chapter 3 determinants linear algebra. ch03_2 3.1 introduction to determinants definition the...
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Chapter 3
Determinants
Linear Algebra
Ch03_2
3.1 Introduction to Determinants
DefinitionThe determinant of a 2 2 matrix A is denoted |A| and is given by
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.
211222112221
1211 aaaaaa
aa
Example 1
13
42A
14122))3(4()12(1342)det( A
Ch03_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_4
10)43()21(2431
120214301
: ofMinor 3232 Ma
3))2(2()11(1221
120214301
: ofMinor 1111
Ma
Example 2
Solution
Determine the minors and cofactors of the elements a11 and a32 of the following matrix A.
120214301
A
3)3()1()1( : ofCofactor 211
111111 MCa
10)10()1()1( : ofCofactor 532
233232 MCa
Ch03_5
DefinitionThe determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors.
These equations are called cofactor expansions of |A|.
nnCaCaCaCaAnnA
CaCaCaCaAA
CaCaCaAA
11131312121111
1414131312121111
131312121111
, is If
4,4 is If
3,3 is If
Ch03_6
Example 3Evaluate the determinant of the following matrix A.
124103121
A
Solution
2403)1)(1(14
13)1(21210)1(1 432
131312121111
CaCaCaA
6622
)]40()23[()]41()13[(2)]21()10[(
Ch03_7
Theorem 3.1The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors.
ith row expansion:
jth column expansion: njnjjjjj
ininiiii
CaCaCaA
CaCaCaA
2211
2211
Example 4Find the determinant of the following matrix using the second row.
124103121
ASolution
66012)]42()21[(1)]41()11[(0)]21()12[(3
2421114
11012123
232322222121
CaCaCaA
Ch03_8
Example 5Evaluate the determinant of the following 4 4 matrix.
3010532720104012
Solution
6)23(63121)2(3
310210412
3
)(0)(3)(0)(0 43332313
4343333323231313
CCCCCaCaCaCaA
Ch03_9
Example 6Solve the following equation for the variable x.
7211
xxx
Solution
Expand the determinant to get the equation
7)1)(1()2( xxx
Proceed to simplify this equation and solve for x.
3or 20)3)(2(06
7122
2
xxxxx
xxx
There are two solutions to this equation, x = – 2 or 3.
Ch03_10
Computing Determinants of 2 2 and 3 3 Matrices
2221
1211
aaaa
A21122211A aaaa
333231
232221
131211
aaaaaaaaa
A
3231
2221
1211
333231
232221
131211
aaaaaa
aaaaaaaaa
left) right to from products (diagonal
right) left to from products (diagonal
A
332112322311312213
322113312312332211
aaaaaaaaa
aaaaaaaaa
Ch03_11
Homework
Exercise 3.1 pages161-162:3, 6, 9, 11, 13, 14
Ch03_12
Let A be an n n matrix and c be a nonzero scalar.
(a) If then |B| = c|A|.
(b) If then |B| = –|A|.
(c) If then |B| = |A|.
3.2 Properties of DeterminantsTheorem 3.2
Proof (a) |A| = ak1Ck1 + ak2Ck2 + … + aknCkn
|B| = cak1Ck1 + cak2Ck2 + … + caknCkn |B| = c|A|.
BAcRk
BARjRi
BAcRjRi
Ch03_13
Example 1
.
392
361
243
Solution
21 31
23 )3(
332
301
203
392
361
243
C32C2
Evaluate the determinant
Ch03_14
Example 2
If |A| = 12 is known.
Evaluate the determinants of the following matrices.
,1042
520341
A
1640
520
341
)c(
520
1042
341
(b)
10122
560
3121
)a( 321 BBB
Solution
(a) Thus |B1| = 3|A| = 36.
(b) Thus |B2| = – |A| = –12.
(c) Thus |B3| = |A| = 12.
123BA
C
232BA
RR
3123BA
RR
Ch03_15
Theorem 3.3Let 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.
0000 212211 knkkknknkkkk CCCCaCaCaA
(c) If Ri=cRj, then , row i of B is [0 0 … 0].
|A|=|B|=0
BAcRjRi
DefinitionA square matrix A is said to be singular if |A|=0.
A is nonsingular if |A|0.
Ch03_16
Example 3Show that the following matrices are singular.
842
421
312
(b)
904
103
702
)(a BA
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_17
Theorem 3.4Let A and B be n n matrices and c be a nonzero scalar.
(a) |cA| = cn|A|.
(b) |AB| = |A||B|.
(c) |At| = |A|.
(d) (assuming A–1 exists) AA
11
Proof
(a)AccAcAA n
cRncRcR
..., ,2 ,1
(d)
AAIAAAA
1 1 111
Ch03_18
Example 4If 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.
(c) |5AtA–1| = (52)|AtA–1| = 25|At||A–1| .251
25 A
A
Example 5
Prove that |A–1AtA| = |A|
Solution
AAAA
AAAAAAAAAAAA tttt 1)( 1111
Ch03_19
Example 6Prove 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_20
Homework
Exercise 3.2 pp. 170-171: 4, 5, 9, 10, 11, 13, 19
Exercise 11
Prove the following identity without evaluating the determinants.
wvu
rqp
fdb
wvu
rqp
eca
wvu
rqp
fedcba
Solution
vu
qpfe
wu
rpdc
wv
rqba
wvu
rqp
fedcba
)()()(
Ch03_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.
triangularupper
1000
9000
5320
7041
,
900
510
283
triangularlower
1854
0207
0041
0008
,
893
012
007
Ch03_22
. find ,
500
430
912
Let AA
Theorem 3.5
The determinant of a triangular matrix is the product of its diagonal elements.
Example 1
Proof
nn
nn
n
n
nn
n
n
nn
n
n
aaa
a
aa
aaa
aa
a
aa
aaa
a
a
aa
aaa
2211444
33433
2211333
22322
11222
11211
00
0
00
0
00
0
.30)5(32 ol. AS
Numerical Evaluation of a Determinant
Ch03_23
Numerical Evaluation of a Determinant
Example 2
Evaluation the determinant.
1094452142
810
510
142
R1)2(3R
R1R2
1094
452
142
1300
510
142
2R3
R2613)1(2
Solution (elementary row operations)
Ch03_24
Example 3
Evaluation the determinant.
1201300101121201
Solution
2400
2200
2310
1201
R1R4
R1)1(R3
R1)2(R2
1201
3001
0112
1201
6000
2200
2310
1201
2R3R4
126)2()1(1
Ch03_25
Example 4
Evaluation the determinant.1122521421
Solution
320
100
421
1R)1(3R
R1R2
1122
521
421
100
320
421
)1( R3R2
2)1(21)1(
Ch03_26
Example 5
Evaluation the determinant.
1566432232112011
Solution
11500
0300
5200
2011
R1)6(R4
R1)2(R3
R1R2
1566
4322
3211
2011
diagonal element is zero and all elements below this diagonal element are zero.
0
Ch03_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 cofactor of A.
The transpose of this matrix is called the adjoint of A and is denoted adj(A).
cofactor ofmatrix 21
22221
11211
nnnn
n
n
CCC
CCC
CCC
matrixadjoint 21
22221
11211t
CCC
CCC
CCC
nnnn
n
n
Ch03_28
Example 1Give the matrix of cofactors and the adjoint matrix of the following matrix A.
531241302
A
Solution The cofactors of A are as follows.
145324
11 C 351
2112 C 131
4113
C
95330
21 C 75132
22 C 63102
23 C
122430
31 C 121
3232
C 841
0233 C
The matrix of cofactors of A is
861173
12914)(adj A
8112
679
1314The adjoint of A is
Ch03_29
Theorem 3.6Let A be a square matrix with |A| 0. A is invertible with
)(adj11 AA
A
Proof
Consider the matrix product Aadj(A). The (i, j)th element of this product is
jninjiji
jn
j
j
inii
CaCaCa
C
CC
aaa
AjAiji
2211
2
1
21
))adj( of (column ) of (row element th ),(
Ch03_30
Therefore
ji
jiAji
if 0
if elementth ).( A adj(A) = |A|In
Proof of Theorem 3.6
.2211 ACaCaCa jninjiji If i = j,
If i j, let .by replaced is
BARiRj
.0 So 2211 BCaCaCa jninjiji
Since |A| 0,
nIAA
A
)(adj
1
Similarly, . nIAAA
)(adj
1Thus )(adj
11 AA
A
row i = row j in B
Matrices A and B have the same cofactorsCj1, Cj2, …, Cjn.
Ch03_31
Theorem 3.7A 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_32
Example 2Use a determinant to find out which of the following matrices are invertible.
082211121
1017124342
1224 23
11 DCBA
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_33
Example 3Use the formula for the inverse of a matrix to compute the inverse of the matrix
531241302
A
Solution
|A| = 25, so the inverse of A exists.We found adj(A) in Example 1
861173
12914)(adj A
258
256
251
251
257
253
2512
259
2514
861
173
12914
251)adj(11 A
AA
Ch03_34
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_35
Theorem 3.8Let 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_36
Example 4Determine whether or not the following system of equations has an unique solution.
9 47
53 4
2 233
321
321
321
xxx
xxx
xxx
Solution
Since
0147314233
Thus the system does not have an unique solution.
Ch03_37
Theorem 3.9 Cramer’s RuleLet AX = B be a system of n linear equations in n variables such that |A| 0. The system has a unique solution given by
Where Ai is the matrix obtained by replacing column i of A with B.
A
Ax
A
Ax
A
Ax n
n , ... , , 22
11
Proof
|A| 0 the solution to AX = B is unique and is given by
BAA
BAX
)(adj1
1
Ch03_38
xi, the ith element of X, is given by
)(1
1
)](adj of row[1
2211
2
1
21
ninii
n
niii
i
CbCbCbA
b
bb
CCCA
BAiA
x
ThusA
Ax i
i
Proof of Cramer’s Rule
the cofactor expansion of |Ai| in terms of the ith column
Ch03_39
Example 5Solving the following system of equations using Cramer’s rule.
6 32
5 5 2
2 3
321
321
321
xxx
xxx
xxx
Solution
The matrix of coefficients A and column matrix of constants B are
652
and 321152131
BA
It is found that |A| = –3 0. Thus Cramer’s rule be applied. We get
621552231
361152121
326155132
321 AAA
Ch03_40
Giving
Cramer’s rule now gives
9 ,6 ,3 321 AAA
339
,23
6 ,1
33 3
32
21
1
A
Ax
A
Ax
A
Ax
The unique solution is .3 ,2 ,1 321 xxx
Ch03_41
Example 6Determine values of for which the following system of equations has nontrivial solutions.Find the solutions for each value of .
0)1(2
0)4()2(
21
21
xx
xx
Solution
homogeneous system
x1 = 0, x2 = 0 is the trivial solution.
nontrivial solutions exist many solutions
= – 3 or = 2.
0142
0)3)(2( 062 0)4(2)1)(2(
Ch03_42
= – 3 results in the system
This system has many solutions, x1 = r, x2 = r.022
0
21
21
xx
xx
= 2 results in the system
This system has many solutions, x1 = – 3r/2, x2 = r.032
064
21
21
xx
xx
Ch03_43
Homework
Exercise 3.3 pages 179-180:8, 12, 14, 15, 17.