Row reduction costs

n^{3}

. This section says exactly how each row operation changes the determinant, so you can reduce to triangular form and just multiply the diagonal.

Since row operations are left multiplication by elementary matrices, everything follows from computing $det (E)$ for each type:

Row operation	$det (E)$	Effect on $det (A)$
Type I — swap two rows	$- 1$	$det (E A) = - det (A)$
Type II — scale a row by $α \neq = 0$	$α$	$det (E A) = α det (A)$
Type III — add a multiple of one row to another	$1$	$det (E A) = det (A)$ — free

Because

det (A^{T}) = det (A)

, the identical statements hold for column operations.

Theorem 2.2.2

n \times n

matrix

A

is singular if and only if

det (A) = 0

Sketch: reduce $A$ to row echelon form $U$ . Then $det (A) = \pm α_{1} \dots α_{k} det (U)$ with all $α_{i} \neq = 0$ , and $U$ is triangular. $A$ is singular exactly when $U$ has a zero on the diagonal, i.e. $det (U) = 0$ .

Theorem 2.2.3 · Multiplicativity

det (A B) = det (A) det (B)

for all

n \times n

matrices

A

and

B

Consequences worth memorising: $det (A^{- 1}) = 1/ det (A)$ ; $det (A^{k}) = det (A)^{k}$ ; and $det (S^{- 1} A S) = det (A)$ , so similar matrices share a determinant (§6.1).

!The two determinant lies

det (A + B) = det (A) + det (B)

is false — try

A = B = I_{2}

, giving

4 \neq = 2

. And

det (α A) = α det (A)

is false: scaling

A

scales all

n

rows, so

det (α A) = α^{n} det (A)

Choosing a determinant technique

1
Is the matrix triangular, or does it have a zero row/column, or two equal rows?
yesRead the answer off immediately — product of the diagonal, or $0$ .
noContinue.
2
Is it $2 \times 2$ or $3 \times 3$ with a zero-rich row or column?
yesCofactor expansion along that row/column.
noContinue.
3
Otherwise: row reduce to triangular form
Use Type III operations freely (they cost nothing), track each swap as a sign flip, and avoid Type II scaling if you can — it forces you to divide back out at the end.
4
Multiply the diagonal, apply the accumulated sign and scale factors
$det (A) = (- 1)^{# swaps} \cdot (\prod α_{i}) \cdot \prod_{i} u_{ii}$
5
If the question is only 'singular or not?', stop as soon as a zero pivot appears
You don't need the value — a zero on the diagonal of the echelon form settles it.

On an exam, spend five seconds on step 1 before computing anything.

Worked example

Determinant by row reduction

0/4 steps

Compute

det (A)

by reducing to triangular form.

A = 246 12 - 3 314

∞Connection

Add

det (A) \neq = 0

to the running equivalence list from Theorem 1.5.2. By the end of §3.6 the list will also include: columns linearly independent, columns form a basis of

R^{n}

rank (A) = n

N (A) = {0}

. By §6.1:

0

is not an eigenvalue.

Check your work

Determinant & inverse

Reduces to triangular form, tracking every sign flip. Square matrices only.

Square matrix

2	1	3
4	2	1
6	-3	4

det =-60

status

nonsingular

Ax = b

unique solution

A⁻¹

-11/60	13/60	1/12
1/6	1/6	-1/6
2/5	-1/5	0

Check yourself

You reduce

A

4 \times 4

matrix) to triangular form

T

using: two row swaps, one row multiplied by

\frac{1}{2}

, and several 'add a multiple of one row to another' steps. If

det (T) = 6

, then

det (A)

is:

T / F

det (A + B) = det (A) + det (B)

for all

n \times n

matrices.

A

5 \times 5

with

det (A) = 3

, what is

det (2 A)

T / F

det (A) = 0

, then

A

must have a row of zeros.

A

and

B

are

n \times n

with

det (A) = 4

and

det (B) = - 2

. What is

det (A^{- 1} B^{T} A)

2.2Properties of Determinants

What you must be able to do

Choosing a determinant technique

Determinant by row reduction

Determinant & inverse

Check yourself