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Linear Algebra. Chapter 3. Determinants
1. Chapter 3 Determinants
Linear AlgebraChapter 3
Determinants
2. 3.1 Introduction to Determinants
DefinitionThe 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
3.
DefinitionLet 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
4. Example 2
Determine the minors and cofactors of the elements a11 and a32of 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
5.
DefinitionThe 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
6. 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
7. Theorem 3.1
The determinant of a square matrix is the sum of the products ofthe 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 anjCnj
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
8. 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
9. 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
10. Computing Determinants of 2 2 and 3 3 Matrices
Computing Determinants of 2 2and 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 a23a31 a13a21a32
(diagonal products from left to right)
a13a22 a31 a11a23a32 a12 a21a33
(diagonal products from right to left)
Ch03_10
11. Homework
Exercises will be given by the teachers ofthe practical classes.
Ch03_11
12. 3.2 Properties of Determinants
Theorem 3.2Let 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
13. Example 1
4 23
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
14. Example 2
4 31
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
15. Theorem 3.3
DefinitionA 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 ak 1Ck 1 ak 2Ck 2 akn Ckn 0Ck 1 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
16. 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
17. 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
A A
1
cA cA c n A
A A
1
I 1
A
1
1
A
Ch03_17
18. Example 4
If A is a 2 2 matrix with |A| = 4, use Theorem 3.4 to computethe 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
A A A (A A )A A A A A
t
t
t
1
1
A A
AA A
A
t
Ch03_18
19. Example 6
Prove that if A and B are square matrices of the same size, with Abeing 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
20. Homework
Exercises will be given by the teachers ofthe practical classes.
Exercise 11
Prove the following identity without evaluating the determinants.
a b c d e f
a b c d
a c e
q
r
u
v
w
( a b)
q
r
v
w
p
q
r
u
v
w
f
p q r p q r
p
e f
b d
u v w
(c d )
u v w
p
r
u
w
(e f )
p q
u
v
Ch03_20
21. 3.3 Numerical Evaluation of a Determinant
DefinitionA 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.
1 4 0
3 8 2
0 1 5 , 0 2 3
0 0 0
0 0 9
0 0 0
upper triangular
7
5
9
1
8 0 0
7 0 0
2 1 0 , 1 4 0
7 0 2
3 9 8
4 5 8
lower triangular
0
0
0
1
Ch03_21
22.
Numerical Evaluation of a DeterminantTheorem 3.5
The determinant of a triangular matrix is the product of its
diagonal elements.
Proof
a11
a12 a1n
0
a22 a2 n
0
ann
0
a11
a22
a23 a2 n
0
a33 a3n
0
ann
0
a33
a34 a3n
0
a44 a4 n
0
ann
a11a22
0
a11a22 ann
Example 1
2 1 9
Let A 0 3 4 , find A .
0 0 5
Sol.
A 2 3 ( 5) 30.
Ch03_22
23. Numerical Evaluation of a Determinant
Example 24 1
2
Evaluation the determinant. 2 5 4
9 10
4
Solution (elementary row operations
2
4
1
2
4
2
5
4
R2 R1
0
1 5
9 10 R 3 ( 2) R1 0
1 8
4
2
4
R3 R 2 0 1
0
1
1 2 ( 1) 13 26
5
0 13
Ch03_23
24. Example 3
1 02 1
Evaluation the determinant.
1 0
1 0
2
1
0
2
1
0
3
1
2
1
Solution
1
0 2 1
1
0
2 1 1 0 R2 ( 2)R1 0 1 3 2
1
0 0 3 R3 ( 1)R1 0
1
0 2 1
R4 R1
0
1
0
0 2
2
0
2
4
2
1
0 1 3 2
R4 2R3 0
0
0 2
2
0
6
0
1 ( 1) ( 2) 6 12
Ch03_24
25. Example 4
1 24
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 ( 2)R1 0
2 3
R2 R3
1 2
( 1) 0
0
2
4
3
0 1
( 1) 1 2 ( 1) 2
Ch03_25
26. Example 5
1 11
1
Evaluation the determinant.
2 2
6 6
0
2
3
5
2
3
4
1
Solution
1
1 0 2
1 1 0
2
1
1 2 3
R2 R1
0
0 2
2 2 3 4 R3 ( 2)R1 0
0 3
5 0
0
6 6 5 1 R4 ( 6)R1 0
0 5 11
diagonal element is zero and
all elements below this
diagonal element are zero.
Ch03_26
27. 3.4 Determinants, Matrix Inverse, and Systems of Linear Equations
DefinitionLet 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
cofactors 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
matrix of cofactors
C11 C12 C1n
C
C
C
22
2n
21
C
C
C
n2
nn
n1
adjoint matrix
t
Ch03_27
28. Example 1
Give the matrix of cofactors and the adjoint matrix of thefollowing matrix A.
0
3
2
A 1
4 2
5
1 3
Solution The cofactors of A are as follows.
4 1
C11 4 2 14 C12 1 2 3 C13 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
adj( A) 3
7
1
6
6
8
1
12 1 8
Ch03_28
29. Theorem 3.6
Let A be a square matrix with |A| 0. A is invertible with1
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
30.
Proof of Theorem 3.6If i = j, ai1C j1 ai 2C j 2 ainC jn A .
If i j, let A
Rj is replaced by Ri
B.
Matrices A and B have the same cofactors
Cj1, Cj2, …, Cjn.
So ai1C j1 ai 2C j 2 ainC jn B 0.
row i = row j in B
A
Therefore (i. j ) th element
0
1
Since |A| 0, A adj( A) I n
A
1
Similarly, adj( A) A I n .
A
if i j
if i j
A adj(A) = |A|In
1
Thus A adj( A)
A
1
Ch03_30
31. 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
32. Example 2
Use a determinant to find out which of the following matrices areinvertible.
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. A is invertible.
|B| = 0.
B is singular. The inverse does not exist.
|C| = 0.
C is singular. The inverse does not exist.
|D| = 2 0. D is invertible.
Ch03_32
33. Example 3
Use the formula for the inverse of a matrix to compute the inverse0
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
1
1
A 1 adj( A) 3
A
25
1
9
7
6
14
12 25
3
1
25
8 1
25
9
25
7
25
6
25
12
25
1
25
8
25
Ch03_33
34. Homework
Exercises will be given by the teachers ofthe practical classes.
Exercise
Show that if A = A-1, then |A| = 1.
Show that if At = A-1, then |A| = 1.
Ch03_34
35. 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.
Ch03_35
36. Example 4
Determine whether or not the following system of equations hasan unique solution.
3 x1 3 x2 2 x3 2
4 x1 x2 3 x3 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
37. Theorem 3.9 Cramer’s Rule
Let AX = B be a system of n linear equations in n variables suchthat |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
Ch03_37
38.
Proof of Cramer’s Rulexi, 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
39. 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 3 x3 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
40.
Giving A1 3, A2 6, A3 9Cramer’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.
Ch03_40
41. Example 6
Determine values of for which the following system ofequations 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
0
2 1
( 2)( 1) 2( 4) 0 2 6 0 ( 2)( 3) 0
= – 3 or = 2.
Ch03_41
42.
= – 3 results in the systemx1 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 3 x2 0
This system has many solutions, x1 = – 3r/2, x2 = r.
Ch03_42
43. Homework
Exercises will be given by the teachers ofthe practical classes.
Ch03_43