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.