chapter 3 determinants - scut.edu.cn€¦ · from cramer’s rule, we can give a general formula...
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SCUT, Liu Rui
Chapter 3 Determinants
Liu Rui
School of Mathematics
South China University of Technology
2019-11-15
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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) =?
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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).
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Linear transformation
T (c~x) =
∣∣∣∣∣∣∣∣∣∣a11 a12 · · · cx1 · · · a1n
......
......
......
......
an1 an2 · · · cxn · · · ann
∣∣∣∣∣∣∣∣∣∣
= c
∣∣∣∣∣∣∣∣∣∣a11 · · · x1 · · · a1n
......
......
......
an1 · · · xn · · · ann
∣∣∣∣∣∣∣∣∣∣= cT (~x).
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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
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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
∣∣∣∣∣∣∣∣∣∣∣∣
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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).
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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
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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�).
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Application of determinants: area or volume
Determinants as area
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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
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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.
![Page 39: Chapter 3 Determinants - scut.edu.cn€¦ · 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](https://reader030.vdocuments.mx/reader030/viewer/2022040415/5eac691434f23355db41d237/html5/thumbnails/39.jpg)
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).
![Page 40: Chapter 3 Determinants - scut.edu.cn€¦ · 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](https://reader030.vdocuments.mx/reader030/viewer/2022040415/5eac691434f23355db41d237/html5/thumbnails/40.jpg)
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).
![Page 41: Chapter 3 Determinants - scut.edu.cn€¦ · 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](https://reader030.vdocuments.mx/reader030/viewer/2022040415/5eac691434f23355db41d237/html5/thumbnails/41.jpg)
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
![Page 42: Chapter 3 Determinants - scut.edu.cn€¦ · 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](https://reader030.vdocuments.mx/reader030/viewer/2022040415/5eac691434f23355db41d237/html5/thumbnails/42.jpg)
SCUT, Liu Rui
Homework
Homework:
Section 3.3 p. 198: 11, 16, 24, 27;