View
17
Download
0
Category
Preview:
Citation preview
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
SCUT, Liu Rui
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)
SCUT, Liu Rui
A linearity property of the determinant function
Note that,
det(A + B) 6= det(A) + det(B).
SCUT, Liu Rui
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.
SCUT, Liu Rui
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
∣∣∣∣∣∣∣∣∣∣∣∣
SCUT, Liu Rui
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.
SCUT, Liu Rui
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.
SCUT, Liu Rui
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)
SCUT, Liu Rui
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.
SCUT, Liu Rui
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.
�
SCUT, Liu Rui
Example
Example:
SCUT, Liu Rui
Example
Solution: The equivalent matrix equation is A~x = ~b.
Since det(A) = 2, the equation has a unique solution. By
Cramer’s rule:
[�]
SCUT, Liu Rui
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.
SCUT, Liu Rui
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.
SCUT, Liu Rui
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
SCUT, Liu Rui
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.
SCUT, Liu Rui
Adjoint Matrix
Theorem
An invertible n× n matrix A has its inverse:
A−1 =1
det(A)adj(A)
SCUT, Liu Rui
Example
Example:
Find the inverse of the matrix
A =
2 1 3
1 −1 1
1 4 −2
SCUT, Liu Rui
Exercise
Solution:
adj(A) =
C11 C21 C31
C12 C22 C32
C13 C23 C33
=
−2 14 4
3 −7 1
5 −7 −3
SCUT, Liu Rui
Exercise
Solution:
SCUT, Liu Rui
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
SCUT, Liu Rui
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|
SCUT, Liu Rui
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.
SCUT, Liu Rui
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.
SCUT, Liu Rui
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}
SCUT, Liu Rui
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.
SCUT, Liu Rui
Application of determinants: Determinants as areaor volume
Transformation and areas
Approximating T (R) by a union of parallelograms.
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).
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;
Recommended