SCUT, Liu Rui

Chapter 3 Determinants

Liu Rui

School of Mathematics

South China University of Technology

scliurui@scut.edu.cn

2019-11-15

SCUT, Liu Rui

1 § 3.3 Cramer’s Rule (�4%{K), Volume, and Linear Transfor-

mations

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Determinant of a 3× 3 Matrix

Exercise

1. If det(A) = 3, det(A3) =?

2. If det(A3) = 0, det(A) =?

3. If A is a 4× 4 matrix, det(−A) =?

4. If A is a 3× 3 matrix, det(2A) =?

5. If det(A) = 5, det(A−1) =?

SCUT, Liu Rui

A linearity property of the determinant function

For an n × n matrix A, the determinant det(A) can be re-

garded as a map from Rn to R1 as follows:

Suppose that the j-th column of A is allowed to vary (an

unknown vector variable):

A = [~a1 ... ~aj−1 ~x ~aj+1 ... ~an].

Define a transformation T from Rn to R1 by

T (~x) = det(A) = det([~a1 ... ~aj−1 ~x ~aj+1 ... ~an])

Then

T (c~x) = cT (~x)

T (~x + ~y) = T (~x) + T (~y)

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A linearity property of the determinant function

Note that,

det(A + B) 6= det(A) + det(B).

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Linear transformation

T (~x + ~y) =

∣∣∣∣∣∣∣∣∣∣a11 a12 · · · x1 + y1 · · · a1n

......

......

......

......

an1 an2 · · · xn + yn · · · ann

∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣∣∣a11 · · · x1 · · · a1n

......

......

......

an1 · · · xn · · · ann

∣∣∣∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣∣∣∣a11 · · · y1 · · · a1n

......

......

......

an1 · · · yn · · · ann

∣∣∣∣∣∣∣∣∣∣= T (~x) + T (~y).

SCUT, Liu Rui

Linear transformation

T (c~x) =

∣∣∣∣∣∣∣∣∣∣a11 a12 · · · cx1 · · · a1n

......

......

......

......

an1 an2 · · · cxn · · · ann

∣∣∣∣∣∣∣∣∣∣

= c

∣∣∣∣∣∣∣∣∣∣a11 · · · x1 · · · a1n

......

......

......

an1 · · · xn · · · ann

∣∣∣∣∣∣∣∣∣∣= cT (~x).

SCUT, Liu Rui

Cramer’s Rule (���444%%%{{{KKK)

Theorem (Cramer’s Rule (�4%{K) )

Let A be an invertible n× n matrix. For any ~b in Rn, the unique

solution ~x =

x1

...

xn

of A~x = ~b has elements given by

xi =det Ai(~b)

det A, i = 1, 2, ..., n.

Notation: Ai(~b) is a matrix obtained by replacing the i-th

column of A by ~b.

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Cramer’s Rule

Notation: For the coefficient matrix A, denote Ai(~b) as a

matrix by replacing the i-th column of A by a vector ~b.

Before replacing, A = [~a1, ~a2, ..., ~ai, ..., ~an]

↑the i-th column

After replacing, Ai(~b) = [~a1, ~a2, ..., ~b, ..., ~an]

↑the i-th column

SCUT, Liu Rui

Cramer’s Rule

After replacing, Ai(~b) = [~a1, ~a2, ..., ~b, ..., ~an]

↑the i-th column

det(Ai(~b)) =

∣∣∣∣∣∣∣∣∣∣∣∣

a11 a12 · · · b1 · · · a1n

......

......

......

......

an1 an2 · · · bn · · · ann

∣∣∣∣∣∣∣∣∣∣∣∣

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Proof of Cramer’s Rule

Proof:

(1). Denote the columns of A by ~a1, ~a2, ..., ~an and the column-

s of the n× n identity matrix I by ~e1, ~e2, ..., ~en.

(2). Denote Ii(~x) is a matrix obtained by replacing the i-th

column of the identity I by ~x.

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Proof of Cramer’s Rule

If A~x = ~b, the definition of matrix multiplication shows that

where Ai(~b) is a matrix obtained by replacing the i-th column

of A by ~b.

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Proof of Cramer’s Rule

From

AIi(~x) = Ai(~b),

we calculate the determinants from both two sides by the

multiplicative property of determinants

det(A)det(Ii(~x)) = det(AIi(~x)) = det(Ai(~b))

The second determinant det(Ii(~x)) on the leftmost side is xi.

(Question: WHY? Let’s see the matrix in the next slide)

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Proof of Cramer’s Rule

det(Ii(~x)) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 0 · · · 0 x1 0 · · · 0

0 1 · · · 0 x2 0 · · · 0... 0

. . ....

...... · · ·

......

.... . . 1 xi−1

... · · ·...

...... 0 xi 0 · · · 0

......

... xi+1 1. . .

......

......

......

. . . 0

0 0 · · · 0 xn 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣Expand the determinant by the 1st column.

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Proof of Cramer’s Rule

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 · · · 0 x2 0 · · · 0

0. . .

......

... · · ·...

.... . . 1 xi−1

... · · ·...

... 0 xi 0 · · · 0

...... xi+1 1

. . ....

......

......

. . . 0

0 · · · 0 xn 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

= · · · =

∣∣∣∣∣∣∣∣∣∣∣∣

xi 0 · · · 0

xi+1 1. . .

......

.... . . 0

xn 0 · · · 1

∣∣∣∣∣∣∣∣∣∣∣∣

SCUT, Liu Rui

Proof of Cramer’s Rule

Since det(Ii(~x)) = xi,

det(A)det(Ii(~x)) = det(A) · xi = det(Ai(~b)).

Remember A is invertible, hence we have

xi =det(Ai(~b))

det(A), i = 1, 2, ..., n.

�

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Example

Example:

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Example

Solution: The equivalent matrix equation is A~x = ~b.

Since det(A) = 2, the equation has a unique solution. By

Cramer’s rule:

[�]

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A Formula for A−1

From Cramer’s rule, we can give a general formula for the

inverse of an n× n matrix A.

Remember that the j-th column of A−1 is a vector ~x, which

satisfies

A~x = ~ej

where ~ej is the j-th column of the identity matrix I, and the

i-th element of ~x is the element in the (i, j)-position of A−1.

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A Formula for A−1

By using Cramer’s rule, we have

Recall that the minor Aji denotes the submatrix corresponding

with aji, which is formed by deleting row j and column i. A

cofactor expansion down column i of Ai(~ej) shows that

det(Ai(~ej)) = (−1)j+idet(Aji) = Cji

where Cji is a cofactor corresponding with aji (see next page).

SCUT, Liu Rui

A Formula for A−1

det(Ai(~ej)) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a11 · · · a1,i−1 0 a1,i+1 · · · a1n

a21 · · · a2,i−1 0 a2,i+1 a2n

......

......

......

... 0...

...

aj,1 · · · aj,i−1 1 aj,i+1 · · · ajn

...... 0

......

......

......

...

an1 · · · an,i−1 0 an,i+1 · · · ann

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

= Cji

Expand the determinant by the ith column.

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A Formula for A−1

Therefore,

{(i, j)− element of A−1} =Cji

det(A).

Note the subscripts of Cji are just the reverse of (i, j)

Thus

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Adjoint Matrix

Definition

Let A = (aij)nn, matrix

adj(A) =

C11 C21 · · · Cn1

C12 C22 · · · Cn2

......

. . ....

C1n C2n · · · Cnn

ia called the adjoint matrix (��Ý) of A, where Cji is the

cofactor of aji.

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Adjoint Matrix

Theorem

An invertible n× n matrix A has its inverse:

A−1 =1

det(A)adj(A)

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Example

Example:

Find the inverse of the matrix

A =

2 1 3

1 −1 1

1 4 −2

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Exercise

Solution:

adj(A) =

C11 C21 C31

C12 C22 C32

C13 C23 C33

=

−2 14 4

3 −7 1

5 −7 −3

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Exercise

Solution:

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Exercise

Exercise: Compute the adjoint matrix and the reverse matrix

of A

1. A =

1 1 3

2 −2 1

0 1 0

, 2. A =

3 5 4

1 0 1

2 1 1

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Exercise

Solution:

1. A−1 =

−1/5 3/5 7/5

0 0 1

2/5 −1/5 −4/5

,

2. A−1 =

−1/6 −1/6 5/6

1/6 −5/6 1/6

1/6 7/6 −5/6

SCUT, Liu Rui

Application of determinants: area or volume

Determinants as area or volume

In this section, we introduce the geometric interpretation of

determinants.

Definition1 If A is a 2×2 matrix, the area of the parallelogram determined

by the columns of A is |det(A)| (det(A)�ýé�).

2 If A is a 3× 3 matrix, the volume of the parallelepiped deter-

mined by the columns of A is |det(A)| (det(A)�ýé�).

SCUT, Liu Rui

Application of determinants: area or volume

Determinants as area

SCUT, Liu Rui

Application of determinants: area or volume

Determinants as volume

|det(A)| = absolute value of

∣∣∣∣∣∣∣∣a 0 0

0 b 0

0 0 c

∣∣∣∣∣∣∣∣ = |abc|

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Application of determinants: area or volume

Determinants as area

~a1 and ~a2 are nonzero vectors. For any scalar c, the area of

the parallelogram determined by ~a1 and ~a2 equals the area of

the parallelogram determined by ~a1 and ~a2 + c~a1.

det(~a1, ~a2) = det(~a1, ~a2 + c~a1) = area of the rectangle.

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The Matrix of A Linear Transformation

Example (Rotation in R2)

T (−→x ) = A−→x =

cosϕ − sinϕ

sinϕ cosϕ

x1

x2

= x1 cosϕ− x2 sinϕ

x1 sinϕ + x2 cosϕ

SCUT, Liu Rui

Application of determinants: area or volume

Determinants as volume

In the 3× 3 case (a parallelepiped) is similar.

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Application of determinants: area or volume

Linear transformations and the areas

Definition1 Let T : R2 −→ R2 be the linear transformation determined

by a 2× 2 matrix A. If S is a parallelogram in R2, then

{area of T (S)} = |det(A)| · {area of S}

2 Let T : R3 −→ R3 be the linear transformation determined

by a 3× 3 matrix A. If S is a parallelepiped in R3, then

{volumn of T (S)} = |det(A)| · {volumn of S}

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Application of determinants: area or volume

Determinants as area or volume

Approximating a planar region by a union of squares. The

approximation improves as the grid becomes finer.

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Application of determinants: Determinants as areaor volume

Transformation and areas

Approximating T (R) by a union of parallelograms.

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Exercise

Exercise: Calculate the area of the parallelogram determined

by the points (−2,−2), (0, 3), (4,−1) and (6, 4).

Solution: Subtract the vertex (−2,−2) from each of the four

vertices. The new parallelogram has the same area, and its

vertices are (0, 0), (2, 5), (6, 1) and (8, 6).

SCUT, Liu Rui

Exercise

Exercise: Calculate the area of the parallelogram determined

by the points (−2,−2), (0, 3), (4,−1) and (6, 4).

Solution: Subtract the vertex (−2,−2) from each of the four

vertices. The new parallelogram has the same area, and its

vertices are (0, 0), (2, 5), (6, 1) and (8, 6).

SCUT, Liu Rui

Exercise

Exercise: Calculate the area of the parallelogram determined

by the points (−2,−2), (0, 3), (4,−1) and (6, 4).

Then the area is |det(A)| = −

∣∣∣∣∣∣ 2 6

5 1

∣∣∣∣∣∣ = 28

SCUT, Liu Rui

Homework

Homework:

Section 3.3 p. 198: 11, 16, 24, 27;