lecture notes on determinants

lecture notes on determinants

1

Introduction.

The

determinant

operation

is an

algebraic

formula

involving

addition

and

multiplication

that combines the

n

2

entries of an

n

×

n

matrix of numbers into a single number.
The determinant has many useful and surprising

properties

. In
particular the determinant “determines” whether or not a matrix is

singular

. If the determinant is zero, the matrix is singular; if not,
the matrix is

invertible

.

2

Notation.

We can regard an

n

×

n

matrix

A

as an entity in and of itself,
as a

collection

of

n

2

numbers arranged in a table, or as a list of

column vectors

:

A

=

[

a

11

a

12

…

a

1

⁢

n

a

21

a

22

…

a

2

⁢

n

⋮

⋮

⋱

⋮

a

n

⁢

1

a

n

⁢

2

…

a

n

⁢

n

]

=

[

𝐚

1

,

𝐚

2

,

…

,

𝐚

n

]

,

where

𝐚

1

=

[

a

11

⋮

a

n

⁢

1

]

=

a

11

⁢

𝐞

1

+

⋯

⁢

a

n

⁢

1

⁢

𝐞

n

,

𝐚

2

=

[

a

12

⋮

a

n

⁢

2

]

=

a

12

⁢

𝐞

1

+

⋯

⁢

a

n

⁢

2

⁢

𝐞

n

,

etc.

Correspondingly, we employ the following notation for the determinant:

det

⁡

(

A

)

=

|

a

11

…

a

1

⁢

n

⋮

⋱

⋮

a

n

⁢

1

…

a

n

⁢

n

|

=

|

𝐚

1

,

…

,

𝐚

n

|

.

3

Defining properties.

The determinant operation obeys
certain key properties. The correct application of these rules allows
us to evaluate the determinant of a given

square matrix

. For the sake
of simplicity, we describe these rules for the case of the

2

×

2

and

3

×

3

determinants. Determinants of larger sizes obey analogous rules.

1.

Multi-linearity

. The determinant operation is linear
in each of its vector

arguments

.

(a)

The determinant distributes over addition. Thus, for a

3

×

3

matrix

A

and a column vector

𝐛

∈

ℝ

3

, we have

|

𝐚

1

+

𝐛

,

𝐚

2

,

𝐚

3

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

+

|

𝐛

,

𝐚

2

,

𝐚

3

|

|

𝐚

1

,

𝐚

2

+

𝐛

,

𝐚

3

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

+

|

𝐚

1

,

𝐛

,

𝐚

3

|

|

𝐚

1

,

𝐚

2

,

𝐚

3

+

𝐛

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

+

|

𝐚

1

,

𝐚

2

,

𝐛

|

Thus, if

A

,

B

are two

2

×

2

matrices the determinant of the
sum

A

+

B

will have a total of

4

terms (think FOIL):

det

⁡

(

A

+

B

)

=

|

𝐚

1

+

𝐛

1

,

𝐚

2

+

𝐛

2

|

=

|

𝐚

1

,

𝐚

2

|

+

|

𝐚

1

,

𝐛

2

|

+

|

𝐛

1

,

𝐚

2

|

+

|

𝐛

1

,

𝐛

2

|

=

det

⁡

(

A

)

+

|

𝐚

1

,

𝐛

2

|

+

|

𝐛

1

,

𝐚

2

|

+

det

⁡

(

B

)

Warning: the formula

det

⁡

(

A

+

B

)

=

det

⁡

(

A

)

+

det

⁡

(

B

)

is most certainly
wrong; it has the F and the L terms from FOIL, but is missing the
O and the I terms. Remember, the determinant is not linear; it’s
multi-linear!

(b)

Scaling

one column of a matrix, scales the determinant by the
same amount. Thus, for a scalar

k

∈

ℝ

, we have

|

k

⁢

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

k

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

k

⁢

det

⁡

(

A

)

.

Similarly,

|

𝐚

1

,

k

⁢

𝐚

2

,

𝐚

3

|

=

|

𝐚

1

,

𝐚

2

,

k

⁢

𝐚

3

|

=

k

⁢

det

⁡

(

A

)

.

Let’s see what happens to the determinant if we scale the entire
matrix:

det

⁡

(

k

⁢

A

)

=

|

k

⁢

𝐚

1

,

k

⁢

𝐚

2

,

k

⁢

𝐚

3

|

=

k

⁢

|

𝐚

1

,

k

⁢

𝐚

2

,

k

⁢

𝐚

3

|

=

k

2

⁢

|

𝐚

1

,

𝐚

2

,

k

⁢

𝐚

3

|

=

k

3

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

k

3

⁢

det

⁡

(

A

)

.

(c)

A matrix with a zero column has a zero determinant. For
example,

|

𝟎

,

𝐚

2

,

𝐚

3

|

=

|

0

a

12

a

13

0

a

22

a

23

0

a

32

a

33

|

=

0

.

2.

Skew-symmetry.

A multi-variable

function

or formula is
called

symmetric

if it does not depend on the order of the
arguments/

variables

. For example, ordinary addition and
multiplication are symmetric operations. A multi-variable function
is said to be

skew-symmetric

if changing the order of any two
arguments changes the sign of the operation. The 3-dimensional
cross-product is an example of a skew-symmetric operation:

𝐮

×

𝐯

=

-

𝐯

×

𝐮

,

𝐮

,

𝐯

∈

ℝ

3

.

Likewise, the

n

×

n

determinant is a skew-symmetric operation,
albeit one with

n

arguments.

Thus, for a

2

×

2

matrix

A

we have

det

⁡

(

A

)

=

|

𝐚

1

,

𝐚

2

|

=

-

|

𝐚

2

,

𝐚

1

|

.

There
are six possible ways to rearrange the columns of a

3

×

3

matrix. Correspondingly, for a

3

×

3

matrix

A

we have

det

⁡

(

A

)

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

-

|

𝐚

2

,

𝐚

1

,

𝐚

3

|

=

-

|

𝐚

3

,

𝐚

2

,

𝐚

1

|

=

-

|

𝐚

1

,

𝐚

3

,

𝐚

2

|

=

+

|

𝐚

2

,

𝐚

3

,

𝐚

1

|

=

+

|

𝐚

3

,

𝐚

1

,

𝐚

2

|

The determinants in the 3rd line are equal to

det

⁡

(

A

)

because, in
each case, the matrices in question differ from

A

by 2 column
exchanges.

Skew-symmetry of the determinant operation has an important

consequence

: a
matrix with two identical columns has zero determinant. Consider,
for example a

3

×

3

matrix

A

with identical first and second
columns. By skew-symmetry, if we exchange the first two columns we
negate the value of the determinant:

|

𝐚

1

,

𝐚

1

,

𝐚

3

|

=

-

|

𝐚

1

,

𝐚

1

,

𝐚

3

|

.

Therefore,

det

⁡

(

A

)

is equal to its

negation

. This can only mean
that

det

⁡

(

A

)

=

0

.

3.

The

identity

rule.

The determinant of the

identity
matrix

is equal to

1

. Written out in symbols for the

3

×

3

case, this means that

det

⁡

(

I

3

)

=

|

𝐞

1

,

𝐞

2

,

𝐞

3

|

=

1

.

4

Evaluation.

The above properties dictate the rules by
which we evaluate a determinant. Overall, the process is very
reminiscent of binomial expansion — the

algebra

that goes into
expanding an

expression

like

(

x

+

y

)

3

. The essential

difference

in
evaluating determinants is that for scalar algebra the order of the
variables is unimportant:

x

⋅

y

=

y

⋅

x

. However, when we
evaluate determinants, the choice of order can introduce a minus sign,
or result in a zero answer, if some of the arguments are repeated.

Let’s see how evaluation works for

2

×

2

determinants.

|

a

b

c

d

|

=

|

a

⁢

𝐞

1

+

c

⁢

𝐞

2

,

b

⁢

𝐞

1

+

d

⁢

𝐞

2

|

=

a

⁢

b

⁢

|

𝐞

1

,

𝐞

1

|

+

a

⁢

d

⁢

|

𝐞

1

,

𝐞

2

|

+

c

⁢

b

⁢

|

𝐞

2

,

𝐞

1

|

+

c

⁢

d

⁢

|

𝐞

2

,

𝐞

2

|

=

0

+

a

⁢

d

⁢

|

𝐞

1

,

𝐞

1

|

-

b

⁢

c

⁢

|

𝐞

1

,

𝐞

2

|

+

0

=

(

a

⁢

d

-

b

⁢

c

)

⁢

|

𝐞

1

,

𝐞

2

|

=

a

⁢

d

-

b

⁢

c

In the above evaluation we used skew-symmetry 3 times:

|

𝐞

1

,

𝐞

1

|

=

0

,

|

𝐞

2

,

𝐞

2

|

=

0

,

|

𝐞

2

,

𝐞

1

|

=

-

|

𝐞

1

,

𝐞

2

|

.

At the very end, we also used the identity rule.
It is useful to contrast the above manipulations with the expansion of

(

a

⁢

x

+

c

⁢

y

)

⁢

(

b

⁢

x

+

d

⁢

y

)

=

a

⁢

b

⁢

x

2

+

(

a

⁢

c

+

b

⁢

d

)

⁢

x

⁢

y

+

c

⁢

d

⁢

y

2

.

The distributive step is the same. The different answers arise because
ordinary multiplication is a symmetric operation.

Next, let’s see how to evaluate a

3

×

3

determinant.

|

a

1

b

1

c

1

a

2

b

2

c

2

a

3

b

3

c

3

|

=

|

a

1

⁢

𝐞

1

+

a

2

⁢

𝐞

2

+

a

3

⁢

𝐞

3

,

b

1

⁢

𝐞

1

+

b

2

⁢

𝐞

2

+

b

3

⁢

𝐞

3

,

c

1

⁢

𝐞

1

+

c

2

⁢

𝐞

2

+

c

3

⁢

𝐞

3

|

The symmetric analogue would be an expansion of the form

(

a

1

⁢

x

+

a

2

⁢

y

+

a

3

⁢

z

)

⁢

(

b

1

⁢

x

+

b

2

⁢

y

+

b

3

⁢

z

)

⁢

(

c

1

⁢

x

+

c

2

⁢

y

+

c

3

⁢

z

)

=

a

1

⁢

b

1

⁢

c

1

⁢

x

3

+

⋯

In both cases, if we were to expand
fully, we would get an expression involving

3

⋅

3

⋅

3

=

27

terms. However, skew-symmetry tells us that a determinant that
involves repeated standard vectors will equal to zero. Thus, from the
27 terms we only keep the 6 terms that involve all three standard
standard vectors:

|

a

1

b

1

c

1

a

2

b

2

c

2

a

3

b

3

c

3

|

=

a

1

⁢

b

2

⁢

c

3

⁢

|

𝐞

1

,

𝐞

2

,

𝐞

3

|

+

a

2

⁢

b

3

⁢

c

1

⁢

|

𝐞

2

,

𝐞

3

,

𝐞

1

|

+

a

3

⁢

b

1

⁢

c

2

⁢

|

𝐞

3

,

𝐞

1

,

𝐞

2

|

+

+

a

1

⁢

b

3

⁢

c

2

⁢

|

𝐞

1

,

𝐞

3

,

𝐞

2

|

+

a

2

⁢

b

1

⁢

c

3

⁢

|

𝐞

2

,

𝐞

1

,

𝐞

3

|

+

a

3

⁢

b

2

⁢

c

1

⁢

|

𝐞

3

,

𝐞

2

,

𝐞

1

|

Using skew symmetry once again (one flip gives a minus sign; two flips
give a

plus sign

), and the identity rule we obtain

|

a

1

b

1

c

1

a

2

b

2

c

2

a

3

b

3

c

3

|

=

a

1

⁢

b

2

⁢

c

3

+

a

2

⁢

b

3

⁢

c

1

+

a

3

⁢

b

1

⁢

c

2

-

a

1

⁢

b

3

⁢

c

2

-

a

2

⁢

b

1

⁢

c

3

-

a

3

⁢

b

2

⁢

c

1

Notice that each of the 6 terms in the above expression involves an
entry from all 3 rows and from all 3 columns; it’s a sudoku kind of
thing. The choice of sign depends on the number of column flips
that takes the corresponding list of standard vectors to the identity
matrix; an

even number

of flips gives a

+

, an

odd number

a

-

1

.

5

Row operations.

Above, we observed that the terms in the final expression treat rows
and columns on an equal footing.

Theorem 1

.

Let

A

be a square matrix. Then,

det

⁡

(

A

)

=

det

⁡

(

A

T

)

.

In other words, transposing a matrix does not alter the value of its
determinant. One consequence of the row-column symmetry is that
everything one can say about determinants in terms of columns, one can
say using rows. In particular, we can state some useful facts about
the effect of

row operations

on the value of a determinant.

Again, for the sake of simplicity we focus on the

3

×

3

case.
Let

A

be a

3

×

3

matrix, expressed as a column of

row vectors

.
The hats above the symbols are there to remind us that we are dealing
with row rather than column vectors.

A

=

[

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

]

=

[

𝐚

^

1

𝐚

^

2

𝐚

^

3

]

;

𝐚

^

1

=

[

a

11

,

a

12

,

a

13

]

,

𝐚

^

2

=

[

a

21

,

a

22

,

a

23

]

,

𝐚

^

3

=

[

a

31

,

a

32

,

a

33

]

.

Below we list the effect of each of the 3 types of

elementary row
operations

on the determinant of a matrix, and a provide an explanation.

1.

Row replacements do not affect the value
of the determinant. Consider, for example the effect of a row
replacement operation

R

2

→

R

2

+

k

⁢

R

1

. Here

k

is a scalar and

E

is an

elementary matrix

that encodes the row operation in question:

E

=

[

1

0

0

k

1

0

0

0

1

]

,

det

⁡

(

E

⁢

A

)

=

|

𝐚

^

1

𝐚

^

2

+

k

⁢

𝐚

^

1

𝐚

^

3

|

=

|

𝐚

^

1

𝐚

^

2

𝐚

^

3

|

+

k

⁢

|

𝐚

^

1

𝐚

^

1

𝐚

^

3

|

=

det

⁡

(

A

)

.

2.

Row scaling operations scale the determinant by the same
factor. Consider, for example, the operation

R

2

→

k

⁢

R

2

:

E

=

[

1

0

0

0

k

0

0

0

1

]

,

det

⁡

(

E

⁢

A

)

=

|

𝐚

^

1

k

⁢

𝐚

^

2

𝐚

^

3

|

=

k

⁢

|

𝐚

^

1

𝐚

^

2

𝐚

^

3

|

=

k

⁢

det

⁡

(

A

)

3.

A row exchange operation negates the determinant. Consider, for
example,the exchange of rows

1

and

3

:

E

=

[

0

0

1

0

1

0

1

0

0

]

,

det

⁡

(

E

⁢

A

)

=

|

𝐚

^

3

𝐚

^

2

𝐚

^

1

|

=

-

|

𝐚

^

1

𝐚

^

2

𝐚

^

3

|

=

-

det

⁡

(

A

)

Above, if we take

A

to be the identity matrix, we obtain the value
for the determinant of an elementary matrix. We summarize as follows.

Theorem 2

.

Let

E

be an elementary matrix. Then,

det

⁡

(

E

⁢

A

)

=

det

⁡

(

E

)

⁢

det

⁡

(

A

)

where

det

⁡

(

E

)

=

{

1

if

E

encodes a row replacement;

k

if

E

encodes a scaling of a particular row by a factor
of

k

≠

0

;

-

1

if

E

encodes the exchange of two rows.

Of course, a

sequence

of elementary row operations can be used to
transform every matrix into

reduced echelon form

. This is a useful
observation, because it gives us an alternative method for computing
determinants; namely, we can compute the determinant of an

n

×

n

square matrix

A

by row reducing

A

to row-echelon form. Let’s
summarize this process by writing

E

k

⁢

⋯

⁢

E

1

⁢

A

=

U

,

det

⁡

(

E

k

)

⁢

⋯

⁢

det

⁡

(

E

1

)

⁢

det

⁡

(

A

)

=

det

⁡

(

U

)

,

where

E

1

,

…

,

E

k

is a sequence of elementary row operations, and
where

U

is a matrix in reduced echelon form. If

A

is singular,
then the bottom row of

U

will
consist of zeros, and hence

det

⁡

(

U

)

=

0

. Since the determinant of an
elementary matrix is never zero, this implies that

det

⁡

(

A

)

=

0

,
also.

If

A

is invertible, then

U

is the identity matrix, and hence

det

⁡

(

E

k

)

⁢

⋯

⁢

det

⁡

(

E

1

)

⁢

det

⁡

(

A

)

=

1

.

Since the determinant of an elementary matrix is explicitly known (see

Theorem

2

), this
gives us a way of calculating

det

⁡

(

A

)

.
We are also in a position to prove some important theorems.

Theorem 3

.

Let

A

be a square matrix. Then,

det

⁡

(

A

)

=

0

if
and only if

A

is singular.

Theorem 4

.

Let

A

,

B

be

n

×

n

matrices. Then,

det

⁡

(

A

⁢

B

)

=

det

⁡

(

A

)

⁢

det

⁡

(

B

)

.

Proof.

As above, Let

E

1

,

…

,

E

k

be the elementary
matrices that row reduce

A

to reduced echelon form;

E

k

⁢

⋯

⁢

E

1

⁢

A

=

U

.

Above, we showed that

det

⁡

(

U

⁢

B

)

=

det

⁡

(

E

k

⁢

⋯

⁢

E

1

⁢

A

⁢

B

)

=

det

⁡

(

E

k

)

⁢

⋯

⁢

det

⁡

(

E

1

)

⁢

det

⁡

(

A

⁢

B

)

.

If

A

is singular, then the bottom row of

U

will be zero, and so
will the bottom row of

U

⁢

B

. Hence, in this case,

det

⁡

(

A

⁢

B

)

=

0

=

det

⁡

(

A

)

⁢

det

⁡

(

B

)

, because

det

⁡

(

A

)

=

0

, also. Suppose then that

A

is
invertible. This means that

U

is the identity matrix, and hence

det

⁡

(

B

)

=

det

⁡

(

E

k

)

⁢

⋯

⁢

det

⁡

(

E

1

)

⁢

det

⁡

(

A

⁢

B

)

.

However,

det

⁡

(

E

k

)

⁢

⋯

⁢

det

⁡

(

E

1

)

=

1

/

det

⁡

(

A

)

,

and the desired

conclusion

follows.

Theorem 5

.

Let

A

be an

invertible matrix

. Then

det

⁡

(

A

-

1

)

=

1

/

det

⁡

(

A

)

.

Proof.

By the above theorem,

1

=

det

⁡

(

I

)

=

det

⁡

(

A

⁢

A

-

1

)

=

det

⁡

(

A

)

⁢

det

⁡

(

A

-

1

)

.

6

Cofactor expansion.

Cofactor expansion

is another method for evaluating determinants. It
organizes the

computation

of larger determinants, and can be useful in
calculating the determinants of matrices containing zero entries.
At this point, we introduce some useful jargon. Given an

n

×

n

matrix

A

, we call the

(

n

-

1

)

×

(

n

-

1

)

matrix obtained by deleting
row

i

and column

j

the

i

⁢

j

minor of

A

, and denote it by

A

i

⁢

j

. We also set

C

i

⁢

j

=

(

-

1

)

i

+

j

⁢

det

⁡

(

A

i

⁢

j

)

and call

C

i

⁢

j

the

i

⁢

j

signed

cofactor

of

A

.
We are going to prove the following.

Theorem 6

.

Consider a

3

×

3

matrix

A

=

[

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

]

.

The determinant of

A

can be
obtained by means of cofactor expansion along the first column:

det

⁡

(

A

)

=

a

11

⁢

C

11

+

a

21

⁢

C

21

+

a

31

⁢

C

31

=

a

11

⁢

|

a

22

a

23

a

32

a

33

|

-

a

21

⁢

|

a

12

a

13

a

32

a

33

|

+

a

31

⁢

|

a

12

a

13

a

22

a

23

|

;

or, along the first row:

det

⁡

(

A

)

=

a

11

⁢

C

11

+

a

12

⁢

C

12

+

a

13

⁢

C

13

=

a

11

⁢

|

a

22

a

23

a

32

a

33

|

-

a

12

⁢

|

a

21

a

23

a

31

a

33

|

+

a

13

⁢

|

a

21

a

22

a

31

a

32

|

.

More generally, the determinant of an

n

×

n

matrix can be
obtained by cofactor expansion along any column

j

:

det

⁡

(

A

)

=

a

1

⁢

j

⁢

C

1

⁢

j

+

a

2

⁢

j

⁢

C

2

⁢

j

+

⋯

+

a

n

⁢

j

⁢

C

n

⁢

j

,

j

=

1

,

2

,

…

,

n

;

or, along any row

i

:

det

⁡

(

A

)

=

a

i

⁢

1

⁢

C

i

⁢

1

+

a

i

⁢

2

⁢

C

i

⁢

2

+

⋯

+

a

i

⁢

n

⁢

C

i

⁢

n

,

i

=

1

,

2

,

…

,

n

.

The proof works by writing

A

=

[

𝐚

1

,

𝐚

2

,

𝐚

3

]

,

𝐚

1

=

a

11

⁢

𝐞

1

+

a

21

⁢

𝐞

2

+

a

31

⁢

𝐞

3

,

and then using multi-linearity:

|

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

a

11

⁢

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

+

a

21

⁢

|

𝐞

2

,

𝐚

2

,

𝐚

3

|

+

a

31

⁢

|

𝐞

3

,

𝐚

2

,

𝐚

3

|

=

a

11

⁢

|

1

a

12

a

13

0

a

22

a

23

0

a

32

a

33

|

+

a

21

⁢

|

0

a

12

a

13

1

a

22

a

23

0

a

32

a

33

|

+

a

31

⁢

|

0

a

12

a

13

0

a

22

a

23

1

a

32

a

33

|

=

a

11

⁢

|

a

22

a

23

a

32

a

33

|

-

a

21

⁢

|

a

12

a

13

a

32

a

33

|

+

a

31

⁢

|

a

12

a

23

a

13

a

23

|

=

a

11

⁢

C

11

+

a

21

⁢

C

21

+

a

31

⁢

C

31

.

The intermediate steps, namely

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

=

C

11

,

|

𝐞

2

,

𝐚

2

,

𝐚

3

|

=

C

21

,

|

𝐞

3

,

𝐚

2

,

𝐚

3

|

=

C

31

.

need to be explained.

Theorem 7

.

Let

A

=

[

a

1

,

a

2

,

a

3

,

…

,

a

n

]

be an

n

×

n

matrix.
Then,

|

𝐞

i

,

𝐚

2

,

𝐚

3

⁢

…

,

𝐚

n

|

=

C

i

⁢

1

,

|

𝐚

1

,

𝐞

i

,

𝐚

3

,

…

,

𝐚

n

|

=

C

i

⁢

2

,

|

𝐚

1

,

𝐚

2

,

𝐞

i

,

…

,

𝐚

n

|

=

C

i

⁢

3

,

⁢

𝑒𝑡𝑐

Proof.

Expanding

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

we obtain 9 terms:

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

=

|

𝐞

1

,

a

12

⁢

𝐞

1

+

a

22

⁢

𝐞

2

+

a

32

⁢

𝐞

3

,

a

13

⁢

𝐞

1

+

a

23

⁢

𝐞

2

+

a

33

⁢

𝐞

3

|

.

However, there can’t be any terms with a double

occurrence

of

𝐞

1

,
and so we end up evaluating a

2

×

2

determinant:

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

=

|

𝐞

1

,

a

22

⁢

𝐞

2

+

a

32

⁢

𝐞

3

,

a

23

⁢

𝐞

2

+

a

33

⁢

𝐞

3

|

=

a

22

⁢

a

33

⁢

|

𝐞

1

,

𝐞

2

,

𝐞

3

|

+

a

22

⁢

a

23

⁢

|

𝐞

1

,

𝐞

2

,

𝐞

2

|

+

a

32

⁢

a

23

⁢

|

𝐞

1

,

𝐞

3

,

𝐞

2

|

+

a

32

⁢

a

33

⁢

|

𝐞

1

,

𝐞

3

,

𝐞

3

|

=

|

a

22

a

23

a

32

a

33

|

=

C

11

Similarly,

|

𝐞

2

,

𝐚

2

,

𝐚

3

|

=

-

|

𝐚

2

,

𝐞

2

,

𝐚

3

|

=

-

|

a

12

0

a

13

a

22

1

a

23

a

32

0

a

33

|

=

-

|

a

12

a

13

a

32

a

33

|

=

C

21

.

|

𝐞

3

,

𝐚

2

,

𝐚

3

|

=

-

|

𝐚

2

,

𝐞

3

,

𝐚

3

|

=

+

|

𝐚

2

,

𝐚

3

,

𝐞

3

|

=

+

|

a

12

a

13

0

a

22

a

23

0

a

32

a

33

1

|

=

|

a

12

a

13

a

22

a

23

|

=

C

31

.

This argument generalizes to matrices of arbitrary size.

Next, by way of example, let’s consider the expansion of a

3

×

3

matrix along the 2nd column:

|

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

|

𝐚

1

,

a

12

⁢

𝐞

1

+

a

22

⁢

𝐞

2

+

a

32

⁢

𝐞

3

,

𝐚

3

|

=

a

12

⁢

|

𝐚

1

,

𝐞

1

,

𝐚

3

|

+

a

22

⁢

|

𝐚

1

,

𝐞

2

,

𝐚

3

|

+

a

32

⁢

|

𝐚

1

,

𝐞

3

,

𝐚

3

|

=

a

12

⁢

|

a

11

1

a

13

a

21

0

a

23

a

31

0

a

33

|

+

a

22

⁢

|

a

11

0

a

13

a

21

1

a

23

a

31

0

a

33

|

+

a

32

⁢

|

a

11

0

a

13

a

21

0

a

23

a

31

1

a

33

|

=

a

12

⁢

|

a

21

a

23

a

31

a

33

|

-

a

22

⁢

|

a

11

a

13

a

31

a

33

|

+

a

32

⁢

|

a

11

a

13

a

21

a

23

|

=

a

12

⁢

C

12

+

a

22

⁢

C

22

+

a

32

⁢

C

32

.

The above argument generalizes to expansions along any column, and
indeed to expansions of a matrix of arbitrary size.
For
example, for a

4

×

4

matrix we write

𝐚

1

=

a

11

⁢

𝐞

1

⁢

a

21

⁢

𝐞

2

+

a

31

⁢

𝐞

3

+

a

41

⁢

𝐞

4

and use multi-linearity to obtain

|

a

11

a

12

a

13

a

14

a

21

a

22

a

23

a

24

a

31

a

32

a

33

a

34

a

41

a

42

a

43

a

44

|

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

,

𝐚

4

|

=

a

11

⁢

|

𝐞

1

,

𝐚

2

,

𝐚

3

,

𝐚

4

|

+

a

21

⁢

|

𝐞

2

,

𝐚

2

,

𝐚

3

,

𝐚

4

|

+

a

31

⁢

|

𝐞

3

,

𝐚

2

,

𝐚

3

,

𝐚

4

|

+

a

41

⁢

|

𝐞

4

,

𝐚

2

,

𝐚

3

,

𝐚

4

|

=

a

11

⁢

C

11

+

a

21

⁢

C

21

+

a

31

⁢

C

31

+

a

41

⁢

C

41

.

Working with row vectors, the same argument also establishes the

validity

of cofactor expansion along rows. For example, here is the

derivation

of
cofactor expansion along the 2nd row of a

3

×

3

matrix:

|

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

|

=

|

𝐚

^

1

𝐚

^

2

𝐚

^

3

|

=

|

𝐚

^

1

a

21

⁢

𝐞

^

1

+

a

22

⁢

𝐞

^

2

+

a

23

⁢

𝐞

^

3

𝐚

^

3

|

=

a

21

⁢

|

𝐚

^

1

𝐞

^

1

𝐚

^

3

|

+

a

22

⁢

|

𝐚

^

1

𝐞

^

2

𝐚

^

3

|

+

a

23

⁢

|

𝐚

^

1

𝐞

^

3

𝐚

^

3

|

=

a

21

⁢

C

21

+

a

22

⁢

C

22

+

a

23

⁢

C

23

.

Here

𝐞

^

i

=

𝐞

i

T

denote the

elementary

row vectors.

Here is a useful theorem about determinants that can be proved using
cofactor expansions.

Theorem 8

.

The determinant of an

upper
triangular matrix

is the

product

of the

diagonal

entries.

Consider, for example the determinant of the following

4

×

4

upper triangular matrix; the stars indicate an arbitrary number. We
repeatedly use cofactor expansion along the first column. The

multiple

zeros mean that, each time, the cofactor expansion has only one term.

|

a

*

*

*

0

b

*

*

0

0

c

*

0

0

0

d

|

=

a

⁢

|

b

*

*

0

c

*

0

0

d

|

=

a

⁢

b

⁢

|

c

*

0

d

|

=

a

⁢

b

⁢

c

⁢

d

7

The adjugate matrix.

The cofactors

C

i

⁢

j

of an

n

×

n

matrix

A

can be arranged
into an

n

×

n

matrix, called

adj

⁡

A

, the adjugate of

A

. In
the

3

×

3

case we define

adj

⁡

A

=

[

C

11

C

21

C

31

C

12

C

22

C

32

C

13

C

23

C

33

]

.

Note that the entries of

adj

⁡

A

are indexed differently than the
entries of

A

. For

A

, the entry in the

i

-th row and

j

-th
column is denoted by

a

i

⁢

j

. However the cofactor

C

i

⁢

j

is
placed in row

j

and column

i

. Remarkably, the matrix of cofactors

adj

⁡

A

is closely related to the

inverse

of

A

.

Theorem 9

.

Let

A

be an

n

×

n

matrix. Then,

A

⁢

adj

⁡

A

=

det

⁡

(

A

)

⁢

I

.
Furthermore, if

A

is invertible, then

A

-

1

=

(

1

/

det

⁡

(

A

)

)

⁢

adj

⁡

A

.

Let’s consider the proof for the

3

×

3

case. We aim
to show that

[

C

11

C

21

C

31

C

12

C

22

C

32

C

13

C

23

C

33

]

⁢

[

a

11

a

12

a

13

a

21

a

22

a

23

a

31

a

32

a

33

]

=

[

det

⁡

(

A

)

0

0

0

det

⁡

(

A

)

0

0

0

det

⁡

(

A

)

]

Writing

A

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

, using the properties of the determinant and Theorem

7

det

⁡

(

A

)

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

a

11

⁢

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

+

a

21

⁢

|

𝐞

2

,

𝐚

2

,

𝐚

3

|

+

a

31

⁢

|

𝐞

3

,

𝐚

2

,

𝐚

3

|

=

a

11

⁢

C

11

+

a

21

⁢

C

21

+

a

31

⁢

C

31

.

0

=

|

𝐚

1

,

𝐚

1

,

𝐚

3

|

=

a

11

⁢

|

𝐚

1

,

𝐞

1

,

𝐚

3

|

+

a

21

⁢

|

𝐚

1

,

𝐞

2

,

𝐚

3

|

+

a

31

⁢

|

𝐚

1

,

𝐞

3

,

𝐚

3

|

=

a

11

⁢

C

12

+

a

21

⁢

C

22

+

a

31

⁢

C

32

.

0

=

|

𝐚

1

,

𝐚

2

,

𝐚

1

|

=

a

11

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

1

|

+

a

21

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

2

|

+

a

31

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

3

|

=

a

11

⁢

C

13

+

a

21

⁢

C

23

+

a

31

⁢

C

33

.

This gives us the first column of the multiplication. To obtain the
second column, we observe

0

=

|

𝐚

2

,

𝐚

2

,

𝐚

3

|

=

a

12

⁢

|

𝐞

1

,

𝐚

2

,

𝐚

3

|

+

a

22

⁢

|

𝐞

2

,

𝐚

2

,

𝐚

3

|

+

a

32

⁢

|

𝐞

3

,

𝐚

2

,

𝐚

3

|

=

a

12

⁢

C

11

+

a

22

⁢

C

21

+

a

32

⁢

C

31

.

det

⁡

(

A

)

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

a

12

⁢

|

𝐚

1

,

𝐞

1

,

𝐚

3

|

+

a

22

⁢

|

𝐚

1

,

𝐞

2

,

𝐚

3

|

+

a

32

⁢

|

𝐚

1

,

𝐞

3

,

𝐚

3

|

=

a

12

⁢

C

12

+

a

22

⁢

C

22

+

a

32

⁢

C

32

.

0

=

|

𝐚

1

,

𝐚

2

,

𝐚

2

|

=

a

12

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

1

|

+

a

22

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

2

|

+

a

32

⁢

|

𝐚

1

,

𝐚

2

,

𝐞

3

|

=

a

12

⁢

C

13

+

a

22

⁢

C

23

+

a

31

⁢

C

33

.

The values in the 3rd column are established in a

similar

fashion.

8

Cramer’s rule

We can also use determinants and cofactors to solve a

linear system

A

⁢

𝐱

=

𝐛

, where

A

is an invertible, square matrix.
Of course, if

A

is invertible, then a solution exists and is
unique; indeed,

𝐱

=

A

-

1

⁢

𝐛

. However, Cramer’s rule allows us
to

calculate

𝐱

directly, without first calculating

A

-

1

and
then performing a matrix-vector multiplication.

Let’s
see how this works for the case of a

3

×

3

matrix

A

=

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

. Given a

𝐛

∈

ℝ

3

, we are

searching

for the

3

numbers

x

1

,

x

2

,

x

3

such that

𝐛

=

x

1

⁢

𝐚

1

+

x

2

⁢

𝐚

2

+

x

3

⁢

𝐚

3

.

Substituting this into the
following determinant and expanding produces a useful equation:

|

𝐛

,

𝐚

2

,

𝐚

3

|

=

|

x

1

⁢

𝐚

1

+

x

2

⁢

𝐚

2

+

x

3

⁢

𝐚

3

,

𝐚

2

,

𝐚

3

|

=

x

1

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

+

x

2

⁢

|

𝐚

2

,

𝐚

2

,

𝐚

3

|

+

x

3

⁢

|

𝐚

3

,

𝐚

2

,

𝐚

3

|

=

x

1

⁢

det

⁡

(

A

)

+

0

+

0

.

Similarly

|

𝐚

1

,

𝐛

,

𝐚

3

|

=

x

1

⁢

|

𝐚

1

,

𝐚

1

,

𝐚

3

|

+

x

2

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

+

x

3

⁢

|

𝐚

1

,

𝐚

3

,

𝐚

3

|

=

x

2

⁢

det

⁡

(

A

)

|

𝐚

1

,

𝐚

2

,

𝐛

|

=

x

1

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

1

|

+

x

2

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

2

|

+

x

3

⁢

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

=

x

3

⁢

det

⁡

(

A

)

Therefore, the desired solution can be obtained as follows:

x

1

=

|

𝐛

,

𝐚

2

,

𝐚

3

|

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

x

2

=

|

𝐚

1

,

𝐛

,

𝐚

3

|

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

x

3

=

|

𝐚

1

,

𝐚

2

,

𝐛

|

|

𝐚

1

,

𝐚

2

,

𝐚

3

|

We generalize and summarize as follows.

Theorem 10

.

Let

A

be an invertible

n

×

n

matrix, and

b

∈

R

n

a
vector. Then the unique solution to the linear equation

A

⁢

x

=

b

is given by

x

i

=

det

⁡

A

i

⁢

(

𝐛

)

det

⁡

A

,

i

=

1

,

2

,

…

,

n

,

where

A

i

⁢

(

b

)

denotes the matrix obtained by replacing column

i

of

A

with

b

.

Title

lecture notes on determinants

Canonical

name

LectureNotesOnDeterminants

Date of creation

2013-03-22 17:33:37

Last modified on

2013-03-22 17:33:37

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

6

Author

rmilson (146)

Entry type

Topic

Classification

msc 15A15

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