[PPT] - Determinants Artem Los (arteml@kth.se) February 6th, 2017 Artem PowerPoint Presentation

SLIDE 1

Determinants

Artem Los (arteml@kth.se) February 6th, 2017

Artem Los (arteml@kth.se) Determinants February 6th, 2017 1 / 16

SLIDE 2

Overview

1

What is a Determinant?

2

Rules and Theorems

3

Area of a Parallelogram

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SLIDE 3

What is a Determinant?

Artem Los (arteml@kth.se) Determinants February 6th, 2017 3 / 16

SLIDE 4

Definition

Determinant

In linear algebra, the determinant is a useful value that can be computed from the elements of a square matrix.

(from https://en.wikipedia.org/wiki/Determinant) Artem Los (arteml@kth.se) Determinants February 6th, 2017 4 / 16

SLIDE 5

Definition

Determinant

In linear algebra, the determinant is a useful value that can be computed from the elements of a square matrix.

(from https://en.wikipedia.org/wiki/Determinant)

For example, Deducing if matrix A is invertible, i.e. ∃A−1 Finding area of a parallelogram Calculating eigenvalues (next week)

Artem Los (arteml@kth.se) Determinants February 6th, 2017 4 / 16

SLIDE 6

Computing determinant

The neat property of determinants is that we can always express a determinant of matrix n × n using determinants of size (n − 1) × (n − 1).

Artem Los (arteml@kth.se) Determinants February 6th, 2017 5 / 16

SLIDE 7

Computing determinant

The neat property of determinants is that we can always express a determinant of matrix n × n using determinants of size (n − 1) × (n − 1). Base case. 2 × 2

a

b c d

= ad − bc

Artem Los (arteml@kth.se) Determinants February 6th, 2017 5 / 16

SLIDE 8

Computing determinant

The neat property of determinants is that we can always express a determinant of matrix n × n using determinants of size (n − 1) × (n − 1). Base case. 2 × 2

a

b c d

= ad − bc

Increasing dimension. 3 × 3

a

b c d e f j h i

= a
e

f h i

− b
d

f j i

+ c
d

e j h

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 5 / 16

SLIDE 9

Computing determinant

The neat property of determinants is that we can always express a determinant of matrix n × n using determinants of size (n − 1) × (n − 1). Base case. 2 × 2

a

b c d

= ad − bc

Increasing dimension. 3 × 3

a

b c d e f j h i

= a
e

f h i

− b
d

f j i

+ c
d

e j h

a

b c d e f j h i

= a
e

f h i

− d
b

c h i

+ j
b

c e f

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 5 / 16

SLIDE 10

Determine sign of the coefficients

Question. In the example below, why is a positive but b negative?
a

b c d e f j h i

= a
e

f h i

− b
d

f j i

+ c
d

e j h

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 6 / 16

SLIDE 11

Determine sign of the coefficients

Question. In the example below, why is a positive but b negative?
a

b c d e f j h i

= a
e

f h i

− b
d

f j i

+ c
d

e j h

For 3 × 3, we have that:

  + − + − + − + − +  

Artem Los (arteml@kth.se) Determinants February 6th, 2017 6 / 16

SLIDE 12

Determine sign of the coefficients

Question. In the example below, why is a positive but b negative?
a

b c d e f j h i

= a
e

f h i

− b
d

f j i

+ c
d

e j h

For 3 × 3, we have that:

  + − + − + − + − +   In general, the coefficient is (−1)i+j for element in ith row and jth column

Artem Los (arteml@kth.se) Determinants February 6th, 2017 6 / 16

SLIDE 13

Example

Problem. Compute the determinant of A, defined as:

A =   1 2 1 1 3 3 6  

Artem Los (arteml@kth.se) Determinants February 6th, 2017 7 / 16

SLIDE 14

Example

Problem. Compute the determinant of A, defined as:

A =   1 2 1 1 3 3 6   Step 1: ”Golden rule of determinant calculations: be lazy”. Pick the row/column with most zeros. Why? A =

1

2 1 1 3 3 6

= 6
1

2 1 3

= 6(3 − 2) = 6

Artem Los (arteml@kth.se) Determinants February 6th, 2017 7 / 16

SLIDE 15

Rules and Theorems

Artem Los (arteml@kth.se) Determinants February 6th, 2017 8 / 16

SLIDE 16

Rules (idea)

Idea. Determinant matrix reductions are carried out to make it easier to

figure out the determinant. The rules differ from elementary row

perations.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 9 / 16

SLIDE 17

Rules (idea)

Idea. Determinant matrix reductions are carried out to make it easier to

figure out the determinant. The rules differ from elementary row

perations.
Goal. Get as many zeros as possible, for then we get less terms to

compute.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 9 / 16

SLIDE 18

Rules (continued)

Constant term factorization. det B = c det A

c ∗ 1

c ∗ 2 c ∗ 3 4 5 6 7 8 9

= c
1

2 3 4 5 6 7 8 9

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 10 / 16

SLIDE 19

Rules (continued)

Constant term factorization. det B = c det A

c ∗ 1

c ∗ 2 c ∗ 3 4 5 6 7 8 9

= c
1

2 3 4 5 6 7 8 9

Swapping two rows.

det B = − det A

1

2 3 4 5 6 7 8 9

= −
4

5 6 1 2 3 7 8 9

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 10 / 16

SLIDE 20

Rules (continued)

Constant term factorization. det B = c det A

c ∗ 1

c ∗ 2 c ∗ 3 4 5 6 7 8 9

= c
1

2 3 4 5 6 7 8 9

Swapping two rows.

det B = − det A

1

2 3 4 5 6 7 8 9

= −
4

5 6 1 2 3 7 8 9

Adding two rows/columns.

det B = det A

1

1 3 1 1 6 7 8 9

= {R2 : R2 − R1} =
1

1 3 6 7 8 9

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 10 / 16

SLIDE 21

Theorems

Theorem. Let A be a square matrix, i.e. n × n. Then:

det A = det AT

Artem Los (arteml@kth.se) Determinants February 6th, 2017 11 / 16

SLIDE 22

Theorems

Theorem. Let A be a square matrix, i.e. n × n. Then:

det A = det AT det A−1 =

1 det A

Artem Los (arteml@kth.se) Determinants February 6th, 2017 11 / 16

SLIDE 23

Theorems

Theorem. Let A be a square matrix, i.e. n × n. Then:

det A = det AT det A−1 =

1 det A

det AB = det A det B

Artem Los (arteml@kth.se) Determinants February 6th, 2017 11 / 16

SLIDE 24

Theorems

Theorem. Let A be a square matrix, i.e. n × n. Then:

det A = det AT det A−1 =

1 det A

det AB = det A det B If two rows are equal = ⇒ det A = 0

Artem Los (arteml@kth.se) Determinants February 6th, 2017 11 / 16

SLIDE 25

Theorems

Theorem. Let A be a square matrix, i.e. n × n. Then:

det A = det AT det A−1 =

1 det A

det AB = det A det B If two rows are equal = ⇒ det A = 0

Theorem. Given that A is a square matrix, i.e. n × n, the following

statements are equivalent: det A = 0 rank(A) = n A is invertible

Artem Los (arteml@kth.se) Determinants February 6th, 2017 11 / 16

SLIDE 26

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4    

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 27

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 28

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 29

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=
1

1 2 −1 1 1 1 −4 1 4

=

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 30

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=
1

1 2 −1 1 1 1 −4 1 4

= −
1

1 2 1 −1 1 −3 1 1 4

=

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 31

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=
1

1 2 −1 1 1 1 −4 1 4

= −
1

1 2 1 −1 1 −3 1 1 4

=

= −

1

−1 1 −3 1 1 4

=

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 32

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=
1

1 2 −1 1 1 1 −4 1 4

= −
1

1 2 1 −1 1 −3 1 1 4

=

= −

1

−1 1 −3 1 1 4

= −
−3

1 1 4

=

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 33

Example

Problem. Find det ATA given that A is defined as shown below:

A =     1 1 2 1 3 1 2 3 1 4     Step 1: (using known theorem): det (ATA) = (det A)2. Step 2: Find det A.

1

1 2 1 3 1 2 3 1 4

=
1

1 2 −1 1 1 1 −4 1 4

= −
1

1 2 1 −1 1 −3 1 1 4

=

= −

1

−1 1 −3 1 1 4

= −
−3

1 1 4

= − (−3 × 4 − 1 × 1) = 13

Artem Los (arteml@kth.se) Determinants February 6th, 2017 12 / 16

SLIDE 34

Area of a Parallelogram

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SLIDE 35

Finding area given two spanning vectors

Area y x

u
v

Given a vectors u, v that span a parallelogram, the area is given by: Area =

det

u1 u2 v1 v2

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 14 / 16

SLIDE 36

Finding area given two spanning vectors

Area y x

u
v

Given a vectors u, v that span a parallelogram, the area is given by: Area =

det

u1 u2 v1 v2

In general, if vectors

u, v don’t lie in the same plane (eg. u, v ∈ R3) Area = || u × v|| =

det

  i j k u1 u2 u3 v1 v2 v3  

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 14 / 16

SLIDE 37

Example

Problem. Given that a parallelogram is spanned by vectors u = (1, 2) and v = (3, 4), find its area.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 15 / 16

SLIDE 38

Example

Problem. Given that a parallelogram is spanned by vectors u = (1, 2) and v = (3, 4), find its area. Step 1: Plug vectors into a matrix:

1

2 3 4

= 1 ∗ 4 − 2 ∗ 3 = −2

Artem Los (arteml@kth.se) Determinants February 6th, 2017 15 / 16

SLIDE 39

Example

Problem. Given that a parallelogram is spanned by vectors u = (1, 2) and v = (3, 4), find its area. Step 1: Plug vectors into a matrix:

1

2 3 4

= 1 ∗ 4 − 2 ∗ 3 = −2

Step 2: Take the absolute value, since it’s an area: | − 2| = 2

Artem Los (arteml@kth.se) Determinants February 6th, 2017 15 / 16

SLIDE 40

Example

Problem. Given that a parallelogram is spanned by vectors u = (1, 2) and v = (3, 4), find its area. Step 1: Plug vectors into a matrix:

1

2 3 4

= 1 ∗ 4 − 2 ∗ 3 = −2

Step 2: Take the absolute value, since it’s an area: | − 2| = 2 ∴ Area is equal to 2 area units.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 15 / 16

SLIDE 41

Volume of parallelepiped

Given vectors u, v, w ∈ R3, the volume of the parallelepiped that they span up is: Volume = | w · ( u × v)| =

det

  u1 u2 u3 v1 v2 v3 w1 w2 w3  

Artem Los (arteml@kth.se)

Determinants February 6th, 2017 16 / 16

SLIDE 42

Volume of parallelepiped

Given vectors u, v, w ∈ R3, the volume of the parallelepiped that they span up is: Volume = | w · ( u × v)| =

det

  u1 u2 u3 v1 v2 v3 w1 w2 w3  

Although the geometrical fact is useful, keep in mind the trick of

converting | w · ( u × v)| to the determinant form.

Artem Los (arteml@kth.se) Determinants February 6th, 2017 16 / 16