Transcript Chapter 3

Linear Algebra
Chapter 3
Determinants
3.1 Introduction to Determinants
Definition
The determinant of a 2  2 matrix A is denoted |A| and is given
by
a11
a12
a21
a22
 a11a22  a12 a21
Observe that the determinant of a 2  2 matrix is given by the
different of the products of the two diagonals of the matrix.
The notation det(A) is also used for the determinant of A.
Example 1
A   2 4
 3 1
det( A)  2 4  (2 1)  (4  (3))  2  12  14
3 1
Ch03_2
Definition
Let A be a square matrix.
The minor of the element aij is denoted Mij and is the determinant
of the matrix that remains after deleting row i and column j of A.
The cofactor of aij is denoted Cij and is given by
Cij = (–1)i+j Mij
Note that Cij = Mij or Mij .
Ch03_3
Example 2
Determine the minors and cofactors of the elements a11 and a32
of the following matrix A.
0 3
1
A   4  1 2
0  2 1
Solution
1
0 3
Minor of a11 : M 11  4  1 2   1 2  (11)  (2  (2))  3
0 2 1 2 1
Cofactor of a11 : C11  (1)11 M 11  (1) 2 (3)  3
1
0 3
Minor of a32 : M 32  4  1 2  1 3  (1 2)  (3  4)  10
0 2 1 4 2
Cofactor of a32 : C32  (1)32 M 32  (1)5 (10)  10
Ch03_4
Definition
The determinant of a square matrix is the sum of the products
of the elements of the first row and their cofactors.
If A is 3  3, A  a11C11  a12C12  a13C13
If A is 4  4, A  a11C11  a12C12  a13C13  a14C14

If A is n  n, A  a11C11  a12C12  a13C13    a1nC1n
These equations are called cofactor expansions of |A|.
Ch03_5
Example 3
Evaluate the determinant of the following matrix A.
 1 2  1
A  3 0
1
1
4 2
Solution
A  a11C11  a12C12  a13C13
 1(1) 2 0 1  2(1)3 3 1  (1)(1) 4 3 0
2 1
4 1
4 2
 [(0 1)  (1 2)]  2[(3 1)  (1 4)]  [(3  2)  (0  4)]
 2  2  6
 6
Ch03_6
Theorem 3.1
The determinant of a square matrix is the sum of the products of
the elements of any row or column and their cofactors.
A  ai1Ci1  ai 2Ci 2    ain Cin
ith row expansion:
jth column expansion: A  a1 j C1 j  a2 j C2 j    anj Cnj
Example 4
Find the determinant of the following matrix using the second row.
 1 2  1
A  3 0
1
1
4 2
Solution
A  a21C21  a22C22  a23C23
 3 2  1  0 1  1  1 1 2
2
1
4
1 4 2
 3[(2 1)  (1 2)]  0[(11)  (1 4)]  1[(1 2)  (2  4)]
 12  0  6  6
Ch03_7
Example 5
Evaluate the determinant of the following 4  4 matrix.
1
2
0  1
7  2
1
0
0
4
0
2
3
5
0  3
Solution
A  a13C13  a23C23  a33C33  a43C43
 0(C13 )  0(C23 )  3(C33 )  0(C43 )
2
1
4
 3 0 1
2
0
1 3
2  6(3  2)  6
 3(2)  1
1 3
Ch03_8
Example 6
Solve the following equation for the variable x.
x x 1  7
1 x  2
Solution
Expand the determinant to get the equation
x( x  2)  ( x  1)(1)  7
Proceed to simplify this equation and solve for x.
x2  2x  x 1  7
x2  x  6  0
( x  2)( x  3)  0
x  2 or 3
There are two solutions to this equation, x = – 2 or 3.
Ch03_9
Computing Determinants of 2  2
and 3  3 Matrices
a11

A
a21
a12 
a22   A  a11a22  a12 a21
 a11
A  a21
a
 31
a12
a22
a32
a13 
 a11
a23   a21
a
a33 
 31
a12
a22
a32
a13  a11
a23  a21
a33  a31
a12
a22
a32
 A  a11a22 a33  a12 a23 a31  a13 a21a32
(diagonalproductsfromleft toright)
 a13 a22 a31  a11a23 a32  a12 a21a33
(diagonalproductsfrom right toleft)
Ch03_10
Homework
Exercise 3.1 pages161-162:
3, 6, 9, 11, 13, 14
Ch03_11
3.2 Properties of Determinants
Theorem 3.2
Let A be an n  n matrix and c be a nonzero scalar.
(a) If A  B then |B| = c|A|.
cRk
(b) If A  B then |B| = –|A|.
Ri Rj
(c) If A  B then |B| = |A|.
Ri cRj
Proof (a)
|A| = ak1Ck1 + ak2Ck2 + … + aknCkn
|B| = cak1Ck1 + cak2Ck2 + … + caknCkn
|B| = c|A|.
Ch03_12
Example 1
4 2
3
Evaluate the determinant  1  6
3.
9 3
2
Solution
3
4 2
1  6
2
3
9 3
3 0 2

C2  2 C3
1 0
3  (3)
2 3 3
3
2
1
3
 21
Ch03_13
Example 2
4 3
 1
2 5, |A| = 12 is known.
If A   0
 2  4 10
Evaluate the determinants of the following matrices.
4 3
 1 12 3
 1
 1 4 3
(a ) B1   0
6 5 (b) B2   2  4 10 (c) B3  0 2 5
 2  12 10
 0
0 4 16
2 5
Solution
(a) A  B1
Thus |B1| = 3|A| = 36.
3C 2
(b) A
(c)
 B2 Thus |B2| = – |A| = –12.
R 2 R 3
A  B3 Thus |B3| = |A| = 12.
R 3 2 R1
Ch03_14
Definition
A square matrix A is said to be singular if |A|=0.
A is nonsingular if |A|0.
Theorem 3.3
Let A be a square matrix. A is singular if
(a) all the elements of a row (column) are zero.
(b) two rows (columns) are equal.
(c) two rows (columns) are proportional. (i.e., Ri=cRj)
Proof
(a) Let all elements of the kth row of A be zero.
A  ak1Ck1  ak 2Ck 2    aknCkn  0Ck1  0Ck 2    0Ckn  0
(c) If Ri=cRj, then
 |A|=|B|=0
A  B , row i of B is [0 0 … 0].
Ri cRj
Ch03_15
Example 3
Show that the following matrices are singular.
 2 0  7
2  1 3
(a ) A   3 0
1 (b) B   1 2 4
 4 0
2 4 8
9
Solution
(a) All the elements in column 2 of A are zero. Thus |A| = 0.
(b) Row 2 and row 3 are proportional. Thus |B| = 0.
Ch03_16
Theorem 3.4
Let A and B be n  n matrices and c be a nonzero scalar.
(a) |cA| = cn|A|.
(b) |AB| = |A||B|.
(c) |At| = |A|.
1
1
(d) A  A (assuming A–1 exists)
Proof
(a)
(d)
A

cR1, cR 2 , ...,cRn
cA  cA  c n A
A  A1  A  A1  I  1
 A1 
1
A
Ch03_17
Example 4
If A is a 2  2 matrix with |A| = 4, use Theorem 3.4 to compute
the following determinants.
(a) |3A|
(b) |A2|
(c) |5AtA–1|, assuming A–1 exists
Solution
(a) |3A| = (32)|A| = 9  4 = 36.
(b) |A2| = |AA| =|A| |A|= 4  4 = 16.
1
(c) |5AtA–1| = (52)|AtA–1| = 25|At||A–1|  25 A  25.
A
Example 5
Prove that |A–1AtA| = |A|
Solution
1
1
1
1
A A A  (A A )A  A A A  A
t
t
t
1
A A
AA A
A
t
Ch03_18
Example 6
Prove that if A and B are square matrices of the same size, with A
being singular, then AB is also singular. Is the converse true?
Solution
()
()
|A| = 0  |AB| = |A||B| = 0
Thus the matrix AB is singular.
|AB| = 0  |A||B| = 0  |A| = 0 or |B| = 0
Thus AB being singular implies that either A or B is singular.
The inverse is not true.
Ch03_19
Homework
Exercise 3.2 pp. 170-171: 4, 5, 9, 10, 11, 13, 19
Exercise 11
Prove the following identity without evaluating the determinants.
ab cd e f
p
u
ab cd
p
u
q
v
q
v
e f
r
w
 ( a  b)
a c e
b d
f
r  p q rp q r
w
u v w u v w
q
r
v
w
 (c  d )
p
r
u
w
 (e  f )
p q
u
v
Ch03_20
3.3 Numerical Evaluation of a
Determinant
Definition
A square matrix is called an upper triangular matrix if all the
elements below the main diagonal are zero.
It is called a lower triangular matrix if all the elements above the
main diagonal are zero.
3
0

0
1
0

0

0
4 0 7
8 2

2
3
5

1 5,
0 0 9
0 9

0 0 1
upper  t riangular
8
1

7

4
0 0 0
7 0 0 

4
0
0
 2 1 0 ,



0 2 0
 3 9 8

5 8 1
lower  t riangular
Ch03_21
Numerical Evaluation of a Determinant
Theorem 3.5
The determinant of a triangular matrix is the product of its
diagonal elements.
Proof
a11
a12  a1n
0

a22  a2 n
0
 a11
  

a33  a3n
0
 a11a22
  

a44  a4 n
   a11a22  ann
  
0



0
ann
a22
0
a23  a2 n
0
ann
a33
0
a34  a3n
0
ann
Example 1
 2  1 9
Let A  0 3  4, find A .
0 0  5
Sol.
A  2  3 (5)  30.
Ch03_22
Numerical Evaluation of a Determinant
Example 2
4 1
 2
Evaluation the determinant.  2  5 4
9 10
 4
Solution (elementary row operations）
2
2
4
4
1

2
 5 4 R2  R1 0
9 10 R 3  ( 2) R1 0

4
1
1 5
1 8
1  2  (1) 13  26
R3  R 2 0  1 5
0 0 13
2
4
Ch03_23
Example 3
1 0
2 1
Evaluation the determinant.
1 0
1 0
2
1
0
2
1
0
3
1
Solution
1 0
2 1
1 0
1 0
2
1
0
2
1

1 0
2
1
0 R2  (2)R1 0  1  3  2
3 R3  (1)R1 0 0  2
2
1 R4  R1 0 0
4
2
1 0
2
1

0 1  3  2
R4  2R3 0 0  2
2
0 0
0
6
 1 (1)  (2)  6  12
Ch03_24
Example 4
1 2
4
2 5
Evaluation the determinant.  1
2  2 11
Solution
1 2
4

1 2 4
1
2  5 R2  R1 0
0 1
2  2 11 R 3  (1)R1 0
2 3

1 2
( 1) 0
R2  R3
0
4
2
3
0 1
 (1) 1 2  (1)  2
Ch03_25
Example 5
1 1
1
1
Evaluation the determinant.
2 2
6 6
0
2
3
5
2
3
4
1
Solution
1 1
1
1
2 2
6 6
0
2
3
5
2

3
R2  R1
4 R3  (2)R1
1 R4  (6)R1
1 1 0
2
0 0 2
5 0
0 0 3
0
0 0 5  11
diagonal element is zero and
all elements below this
diagonal element are zero.
Ch03_26
3.4 Determinants, Matrix Inverse,
and Systems of Linear Equations
Definition
Let A be an n  n matrix and Cij be the cofactor of aij.
The matrix whose (i, j)th element is Cij is called the matrix of
cofactor of A.
The transpose of this matrix is called the adjoint of A and is
denoted adj(A).
 C11 C12  C1n 
C

C

C
22
2n 
 21
 




C
C

C
n2
nn 
 n1
matrixof cofactor
 C11 C12  C1n 
C

C

C
22
2n 
 21
 




C
C

C
n2
nn 
 n1
adjoint matrix
t
Ch03_27
Example 1
Give the matrix of cofactors and the adjoint matrix of the
following matrix A.
0
3
 2
A   1
4  2
5
 1  3
Solution The cofactors of A are as follows.
C11  4  2  14 C12    1  2  3 C13   1 4  1
3
5
1
5
1 3
0 6
C21   0 3  9 C22  2 3  7
C23   2
3 5
1 5
1 3
2
3
0
3
2 0 8
C31 
 12 C32  
C

1
33
4 2
1 4
1  2
The matrix of cofactors
The adjoint of A is
of A is  14 3  1
 14  9  12
  9 7 6
adj( A)   3
7
1


6
8
 1
 12 1 8


Ch03_28
Theorem 3.6
Let A be a square matrix with |A|  0. A is invertible with
1
1
A  adj( A)
A
Proof
Consider the matrix product Aadj(A). The (i, j)th element of this
product is
(i, j ) th element  (row i of A)  (column j of adj( A))
 C j1 
C j 2 
 ai1 ai 2  ain  
  
C jn 
 ai1C j1  ai 2C j 2    ain C jn
Ch03_29
Proof of Theorem 3.6
If i = j, ai1C j1  ai 2C j 2   ainC jn  A.
If i  j, let A

Rj is replacedby Ri
B. Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.
So ai1C j1  ai 2C j 2   ainC jn  B  0.
 A if i  j
Therefore (i. j ) th element  
0 if i  j
1

Since |A|  0, A adj( A)   I n
A



1

Similarly,  adj( A)  A  I n .
A

row i = row j in B
 A adj(A) = |A|In
1
Thus A  adj( A)
A
1
Ch03_30
Theorem 3.7
A square matrix A is invertible if and only if |A|  0.
Proof
() Assume that A is invertible.
 AA–1 = In.
 |AA–1| = |In|.
 |A||A–1| = 1
 |A|  0.
() Theorem 3.6 tells us that if |A|  0, then A is invertible.
A–1 exists if and only if |A|  0.
Ch03_31
Example 2
Use a determinant to find out which of the following matrices are
invertible.
 2 4  3
 1 2  1
1

1
4
2




A
B
C   4 12  7 D   1 1 2
3 2
2 1
1
 1 0
 2 8 0
Solution
|A| = 5  0.
|B| = 0.
|C| = 0.
|D| = 2  0.
A is invertible.
B is singular. The inverse does not exist.
C is singular. The inverse does not exist.
D is invertible.
Ch03_32
Example 3
Use the formula for the inverse of a matrix to compute the inverse
0
3
 2
of the matrix
A   1
4  2
5
 1  3
Solution
|A| = 25, so the inverse of A exists.We found adj(A) in Example 1
 14  9  12
adj( A)   3
7
1
6
8
 1
 14
9
12 



14  9 12  25
25
25

  3
1
1 
7
1
1

A  adj( A)   3
7
1  
A
25 
25
25 
  25
6
8  1
6
8
 1



25
25
25


Ch03_33
Exercise 3.3 page 178-179: 4, 7.
Exercise
Show that if A = A-1, then |A| = 1.
Show that if At = A-1, then |A| = 1.
Ch03_34
Theorem 3.8
Let AX = B be a system of n linear equations in n variables.
(1) If |A|  0, there is a unique solution.
(2) If |A| = 0, there may be many or no solutions.
Proof
(1) If |A|  0
 A–1 exists (Thm 3.7)
 there is then a unique solution given by X = A–1B (Thm 2.9).
(2) If |A| = 0
 since A  C implies that if |A|0 then |C|0 (Thm 3.2).
 the reduced echelon form of A is not In.
 The solution to the system AX = B is not unique.
 many or no solutions.
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Example 4
Determine whether or not the following system of equations has
an unique solution.
3 x1  3 x2  2 x3  2
4 x1  x2  3x3  5
7 x1  4 x2  x3  9
Solution
Since
3 3 2
4 1
3 0
7 4
1
Thus the system does not have an unique solution.
Ch03_36
Theorem 3.9 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables such
that |A|  0. The system has a unique solution given by
A1
A2
An
x1 
, x2 
, ... , xn 
A
A
A
Where Ai is the matrix obtained by replacing column i of A with B.
Proof
|A|  0  the solution to AX = B is unique and is given by
X  A1 B
1
 adj( A) B
A
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Proof of Cramer’s Rule
xi, the ith element of X, is given by
1
xi  [row i of adj( A)]  B
A
 b1 
1
 
 C1i C2i Cni b2 

A
b 
 n
1
 (b1C1i  b2C2i    bnCni )
A
Ai
Thus xi 
A
the cofactor expansion of |Ai|
in terms of the ith column
Ch03_38
Example 5
Solving the following system of equations using Cramer’s rule.
x1  3 x2  x3  2
2 x1  5 x2  x3  5
x1  2 x2  3x3  6
Solution
The matrix of coefficients A and column matrix of constants B are
 1 3 1
  2
A  2 5 1 and B    5
 1 2 3
 6
It is found that |A| = –3  0. Thus Cramer’s rule be applied. We
get
 2 3 1
 1  2 1
 1 3  2
A1    5 5 1 A2  2  5 1 A3  2 5  5
6 3
6
 6 2 3
 1
 1 2
Ch03_39
Giving A1  3, A2  6, A3  9
Cramer’s rule now gives
A1  3
A2
A3  9
6
x1 

 1, x2 

 2, x3 

3
A 3
A 3
A 3
The unique solution is x1  1, x2  2, x3  3.
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Example 6
Determine values of  for which the following system of
equations has nontrivial solutions.Find the solutions for each
value of .
(  2) x  (  4) x  0
1
2
2 x1  (  1) x2  0
Solution
homogeneous system
 x1 = 0, x2 = 0 is the trivial solution.
 nontrivial solutions exist  many solutions
 2 4
  1  0
 (  2)(  1)  2(  4)  0  2    6  0  (  2)(  3)  0
  = – 3 or  = 2.
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 = – 3 results in the system
 x1  x2  0
2 x1  2 x2  0
This system has many solutions, x1 = r, x2 = r.
 = 2 results in the system
4 x1  6 x2  0
2 x1  3x2  0
This system has many solutions, x1 = – 3r/2, x2 = r.
Ch03_42
Homework
Exercise 3.3 pages 179-180:
8, 12, 14, 15, 17.
Ch03_43