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Matrices

A matrix is a set of real or complex numbers (or elements) arranged in

rows and columns to form a rectangular array.

A matrix having m rows and n columns is an m x n (read m by n or

m cross n ) matrix and is referred to as having order m x n.

A matrix can be represented explicitly by enclosing the array within

large square brackets.

A matrix is any doubly subscripted array of elements arranged in rowsand columns.

Capital letters A , B ,C , X , Y, Z etc are used for matrix notation .

2 3 7

A3x3 = 8 5 0 ; A is 3 x 3 matrix

3 4 1

3 x3

2 3 0

B4x3 = 5 7 1 ; B is 4 x 3 matrix

3 2 9

4 0 3 4 x 3

In matrix A, 1st row 1st column element is denoted as a11 = 2

1st row 2nd column element is denoted as a12 = 3

1st row 3rd column element is denoted as a13 = 7

2nd row 1st column element is denoted as a21 = 8

2nd

row 2nd

column element is denoted as a22 = 5Developed by Ms. SAROJ MISHRA

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and so on ...........

Types of Matrices

(1) Vectors : A Vector is a special type of matrix in which there is only

one row or one column .(a) Row Vector : If there is only one row and more than one column in any matrix ,called

'Row Vector' .

1 x n matrix [ a1 a2 a3 ...................... an ] is a row vector .

(b) Column Vector : If there is only one column and more than one row in any matrix ,

called ' Column Vector ' .

n x 1 matrix a1

a2

a3

.

.

.

.

.

.

an

(2) Zero or Null Matrix A matrix , with every element zero , is called a null amtrix . It is

denoted by O . It need not be sqaure . In matrix theory it plays the role of zero .

0 0 0

O = 0 0 00 0 0

0 0 0 4 x 3

(3) Square Matrix : A matrix in which number of rows is equal to number of columns.

4 78 2 is a 2 x 2 square matrix .

2x2

3 7 8

9 0 4 is a 3 x 3 square matrix

1 2 5

3x3

Developed by Ms. SAROJ MISHRA

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(4) Diagonal Matrix :A square matrix all of whose elements are zero except those

in the leading diagonal is called a diagonal matrix .

i.e. All elements except diagonal lements are zero called diagonal matrix .

1 0 0

D = 0 5 0 is a 3 x 3 diagonal matrix .

0 0 8 3x3

4 0 0 0

0 9 0 0

D = 0 0 5 0 is a 4 x 4 diagonal matrix .

0 0 0 13

4x4

(5) Upper / Lower Triangular Matrix : A square matrix all of whose elements below the

main diagonal are zero is called upper triangular .

7 3 2

0 4 9 is a 3 x 3 upper diagonal matrix .

0 0 3

3x3

8 3 2 1

0 12 3 4

0 0 9 8 is a 4 x 4 upper diagonal matrix .

0 0 0 1 4x4

Lower Diagonal Matrix : If all elments above the main diagonal are zero it is lower

triangular matrix.

13 0 0

7 5 0 is a 3 x 3 lower diagonal matrix .

6 8 12

3x3

8 0 0 0

13 5 0 0

1 17 9 0 is a 4 x 4 lower diagonal matrix .

7 9 5 1 4x4

(6) Scalar Matrix : If in the diagonal matrix D , diagonal elements are same , it behaves

like a scalar matrix .


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2 0 0

0 2 0 is a 3 x 3 scalar matrix .

0 0 2

3x3

11 0 0 0

0 11 0 0 is a 4 x 4 scalar matrix .

0 0 11 0

0 0 0 11

4x4

(7) The Identity Matrix : An identity matrix has 1 for every diagonal element and

zero elsewhere .

1 0 0

0 1 0 is a 3 x 3 identity matrix .

0 0 1

3x3

1 0 0 0

0 1 0 0

0 0 1 0 is a 4 x 4 identity matrix .

0 0 0 1

4x4

(8) Transpose Matrix : Interchanging the rows into columns or columns into rows in any given

matrix , the new transformed matrix is called as transpose matrix.

A = 3 2 5

6 9 0 2x3

3 6

A'(A Transpose , AT

) = 2 9 is transpose of matrix A

5 03x2

10 7 6

4 3 0

B = 3 8 7

6 13 19

4x3


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10 4 3 6

B' (B Transpose ,BT

)

= 7 3 8 13 is transpose of matrix B .6 0 7 19

3x4

(9) Symmetric Matrix :A square matrix M is symmetric ifM = MT

1 2 5 1 2 5

M = 2 8 9 ,MT

= 2 8 9

5 9 4 5 9 4

MATRIX ALGEBRA

MATRIX ADDITION : For matrix addition , order of matrix should be same .

i.e. To add two matrices A and B:

# of rows in A = # of rows in B

# of columns in A = # of columns in B

2 3 5

A = 5 0 6 can be added with 3x3 matrix only .

7 8 1 3x3

1 7 0

B = 3 4 8

12 5 4

3x3

2 3 5 1 7 0 2 + 1 3 + 7 5 + 0

A + B = 5 0 6 + 3 4 8 = 5 + ( -3) 0 + 4 6 + 87 8 1 12 5 4 7 + 12 8 + 5 1 + 4


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3 10 5

= 2 4 14

5 13 5

Subtraction :

To subtract two matrices A and B:

# of rows in A = # of rows in B

# of columns in A = # of columns in B

Multiplication of Matrices

Scalar Multiplication : 47 9 53 6 3 =28 36 2012 24 12 i.e. k[aij] = [kaij]

If any element is multiplied with any matrix, each element inside the matrix will be

multiplied .

This also means that we can take a common factor out of every element.

Multiplication of two matrices : Two matrices can be multiplied together only when the

number ofcolumns in the matrix on the left equals the number of rows in the matrix on

the right.

A B = C

mn pq = mq for n=p

Regular Multiplication : To multiply two matrices A and B ,

# of columns in A = # of rows in B

Multiply: A (m x n ) by B (n by p)

Matrices A and B can be multiplied if : [r x c] and [s x d]


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1

4

2

3

5

8

6

7+ =

6

12

8

10

A B+ = C

1

4

2

3

5

8

6

7=

4

4

4

4

B A = C

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c = s

The resulting matrix will have the dimensions : [r x c] and [s x d]

r x d

2 5 7 7 2 5 1

Q. Multiply A = , B = 1 0 8 2

3 4 9 3 3 4 4

2x3 3x4

Here , matrix multiplication AB is possible as number of columns in first

matrix A is equal to number of rows in second matrix B , i.e. Matrix A and B satisfies

condition required for matrix multiplication.

2 x 7+ 5 x 1+7 x ( 3) 2x2 + 5 x 0 + 7x3 2 x 5 + 5 x 8 + 7 x 4 2 x ( 1) + 5 x 2 + 7 x( 4)

AB =

3 x 7 + 4 x 1+ 9 x ( 3) 3x2 + 4 x 0 + 9x3 3 x 5 + 4 x 8 + 9 x 4 3 x ( 1) + 4 x 2 + 9 x ( 4)

2 x 4

2 25 78 20

=

2 33 83 31

2 x 4 Ans .

2 1

1 0 2 3 0 2

Q. 2 . Multiply A = 2 1 0 3 and B = 1 4

3 2 2 0 2 0

3x4 4x2

Soln : Here , Product AB is possible because number of columns in A = number of rows in B

i.e. A and B satisfies the condition of multiplication and hence AB is possible .


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1 0 2 3 0 2

Q. 2 . AB = 2 1 0 3 1 4

3 2 2 0 2 0

3x4 4x2

1.2 + 0.0 + 2.1 + 3.2 1.1 + 0.2 + 2.4 + 3.0

= 2.2 1.0 + 0.1 + 3.2 2.1 1.2 + 0.4 + 3.0

3.2 + 2.0 2.1 + 0.2 3.1 + 2.2 2.4 + 0.0

3x2

( Here, . is used for ( x ) multiplication )

10 9

= 10 0

4 1

3x2 Ans .

Find the product of following matrices :

2 4 1 2 10 2

(1)

6 8 2 0 Ans . 22 6

(2) 2 1 2

3 2 Ans . 6 9 14

6 3 4

1 2 3 5 2 1 1 1 9

0 2 4 3 1 7 Ans . 6 2 6

1 3 1 0 1 2 4 0 18

3 4

2 2 1 3 0 1 11 0

(3) 1 0 1 0 1 0 Ans . 4 4

0 1 4 0 2 2 4 1

Cofactor Matrix of a Given Matrix :

1 3 3


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Q. 1 . Find cofactor matrix for the given matrix : 1 4 3

1 3 4

1 3 3Soln : Let A = 1 4 3

1 3 4

In order to find cofactor matrix , we need to find cofactors of each element aij in matrix A .

Let us take notation Aij for the cofactors of each aij respectively .

4 3

Cofactor of a11(=1) is , A11 = + = 7

3 4

1 3

Cofactor of a12(=3) is , A12 = = 11 4

1 4

Cofactor of a13(=3) is , A13 = + = 11 3

3 3

Cofactor of a21(=1) is , A21 = = 33 4

1 3

Cofactor of a22(=4) is , A22 = + = ( 4 3 ) = 11 4

1 3

Cofactor of a23(=3) is , A23 = = 01 3

3 3

Cofactor of a31(=1) is , A31 = + = 3


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4 3

1 3

Cofactor of a32(=3) is , A32 = = 01 3

1 3

Cofactor of a33(=4) is , A33 = + = 31 4

Then . cofactor matrix for the given matrix A is ,

A11 A12 A13

B = A21 A22 A23

A31 A32 A33

7 1 1

= 3 1 0

3 0 1

1 4 2

Q. Given matrix X = 1 2 1 , find cofactor matrix.

1 3 2

In order to find cofactor matrix , we need to find cofactors of each element aij in matrix A .

Let us take notation Xij for the cofactors of each xij respectively .

2 1

Cofactor of x11(=1) is , X11 = + = 13 2

1 1

Cofactor of x12(=4) is , X12 = = 01 2


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1 2

Cofactor of x13(=2) is , X13 = + = 5

1 3

4 2

Cofactor of x21(=1) is , X21 = = 23 2

1 2

Cofactor of x22(=2) is , X22 = + = 0

1 2

1 4

Cofactor of x23(=1) is , X23 = = 11 3

4 2

Cofactor of x31(=1) is , X31 = + = 02 1

1 2

Cofactor of x32(=3) is , X32 = = 31 1

1 4

Cofactor of x33(=2) is , A33 = + = 61 2

Then . cofactor matrix for the given matrix X is ,

X11 X12 X13

Y = X21 X22 X23

X31 X32 X33


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1 3 5

= 2 0 1

0 3 6

1 3 3

Q. Find adjoint matrix for the given matrix : 1 4 3

1 3 4

1 3 3

Soln : Let A = 1 4 3

1 3 4

In order to find adjoint matrix for the given matrix , first we need to find cofactor matrix of the

matrix A .

A11 A12 A13

Step 1 : Cofactor matrix B for the given matrix A = A21 A22 A23

A31 A32 A33

7 1 1 See solution of Q .1

on page 8,9 ,and 10

3 1 0

3 0 1

Step 2 : Transpose matrix of cofactor matrix B will give adjoint matrix A .

T

7 1 1

= 3 1 0

3 0 1


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7 3 3

adj (A) = 1 1 0

1 0 1

Which is the required adjoint matrix for the matrix A denoted by adj A .

1 4 2

Q. Given matrix X = 1 2 1 , find adjoint matrix.

1 3 2


matrix A .

X11 X12 X13

Step 1 : Cofactor matrix B for the given matrix A = X21 X22 X23

X31 X32 X33

1 For cofactor matrix,see page no 10 , 11 , 12

1 3 5

= 2 0 1

0 3 6

Step 2 : Transpose matrix of cofactor matrix Y will give adjoint matrix X .

T

1 3 5

= 2 0 1

0 3 6

1 2 0

adj (X) = 3 0 3

5 1 6

Which is the required adjoint matrix for the matrix X denoted by adj X .


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Condition for evaluating inverse of the matrix : The conditions required to calculate

inverse of the matrix is as follows :1. Matrix shuold be a square matrix

2. Determinant of the matrix should be non zero. If determinat of any matrix becomes zero ,

inverse of that matrix can not be calculated .

1 3 3

Q. Find inverse matrix for the given matrix : 1 4 3

1 3 4

1 3 3

Soln : Let A = 1 4 3

1 3 4

In order to find inverse matrix for the given matrix , first we need to find determinant of the

matrix A .

Step 1 : We find determinant of the given matria as :

1 3 3

A = 1 4 3 = 1(4x4 3x3) 3(1 x4 1 x 3)

1 3 4 + 3 ( 1x3 1x4)

= 1(7) 3(4 3) + 3(3 4)

= 1(7) 3 (1) + 3( 1)

= 1 0

Here , Determinant A , inverse A is possible .

Step 2 : To find cofactor matrix B for the given matrix A

Cofactor matrix B = 7 1 1 See solution of Q .1

on page 8,9 ,and 10

3 1 0

3 0 1

Step 3 : To find transpose of matrix B , which is adjoint A

adj(A) = Transpose cofactor matrix B is given as ,

T

7 1 1

adj (A) = 3 1 0

3 0 1


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7 3 3

adj (A) = 1 1 0

1 0 1

adj(A)

Step 3 : Inverse A = =

A

A11 A12 A13

Cofactor matrix B for the given matrix A = A21 A22 A23

A31 A32 A33

7 1 1 See solution of Q .1

on page 8,9 ,and 10= 3 1 0

3 0 1

Step 2 : Transpose matrix of cofactor matrix B will give adjoint matrix A .

Which is the required adjoint matrix for the matrix A denoted by adj A .

1 4 2

Q. Given matrix X = 1 2 1 , find adjoint matrix.

1 3 2


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matrix A .

X11 X12 X13

Step 1 : Cofactor matrix B for the given matrix A = X21 X22 X23

X31 X32 X33

2 For cofactor matrix,see page no 10 , 11 , 12

1 3 5

= 2 0 1

0 3 6

Step 2 : Transpose matrix of cofactor matrix Y will give adjoint matrix X .

T

1 3 5

= 2 0 1

0 3 6

1 2 0

adj (X) = 3 0 3

5 1 6

Which is the required adjoint matrix for the matrix X denoted by adj X .

Application of Matrices in Industry :

Q. 1 . Use matrix method to solve the system of equations :

x y + z = 1 , 2x + y z = 2 , x 2y z = 4

Solution : The given system of equations can be written in matrix form as

1 1 1 x 1

2 1 1 y = 2


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1 2 1 z 4

This can be represented as AX = B

where

1 1 1 x 1

A = 2 1 1 , X = y , C = 2

1 2 1 z 4

Solving x , y and z we require the formula , X = A1

C

Here , method to calculate A1

is already discussed on the above pages .

3 3 0

A1

= 1/9 1 2 3

5 1 3

Now , X = A1

C = 1/9

3 3 0 1 9 1

A1

= 1/9 1 2 3 2 = 1/9 9 = 1

5 1 3 4 9 1

x 1

y = 1

z 1

x = 1 , y = 1 , z = 1 Ans ..


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Q. The price of three commodities X , Y and Z are x , y and z per unit respectively . A

purchases 4 units of Z and sells 3 units of X and 5 units of Y . B purchases 3 units of Y

and sells 2 units of X and 1 units of Z . C purchases 1 unit of X and slls 4 units of Y and

6 units of Z . In process A , B , C , earn Rs. 6 ,000 Rs. , RS . 5,000 and RS . 13, 000respectively . Using matrices , find the prices per unit of the three

( Note that selling the units is positive earnings and buying the units is negative

earnings . )

Solution : We can formulate the above data in the form of simultaneous equations as :

3x + 5y 4z = 6,000 , 2x 3y + z = 5,000 , x + 4y + 6z = 13,000

The above system of equations can be written in the matrix form as :

3 5 4 x 6,000

2 3 1 y = 5,000

1 4 6 z 13,000

We can represent above marix

AX = B

22 46 7 6,000

X = A1

B = 1/ 151 13 14 11 5,000

5 17 19 13,000

x 4,53,000 3,000

y = (1/151) 1,51.000 = 1,000

z 3,02,000 2,000

Hence the prices of three commodities X , Y and Z are RS. 3,000 , Rs. 1,000 and RS.

2,000 per unit . Ans ...

Q. A company is to employ 60 laborers from either of the party X and Y comprising of

persons in different skills are as under :

Category


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Party Skill I Skill II Skill III

X 25 20 15Y 20 30 10

Rate of labor applicable to categories I , II and III are Rs. 500 , Rs . 700 and Rs.

600 respectively . Using matrices , find which party is economically preferable .

Soln : Hint :Let A be the matrix representing the nmber of labourers of three categories I , II

and III of the party X and Y ; and B be the matrix representing the rate of labour applicable to

categories I , II and III .

Then ,

Category

Rate

I II III I 500

A = 25 20 15 X B = II 700

20 30 10 Y III 600

The total labour charges payable to each party are given by the elements of product of the

matrices A and B , i.e. ,

I II III I 500

12500 + 20000 + 9000

X 25 20 15 II 1000 =

AB =

Y 20 30 10 III 600 10000 + 30000 + 6000

X 41500

=

Y 46000

This shows that the company has to pay Rs. 41,500 to party X and RS . 46,000 to party Y as

labour charges . Since Rs. 46000 < 41500 , therefore party X is economicaly preferable .


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Determinant

A scalar quantity obtained by expanding the elements of a sqaure matrix , say A with respect to

the elements of any row or column is called is called a determinat of matrix A and is denoted by

A or det. A .

Determinat of second order : consider the square matrix of order 2 as :

2 3


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A =

5 7

The product of the elements in the principal diagonal is 2x7 and the product of the lements in

the of diagonal is : 3 x 5 . The difference ( 2 x 7) ( 3 x 5) is called the determinat of A , and is

denoted by A or det. A .

2 4

Illustration : = 2( 8) ( 4) 6 = 16 + 24 = 8

6 8

Determinant of order 3 : Consider the sqaure matrix of order 3 as :

2 3 4

0 7 1 7 1 0 A = 1 0 7 = 2 3 + 4

5 6 4 6 4 5

4 5 6

= 2(0 x 6 7 x 5) 3 ( 1 x6 7 x 4 ) + 4 ( 1 x5 0 x 4)

= 2(0 35 ) 3( 6 28) + 4 ( 5 0 )

= 2 ( 35 ) 3 ( 22 ) + 20

= 70 + 66 + 20

= 4 + 20 = 16 Ans .. ..

What is the relation between matrix and determiants Every determiant is a scalar quantity

( real number) obtained by expanding the elemnts ofsquare matrix .

Properties of matrix :

1. The value of the determinat remains unchanged, if the rows and the columns of

the determiant are interchanged , That is , A = AT

1 2 3 1 4 7 example :

4 5 6 = 2 5 8


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7 8 9 3 6 9

2. If any two adjacent rows( or columns) of the determiant are interchanged , its valueremain unchanged , but its sign is changed , That is , A = A

Changing first two columns ,

1 2 3 2 1 3

4 5 6 = 5 4 6

7 8 9 8 7 9

3 . If two rows( or columns) of a determiant are identical , then the value of the determiant is zero .

That is

1 2 3

A = 1 2 3 = 0 , as first two rows are identical .

4 5 6

1 2 2

B = 4 3 3 = 0 , as last two columns are identica .

5 6 6

4. If any scalar k is multiplied with any determinant , then every element in any one

row ( or column) of the determient gets multiplied by the scalar i.e. ,

1 2 3 k1 2 3

k 4 5 6 = k4 5 6

7 8 9 k7 8 9

5. If to each element of any row ( or column) of a determinant are added ( or

subtracted ) equi multiples of the corresponding elemnts of one or more rows ( or

columns) , the value of the determiant remains same .


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a1 a2 a3 a1 ma3 a2 a3

b1 b2 b3 = b1 mb3 b2 b3

c1 c2 c3 c1 mc3 c2 c3

Cramer's rule for solving unknowns ( variables ) in two / three linear equations :

Let us consider the following linear equations :

ax + by = c

ux + vy = d

Solve for x , y .

We can write the given equatons as follows :

a b x c

u v y d

Let D be the determiant of the coefficients of the variables x and y such that

a b

D = ,

u v

c b

Further , let Dx = be the determiant obtained from D

d v by repalcing the first column by the

elements c , d .

and by repalicng the second column by the elemnts

a c

Dy = be the determinant obtained from D by replacing the second column by

u d the elements c , d .

Thus the values of x and y can be expressed in the form of determiant as

x y 1


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= = , provided D 0

Dx Dy D

1

x = x Dx , provided D 0

D

1

y = X Dy

D

Solve the following linear equations using cramers rule .

Q . Solve the following system of euations using Cramer's Method ,

x + y + z = 6 ; x y + z = 2 ; 2x + y z = 1

Solution : The determinant of the equations of x , y and z is given by ,

1 1 1

D = 1 1 1 = 1(1 1) 1(1 2) + 1(1 + 2 )

2 1 1

= 6 ( 0 )

The solution is given by

x y z 1

= = = , provided D 0

Dx Dy Dz D

6 1 1 1 1 2 1 2 1

Dx = 2 1 1 = 6 1 + 1

1 1 1 1 1 1 1 1 1

= 6 ( 1 1) 1( 2 1) + 1 ( 2 + 1)

= 6 x 0 1 ( 3) + 1 ( 3) = 0 + 3 + 3 = 6 Ans ..


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1 6 1

2 1 1 1 1 2

Similarly , Dy = 1 2 1 = 1 6 +1

1 1 2 1 2 1

2 1 1

= 1( 2 1) 6( 1 2) + 1(1 4)

= 1( 3) 6( 3) + 1( 3)

= 3 + 18 3

= 6 + 18 = 12 Ans .

1 1 6 1 2 1 2 1 1

Also , Dz = 1 1 2 = 1 1 + 6

2 1 1 1 1 2 1 2 1

= 18

Thus using cramer's rule we have ,

x / 6 = y / 12 = z / 18 = 1 / 6

x = 1 , y = 2 , z = 3 Ans . ....sss.


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