Recap

Row and Reduced Row Echelon Elementary Matrices

If m and n are positive integers, then an m n matrix is a rectangular array in which each entry aij of the matrix is a number. The matrix has m rows and n columns.

a1,1 a1,2 a1,3 a1,na2,1 a2, 2 a2,3 a2,na3,1 a3,2 a3,3 a3, n am ,1 am,2 am,3 am ,n

A real matrix is a matrix all of whose entries are real numbers.

i (j) is called the row (column) subscript.

An mn matrix is said to be of size (or dimension) mn.

If m=n the matrix is square of order n.

The ai,i’s are the diagonal entries.

Given a system of equations we can talk about its coefficient matrix and its augmented matrix.

These are really just shorthand ways of expressing the information in the system.

To solve the system we can now use row operations instead of equation operations to put the augmented matrix in row echelon form.

1. Interchange two rows.

2. Multiply a row by a nonzero constant.

3. Add a multiple of a row to another row.

Two matrices are said to be row equivalent if one can be obtained from the other using elementary row operations.

A matrix is in row-echelon form if:› All rows consisting entirely of zeros are

at the bottom.› In each row that is not all zeros the first

entry is a 1.› In two successive nonzero rows, the

leading 1 in the higher row is further left than the leading 1 in the lower row.

1. Write the augmented matrix of the system.

2. Use elementary row operations to find a row equivalent matrix in row-echelon form.

3. Write the system of equations corresponding to the matrix in row-echelon form.

4. Use back-substitution to find the solutions to this system.

In Gauss-Jordan elimination, we continue the reduction of the augmented matrix until we get a row equivalent matrix in reduced row-echelon form. (r-e form where every column with a leading 1 has rest zeros)1 0 0 a

0 1 0 b

0 0 1 c

A system of linear equations in which all of the constant terms is zero is called homogeneous.

All homogeneous systems have the solutions where all variables are set to zero. This is called the trivial solution.

New Stuff

Using Elementary Matrices

An n by n matrix is called an elementary matrix if it can be obtained from In by a single elementary row operation.

These matrices allow us to do row operations with matrix multiplication.

Theorem: Let E be the elementary matrix obtained by performing an elementary row operation on In. If that same row operation is performed on an m by n matrix A, then the resulting matrix is given by the product EA.

Three types of Elementary Matrices

These correspond to the three types of EROs that we can do:› Interchanging rows of I -> Type I EM› Multiplying a row of I by a constant -> Type

II EM› Adding a multiple of one row to another ->

Type III EM

Type I EM

E1 =

How is this created? Eg. 1 Suppose

A =

E1A =

=

What is AE1?

100

001

010

333231

232221

131211

aaa

aaa

aaa

100

001

010

333231

232221

131211

aaa

aaa

aaa

333231

131211

232221

aaa

aaa

aaa

Type II EM

E2 =

E2A = =

AE2 = =

300

010

001

333231

232221

221211

333 aaa

aaa

aaa

300

010

001

333231

232221

221211

aaa

aaa

aaa

333231

232221

221211

aaa

aaa

aaa

300

010

001

333231

232221

131211

3

3

3

aaa

aaa

aaa

Type III EM

E3 =

E3A = =

AE3 = =

100

010

301

333231

232221

221211

aaa

aaa

aaa

100

010

301

333231

232221

331332123111 333

aaa

aaa

aaaaaa

333231

232221

221211

aaa

aaa

aaa

100

010

301

33313231

23212221

13111211

3

3

3

aaaa

aaaa

aaaa

Let A and B be m by n matrices. Matrix B is row equivalent to A if there exists a finite number of elementary matrices E1, E2, ... Ek such that

B = EkEk-1 . . . E2E1A.

Break it down

This means that B is row equivalent to A if B can be obtained from A through a series of finite row operations.

If we then take two augmented matrices (A|b) and (B|c) and they are row equivalent, then Ax = b and Bx=c must be equivalent series

Break it down

If A is row equivalent to B, B is row equivalent to A

If A is row equivalent to B and B is row equivalent to C then A is row equivalent to C

Example

Compute the inverse of A for A =

322

021

341

100

010

001

|

322

021

341

102

011

001

|

360

320

341

13121

21

21

21

23

21

|

600

020

041

131

011

001

|

600

320

341

Example

Now, solve the system:

61

21

61

41

41

41

21

21

21

|

100

010

001

13121

21

21

21

21

21

|

600

020

001

8322

122

1234

321

21

321

xxx

xx

xxx

Example

We can employ the format Ax = b so x=A-

1b

We just calculated A-1 and b is the column vector

So we can easily find the values of x by multiplying the two matrices

8

12

12

Determinants

Keys to calculating Inverses

Require square matrices Each square matrix has a determinant

written as det(A) or |A| Determinants will be used to:

› characterize on-singular matrices› express solutions to non-singular systems › calculate dimension of subspaces

 

If A and B are square then

It is not difficult to appreciate that

If A has a row (or column) of zeros then

If A has two identical rows (or columns) then

BAAB

AAT

0A0

353

202

171

21 CC

If B is obtained from A by ERO,

interchanging two rows (or columns) then

If B is obtained from A by ERO where row (or column) of A were multiplied by a scalar k, then

)( jiji CCRR AB

AkB

If B is obtained from A by ERO where a multiple of a row (or column) of A were added to another row (or column) of A then BA

211222112221

1211 aaaaaa

aaA

That Is the determinant is equal to the product of the elements along the diagonal minus the product of the elements along the off-diagonal.

51

23A

0)6)(1()2)(3( A

Note: The matrix A is said to be invertible or non-singular if det(A)≠ 0. If det(A) = 0, then A is singular.

For a 3 by 3 matrix

Using EROs on rows 2 and 3

333231

232221

131211

aaa

aaa

aaa

11

13313311

11

12313211

11

13212311

11

12212211

131211

0

0

a

aaaa

a

aaaaa

aaaa

a

aaaaaaa

For a 3 by 3 matrix

The matrix will be row equivalent to I iff:

0

11

13313311

11

12313211

11

13212311

11

12212211

11

a

aaaa

a

aaaaa

aaaa

a

aaaa

a

For a 3 by 3 matrix

This implies that the

Det(A) =

223113322113233112332112233211332211 aaaaaaaaaaaaaaaaaa

Use EROs to find:

605

245

123

STEP 1: Apply from property 5 this

gives us

STEP 2: Convert matrix to Echelon form

31 CC

AB

506

542

321

133

122

6

2

RRR

RRR

23120

100

321

Therefore is the same

as:

matrix is now in echelon form so we can multiply elements of main diagonal to get determinant

32 RR 100

23120

321

605

245

123

12)1)(12(1

100

23120

321

Factorize the determinants of

What is ?

2

2

2

1

1

1

zz

yy

xx

931

111

421

We see that y – x is a factor of row 2 and z – x is a factor of row 3 so we factor them out from:

And we get:

133

122

RRR

RRR

22

22

2

0

0

1

xzxz

xyxy

xx

))(( xzxy xz

xy

xx

10

10

1 2

233 RRR ))(( xzxy yz

xy

xx

00

10

1 2

The matrix is now in echelon form so we can multiply elements of main diagonal to get determinant and then multiply by factors to get:

= 2

2

2

1

1

1

zz

yy

xx

))()(( yzxzxy

Now, the matrix corresponds to

Since =

931

111

421

3

1

2

z

y

x

2

2

2

1

1

1

zz

yy

xx))()(( yzxzxy

Then

= 931

111

421

12)4)(1(3)13))(2(3))(2(1(

Cofactor expansion is one method used to find the determinant of matrices of order higher than 2.

If A is a square matrix, then the minor Mi,j of the element ai,j of A is the determinant of the matrix obtained by deleting the ith row and the jth column from A.

Consider the matrix .

The minor of the entry “0” is found by deleting the row and the column associated with the entry “0”.

605

245

123

A

The minor of the entry “0” is

Note: Since the 3 x 3 matrix A has 9 elements there would be 9 minors associated with the matrix.

1)5(1)2(325

13

The cofactor Ci,j = (-1)i+jMi,j.

Since we can think of the cofactor of as nothing more than its signed minor.

oddji

evenjiji

,1

,1)1(

Find the minor and cofactor of the entry “2” for

We first need to delete the row and column corresponding to the entry “2”

605

245

123

A

The Minor of 2 is

The minor corresponds to row 1 and column 2 so applying the formula, we have

So the cofactor of the entry “2” is 40.

40)2(5)6(565

25

40)40(140)1( 21 ijc

Theorem: Let A be a square matrix of order n. Then for any i,j,

Columns: and Rows:

det(A) A ai , jCi, jj1

n

det(A) A ai , jCi, j .i1

n

Given find det(A).

Cofactor is found for the first entry in column 1 “-3”

Cofactor is found for the second entry in column 1 “-5”

605

245

123

A

2402460

2411 c

1201260

1221 c

Cofactor is found for the third entry in column 1 “5”

The cofactors are then multiplied by the corresponding entry and summed.

04424

1231 c

312111 55)(3 cccA

12

6072

)0(5)12(5)24(3

Using row 2 - expansion we fix row 2 and find the minors for each entry in row 2 then apply the sign corresponding to each entries position to find the cofactors. The cofactors are then multiplied by the corresponding entry and summed.

 

12

209260

)10(*2)23(*4)12(*5

05

23*2

65

13*4

60

12*5

232222 24)(5 cccA

It is easy to show that

44332211

44

3433

242322

14131211

000

00

0aaaa

a

aa

aaa

aaaa

If A is square and is in Echelon form then is the product of the entries on the (main) diagonal.

00*1*2*1

0000

0100

0020

2001

Using CRAMER’S RULE we can apply this method to finding the solution to a system of linear systems that have the same number of variables as equations

There are two cases to consider

Consider the square system AX = B where A is n x n.

 If then the system has either I) No solution or ii) Many solutions

If A1 is formed from A by replacing column 1 of A with column B and

I. , then the system has NO solution

II. |A| ≠ 0, then the system has a unique solution

0A

01 A

Consider the square system AX = B where A is n x n.

 If |A|≠ 0 then the system has a unique solution

The unique solution is obtained by using the Cramer’s rule.

Where Ai is found from A by replacing column i of A with B.

etcA

AX

A

AX

A

AX

33

22

11

Use Cramer’s rule to write down the solution to the system

354

735

21

21

xx

xx

354

735

21

21

xx

xx371225 A

371225 A

4334

75

2653

37

2

1

A

A

37

43

37

26

22

11

A

AX

A

AX

37

43,

27

26

The adjoint of A (adj.A)) is- : , the transpose of the matrix of cofactors (a matrix of signed minors).

 If then the inverse of A exists and

If , A has no inverse.

TijCadjA )(

0A

adjAA

A11

0A

Find inverse of

Det(A) =

= (3*2*0) – (3*3*1) – (-2*1*0)+(-2*1*1) +(3*1*3) –(3*1*2)

=0 – 9 – 0 – 2 + 9 – 6= -8

031

121

323

A

223113322113233112332112233211332211 aaaaaaaaaaaaaaaaaa

8110

031

893

808

1139

013

21

23

11

33

12

3231

23

01

33

03

3231

21

01

11

03

12

T

T

TijCadjA )(

1110

031

893

8

1

1

1

1

A

adjAA

A

Given that show:

a.b.c.

d.e. Solve where and

542

111

241

A

0342 IAA

0A

IAA 12133

)(3

11 AAIA

bxA Txxxx ),,( 321 Tb )1,1,1(

17168

414

8161

542

111

241

542

111

2412A

0

000

000

000

100

010

001

3

542

111

241

4

17168

414

8161

342

IAA

IIAA

IAA

IAA

3)4(

34

0342

2

Taking the determinant

Implying that and therefore; exists

IIAA

IIAA

34

3)4(

Since, Multiplying through by A

IAA

AIAA

AIAA

IAA

AAA

AAA

1213

31216

3)34(4

)34(

34

034

3

3

3

2

23

23

0342 IAA

Multiplying through by A-1

)4(3

1

43

034

)0(

034)(

034

1

1

1

1

111

2

AIA

AIA

AIA

AA

AAAAAA

IAA

142

131

243

3

1

542

111

241

400

040

004

3

1

)4(3

1

1

1

A

AIA

Since

1

1

1

3

3

3

3

1

1

1

1

142

131

243

3

1

1

x

bAx

bxA