# Chapter 1 matrices, Formulas and forms for Accounting. Universiti Kuala Lumpur

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Matrices

Engineering Mathematics 3 Chapter 1

MATRICES

Page 1 of 15 BTB28304

1.0 MATRICES

A set of m x n numbers or functions, arranged in a rectangular array of m rows and n

column is called a matrix of order m x n, read as m by n.

To indicate the position of an element in a matrix, denote each element by a letter

followed by two suffixes.

A matrix of order m x n with double suffix notation is given by:

   

   

mnmm

n

n

aaa

aaa

aaa

...

....

...

...

21

22221

11211

The suffixes i and j in the element i ja indicate that the element i ja belongs to the i th row

and j th column. Thus 23a belongs to second row and third column.

1.1 TYPES OF MATRICES

a) Row matrix – a matrix that has only one row.

Example:

 741A Order of A: 1 x 3

b) Column matrix – a matrix that has only one column.

Example:

  

  

2

5

7

A Order of A: 3 x 1

Engineering Mathematics 3 Chapter 1

MATRICES

Page 2 of 15 BTB28304

c) Square matrix – a matrix which the numbers of row (m) and column (n) are equal.

The matrix is called a square matrix of order n x n or n.

Example:

   

   

6310

4869

1443

2135

A Order of A: 4 x 4 or 4

d) Diagonal matrix – it is a square matrix of whose elements except those in the

leading diagonal are zero i.e. i ja = 0 when ji  .

Example:

  

  

300

010

002

A

A diagonal matrix whose diagonal elements are all equal is called a scalar matrix.

Example:

  

  

200

020

002

A

A diagonal matrix whose diagonal elements are all equal to unity is called a unit

/identity matrix denoted by I.

Example:

 

  

 

10

01 I and

  

  

100

010

001

I

e) Null /zero matrix – a matrix whose elements are all zero.

Example:

  

  

000

000

000

A

Engineering Mathematics 3 Chapter 1

MATRICES

Page 3 of 15 BTB28304

f) Triangular matrix – a square matrix whose elements either above or below the

Example:

Upper Triangular Matrix

  

  

33

2322

131211

00

0

a

aa

aaa

A

Lower Triangular Matrix

  

  

333131

2221

11

0

00

aaa

aa

a

A

g) MatrixTranspose– a matrix obtained from any given matrix A by changing rows

into columns or columns into rows and denoted by TA .

Example:

If

  

  

887

654

321

A , then

  

  

963

852

741 TA

Meanwhile, if

32 703

412

 

  

 B then

23 74

01

32

   

  



TB

*transpose m x n matrix will change the order of matrix become n x m.

h) Symmetric matrix – if a square matrix A and its transpose TA are identical such that

AAT  .

Example:

  

  



705

034

541

A TA

Engineering Mathematics 3 Chapter 1

MATRICES

Page 4 of 15 BTB28304

i) Skew symmetric matrix – if a square matrix A such that AAT 

Example:

 

  

 

02

20 A AAT 

  

 

  

  

02

20

02

20

j) Singular matrix – a matrix whose determinant A is equal to zero.

If 0A , A is a non-singular matrix.

k) Adjoint matrix – a matrix whose elements are the transpose of cofactor matrix.

l)Equal matrix – Two matrices A and B are said to be equal when they are of the same

order and the element in the corresponding positions are equal.

1.2 MATRIX OPERATIONS

Two matrices A and B can be added /subtracted only if the order of A and B are

the same order (both have the same shape and size).

Example 1:

Given the following matrices:

 

  

 

123

321 A

  



 

212

011 B

  

 

24

31 C

Solution:

 

  

 

131

312 BA

  

 

315

330 BA

The expression CACBCA  ,, and CB  are undefined since they have

different orders.

Engineering Mathematics 3 Chapter 1

MATRICES

Page 5 of 15 BTB28304

Note: In general, matrix addition is commutative that is

ABBA 

CBACBA  )()(

1.2.2 Scalar Multiplication

For any matrix A, we can multiply A by a scalar, to form a new matrix of the same

order as A. This multiplication is performed by multiplying every element of A by

the scalar.

Example 2:

Let

  

  

42

03

21

A . Find A2 and A 3

1  .

Solution:

  

  

  

  

84

06

42

42

03

21

22A

  

  



  

  



3 4

3 2

01 3

2 3

1

42

03

21

3

1

3

1 A

1.2.3 Matrix Multiplication

Two matrices A and B can be multiplied if and only if matrix A has the same

number of columns as B has in its rows (i.e. A is the order of p x n and B is the

order of n x q). The resulting matrix AB will be of order p x q.

Example 3:

If  31C and  

  

 

520

312 D , can the product CD be formed?

Commutative laws: - If changing the order of the

operands does not change the result.

- Swap the numbers over you still get the same result.

Engineering Mathematics 3 Chapter 1

MATRICES

Page 6 of 15 BTB28304

Solution:

C has size 21 . D has size 32 . Since the number of columns in C is the same as

the number of rows in D, we can form the product CD. The resulting matrix will have

size 31 as there is one row in C and three columns in D.

Suppose we wish to find CD when  31C and  

  

 

520

312 D .

Then,  )53()31())2(3()11()03()21( CD

 1852 CD

Example 4:

Given  

  

 

062

421 A ,

  

  



7

1

1

B ,

  

  

51

40

32

C ,    

  

 

31

21 ,21 ED . determine

AB and AC . Solve for BD and CE

Solution:

  

  

 

  

 

7

1

1

062

421 AB

 

  



 

)7(0)1(6)1(2

)7(4)1(2)1(1

 

  

 

4

27

  

  

 

  

 

51

40

32

062

421 AC

 

  



 

)5(0)4(6)3(2)1(0)0(6)2(2

)5(4)4(2)3(1)1(4)0(2)2(1

 

  

 

304

276

*Note:

 Matrix multiplication is not commutative, BAAB

 Matrix multiplication is associative, )()( BCACAB

Associative laws: - The order in which

the operations are performed does not matter as long as the sequence of the operands is not changed.

Engineering Mathematics 3 Chapter 1

MATRICES

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1.3 MINOR AND COFACTORS

1.2.2 Minor

Definition: Let A be an n x n matrix. The minor, ijM , of the element i ja is the

determinant of the matrix obtained by deleting the i-th row vector and j-th column

vector of A.

Example:

If

  

  

333231

232221

131211

aaa

aaa

aaa

A , then 3231

1211 23

aa

aa M

1.2.3 Cofactor

Definition: Let A be an n x n matrix. The cofactor, i jA , of the element i ja is defined by

  ij ji

ij MA

 1 , where ijM is the minor of i ja .

The appropriate sign in the cofactor i jA , is easy to remember, since it alternates in

the following manner:

:::::

...

...

...







1.4 DETERMINANT

Definition: the determinant of an n x n matrix A, denoted )det(A or A , is a scalar

associated with the matrix A that is defined as follows:

1 if ...

1 if )det(

1112121111

11

  



 

nAaAaAa

na A

nn

where    j j

j MA 1 1

1 det1 

 , nj ,...,3,2,1

are the cofactor associated with the entries in the first row of A.

Engineering Mathematics 3 Chapter 1

MATRICES

Page 8 of 15 BTB28304

Example 5:

If

  

  

652

420

132

A , find  Adet .

Solution:

Expand the determinant along the first row:

*Note: we can expand the determinant by any row or column in the same way and

obtain the same result in each case.

1.4.1 Singular Matrix

Theorem: An n x n matrix is nonsingular if and only if 0)det( A .

If   0det A , then A is a singular matrix.

Example 6:

Determine whether the following matrices are nonsingular or singular.

  

  

753

242

311

C and

  

  

374

251

132

D

Solution:

C is a nonsingular matrix because 052)det( C .

D is a singular matrix because 0)det( D

1.4.2 Use the Determinants to find Vector Products.

If a = a1i + a2j + a3k and b = b1i + b2j + b3k, then

a x b =

321

321

bbb

aaa

kji

 

44

42416

))]2(2()50[())]2(4()60[(3)]54()62[(2

52

20 )1(

62

40 3

65

42 2det







 

 A

Engineering Mathematics 3 Chapter 1

MATRICES

Page 9 of 15 BTB28304

Example 7:

If a = 3i + j - 2k and b = 4i + 5k, find a x b.

Solution:

a x b =

504

213 

kji

= 5i - 23j - 4k

1.5 PROPERTIES OF DETERMINANTS

Theorem: If A is an n x n upper or lower triangular matrix, then

  nnaaaaAA  ...det 332211

Example 8:

Let

  

  

700

010

152

A . Find  Adet .

Solution:

Since A is an upper triangular matrix, then       14712det A .

1.6 ELEMENTARY ROW OPERATIONS (ERO)

There are three operations for elementary row operations:

a) Interchange rows

b) Multiply a row by a nonzero constant

c) Add a multiple of one row to another row

Notations:

a) :RR ji  interchange row i and row j

b) ii kRR  : multiply row i by the scalar k

c) jii kRRR  : add k times row j to row i

Engineering Mathematics 3 Chapter 1

MATRICES

Page 10 of 15 BTB28304

1.6.1 Elementary Row Operations and Determinants

Let A be an n x n matrix.

P1: If B is a matrix obtained by interchanging two distinct rows of A, then

  )det(det BA 

P2: If B is the matrix obtained by dividing any row of A by a nonzero scalar k, then

  )det(det BkA

P3: If B is the matrix obtained by adding a multiple of any row of A to a different row

of A then   )det(det BA

Example 9:

Let

  

  

243

421

312

A . Find  Adet by using the properties of determinants.

Solution:

243

312

421

:RR

243

421

312

21 



by [P1]

243

530

421

:2RRR 122 

 by [P3]

10100

530

421

:3RRR 133

 by [P3]

530

10100

421

:RR 32

 by [P1]

Engineering Mathematics 3 Chapter 1

MATRICES

Page 11 of 15 BTB28304

530

110

421

10 :R 10

1 R 22

 by [P2]

200

110

421

10 :R3RR 233

 by [P3]

So, 2021110)det( A

P4: If A is an n x n matrix, then   )det(det AAT  .

P5: Let A and B be n x n matrices. Let naaa ,...,, 21 denote the row vectors of A. If the

i-th row vector of A is the sum of two row vectors, for example iii cba  , then

)det()det()det( CBA 

where

        

        

n

i

i

i

a

a

b

a

a

B

:

:

1

1

1

and

        

        

n

i

i

i

a

a

c

a

a

C

:

:

1

1

1

.

The corresponding property is also true for columns.

P6: If A has a row(or column) of zeros, then   0det A .

P7: If two rows ( or columns) of A are the same, then   0det A .

P8: If A and B are matrices of the same size, then      BAAB detdetdet 

P9: Let A be an n x n matrix and k be a scalar, then    AkkA n detdet 

P10: Let A be an invertible matrix, then   )det(

1 det 1

A A  .

MATRICES CHAPTER 1

Page 12 of 15 BTB28304

Example 10:

Evaluate

1221

3335

2442







x

x

x

.

Solution:

Apply P5 to the first column:

122

333

244

121

335

242





x

x

x

122

333

244

121

335

121

2





x

Since 31 RR  and 21 CC 

Thus, 0)0()0(2

1221

3335

2442









x

x

x

x

by P7.

1.7 MATRIX INVERSE

If 0)det( A , A is a nonsingular matrix. Then, 1A exists.

1.7.1 Steps to find the inverse of any matrix:

 If 0)det( A , 1A does not exist. We stop all calculations.

 If 0)det( A , we proceed to step 2

2. Find matrix of cofactor.

4. Find matrix inverse,   

A A

det

MATRICES CHAPTER 1

Page 13 of 15 BTB28304

Example 11:

Find 1A if

  

  



134

421

203

A .

Solution:

32

134

421

203

Since   032det A , A is a nonsingular matrix so 1A exists.

2. Find the matrix of cofactor:

       

  6144 956

51714

21

03

41

23

42

20

34

03

14

23

13

20

34

21

14

41

13

42







 

 

 

 

 

 

 

So, the matrix of cofactor is

  

  





6144

956

51714

  

  





695

14517

4614

4. Find matrix inverse,   

A A

det

 

32

11A

  

  





695

14517

4614

MATRICES CHAPTER 1

Page 14 of 15 BTB28304

1.8 ROW TRANSFORMATION METHOD

Row transformation method is another method used to determine 1A .

Example 12:

Use row transformation method to find 1A if

  

  



321

842

456

A .

Solution:

  

  

100:321

010:842

001:456

Use ERO:

  

  



001:456

010:842

100:321

:31 RR

  

  

001:456

210:1480

100:321

:2 12 RR

  

  



601:1470

210:1480

100:321

:6 13 RR

  

  

 601:1470

41810:4710

100:321

: 8

1 2R

  

  



417871:4700

41810:4710

100:321

:7 23 RR

  

  



7172174:100

41810:4710

100:321

: 7

4 3R

reduce the LHS to an identity matrix

MATRICES CHAPTER 1

Page 15 of 15 BTB28304

  

  





7172174:100

411:010

100:321

: 4

7 32 RR

  

  



7172174:100

411:010

74423712:021

:3 31 RR

  

  





7172174:100

411:010

7122172:001

:2 21 RR

2

Therefore,

  

  







7172174

411

7122172 1A .

*Note: You should check this result by evaluating 1AA .

the required 1A on the RHS