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MODULE - I

Algebra

Determinants and Their Applications

4DETERMINANTSANDTHEIR

APPLICATIONS

Every square matrix is associated with a unique number called the determinant of the matrix.There are numerous applications of determinants, the most famous of which is a method ofsolving system of n linear equations in nvariables called Cramers rule, named afterGabrielCramer (1704-1752).

In this lesson, we will learn various properties of determinants and also evaluate determinantsby different methods. We will also learn to solve system of linear equations by Cramers rule

using determinant.

After studying this lesson, you will be able to :

define determinant of a square matrix;

define the minor and the cofactor of an element of a matrix;

find the minor and the cofactor of an element of a matrix;

find the value of a given determinant of order not exceeding 3;

state the properties of determinants; evaluate a given determinant of order not exceeding 3 by using expansion method

and using SARRUS diagram;

state the conditions for a system of linear equations in two/three variables to have aunique solution; and

state and apply Cramers rule to solve a system of linear equations.

OBJECTIVES

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Knowledge of solution of equations Knowledge of number system including complex number Four fundamental operations on numbers and expressions

4.1 DETERMINANT OF ORDER 2

Let us consider the following system of linear equations:

a x b y c

a x b y c

1 1 1

2 2 2

+ =

+ =

On solving this system of equations for x and y, we get

2 1 1 2 1 2 2 1

1 2 2 1

1 2 2 1 1 2 2 1

and provided 0b c b c a c a c

x y a b a ba b a b a b a b

= =

The number a b a b1 2 2 1 determines whether the values ofx yand exist or not.

The number a b a b1 2 2 1 is called the value of the determinant, and we write

a a

b b

1 2

1 2

= a b a b1 2 2 1

4.i.e. a11

belongs to the 1st row and 1st column

a12

belongs to the 1st row and 2nd column

a21

belongs to the 2nd row and 1st column

a22

belongs to the 2nd row and 2nd column

4.2 EXPANSION OF A DETERMINANT OF ORDER 2

A formal rule for the expansion of a determinant of order 2 may be stated as follows:

In the determinant,11 12

21 22

a a

a a

write the elements in the following manner :

a a

a a

11 12

21 22

Multiply the elements by the arrow. The sign of the arrow goingdownwards is positive, i.e.,

a11

a22

and the sign of the arrow going upwards is negative, i.e., a21 a12Add these two products, i.e., a

11a

22+( a21.a12) or a11 a22 a21 a12 which is the required

value of the determinant.

EXPECTED BACKGROUND KNOWLEDGE

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Example 4.1 Evaluate :

(i)6 4

8 2(ii)

a b b

a a b

+

+

2

2(iii)

x x x

x x x

2

2

1 1

1 1

+ + +

+

Solution :

(i)6 4

8 2= (6 2) (8 4) = 12 32 = 20

(ii)a b b

a a b

+

+

2

2= (a+b) (a+b) (2a) (2b)

= a2+2ab + b2 4ab= a2+b2 2ab= (a_b)2

(iii) x x x

x x x

2

2

1 1

1 1

+ + +

+ = (x2+x+1) (x_1) (x2_x+1) (x+1)

= (x3_1) (x3+1)

= _2

Example 4.2 Find the value ofx if

(i)x x

x x

+ +=

3

1 36 (ii)

2 1 2 1

1 4 2

x x

x x

+

+ += 0

Solution :

(i) Now,x x

x x

+ +

3

1 3= (x

_3) (x+3) x (x+1)

= (x2_

9)_x2 _ x =

_x _ 9

According to the question,_x _ 9 = 6

x = _15

(ii) Now, 2 1 2 11 4 2

x x

x x

+

+ += (2x_1) (4x+2) _ (x+1) (2x+1)

= 8x2+4x_

4x_

2_

2x2_

x_

2x_

1

= 6x2_

3x_3 = 3(2x2

_x

_1)

According to the equation

3(2x2_

x_

1) = 0

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or, 2x2_

x_

1 = 0

or, 22 2 1 0 x x x + =

or, 2x (x_1) +1(x

_1) = 0

or, (2x+1) (x_1)= 0

or,1

1,2

x=

4.3 DETERMINANT OF ORDER 3

The expression

a b c

a b c

a b c

1 1 1

2 2 2

3 3 3

contains nine quantities a1, b

1, c

1, a

2, b

2, c

2, a

3, b

3and c

3

aranged in 3 rows and 3 columns, is called determinant of order 3 (or a determinant of third

order). A determinant of order 3 has (3)2 = 9 elements.

Using double subscript notations, viz.,a11

, a12

, a13

, a21

, a22

, a23

, a31

, a32

, a33

for the elements

a1, b

1, c

1, a

2, b

2, c

2, a

3, b

3and c

3,we write a determinant of order3 as follows:

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

Usually a determinant, whether of order 2 or 3, is denoted by or |A|, |B| etc. = |a

ij| , where i = 1, 2, 3 and j = 1, 2, 3

4.4 DETERMINANT OF A SQUARE MATRIX

With each square matrix of numbers (we associate) a determinant of the matrix.

With the 1 1 matrix [a], we associate the determinant of order 1 and with the only

element a. The value of the determinant isa.

If A=a a

a a

11 12

21 22

L

N

MO

Q

P be a square matrix of order 2, then the expression

a11

a22

_ a21

a12

is defined as the determinant of order 2. It is denoted by

A =a a

a a

11 12

21 22

L

NM

O

QP = a11a22 _ a21a12

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With the 3 3 matrix

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

L

N

MMM

O

Q

PPP

, we associate the determinant

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

and its value is defined to be

a11

a a

a a

22 23

32 33

+ (_

1) a12

a a

a a

21 23

31 33

+ a13

a a

a a

21 22

31 32

Example 4.3 If A =3 6

1 5

L

NM

O

QP , find |A|

Solution : |A| =3 6

1 5

= 3 5_

1 6 = 15_6 =9

Example 4.4 If A=a b a

b a b

+

L

NM

O

QP , find |A|

Solution : |A| =a b a

b a b

+

= (a+b) (a

_b)

_b a = a2

_b2

_ab

Note : 1. The determinant of a unit matrix I is 1.

2. A square matrix whose determinant is zero, is called the singular matrix.

4.5 EXPANSION OF A DETERMINANT OF ORDER 3

In Section 4.4, we have written

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

= a11

a a

a a

22 23

32 33

+ (_1)a

12 a a

a a

21 23

31 33

+ a13

a a

a a

21 22

31 32

which can be further expanded as

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

=a a a a a a a a a a a a a a a11 22 33 32 23 12 21 33 31 23 13 21 32 22 31( ) ( ) ( ) +

=a a a a a a a a a a a a a a a a a a11 22 33 12 23 31 13 21 32 11 23 32 12 21 33 13 22 31+ +

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We notice that in the above method of expansion, each element of first row is multiplied by

the second order determinant obtained by deleting the row and column in which the element

lies.

Further, mark that the elements a11

, a12

and a13

have been assigned positive, negative and

positive signs, respectively. In other words, they are assigned positive and negative signs,

alternatively, starting with positive sign. If the sum of the subscripts of the elements is an even

number, we assign positive sign and if it is an odd number, then we assign negative sign.

Therefore,a11

has been assigned positive sign.

Note :We can expand the determinant using any row or column. The value of the determinant

will be the same whether we expand it using first row or first column or any row or column,

taking into consideration rule of sign as explained above.

Example 4.5 Expand the determinant, using the first row

1 2 3

2 4 1

3 2 5

Solution : =

1 2 3

2 4 1

3 2 5= 1

4 1

2 5_

2 2 1

3 5+ 3

2 4

3 2

= 1 x (20_

2)_

2 (10_3) + 3 (4

_12)

= 18_14_24

=_20

Example 4.6 Expand the determinant, by using the second column

1 2 3

3 1 2

2 3 1

Solution : =

1 2 3

3 1 2

2 3 1= (

_1) 2

3 2

2 1+1

1 3

2 1+ (

_1) 3

1 3

3 2

=_2 (3

_4) + 1 (1

_6)

_3 (2

_9)

= 2_

5+ 21

= 18

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Solution :

LetMijdenote the minor ofa

ij. Now, a

11occurs in the 1st row and 1st column. Thus

to find the minor ofa11

, we delete the 1st row and 1st column of |A|.

The minor M11

ofa11

is given by

M11

= a aa a

22 23

32 33

= a22

a33

_ a32

a23

Similarly, the minorM12

ofa12

is given by

M12

=a a

a a

21 23

31 33

= a21

a33

_ a23

a31

; M13

=a a

a a

21 22

31 32

= a21

a32

_ a31

a22

M21

=a a

a a

12 13

32 33

= a12

a33

_ a32

a13

; M22

=a a

a a

11 13

31 33

= a11

a33

_ a31

a13

M23

=a a

a a

11 12

31 32

= a11

a32

_ a31

a12

Similarly we can findM31

,M32

andM33

.

4.6.2 Cofactors of aijin |A|

The cofactor of an element aijin a determinant is the minor ofa

ijmultiplied by

(_1)i+j. It is usually denoted by C

ij, Thus,

Cofactor ofaij = Cij = (

_

1)

i+j

Mij

Example 4.8 Find the cofactors of the elements a11

, a12

,and a21

of the determinant

|A| =

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

Solution :

The cofactor of any element aijis (

_1)i+jM

ij, then

C11 = (

_

1)

1+1

M11 = (

_

1)

2

(a22 a33_

a32 a23)= (a

22a

33

_a

32a

23)

C12

= (_1)1+2M

12=

_M

12=

_(a

21a

33

_a

31a

23) = (a

31a

23

_a

21a

33)

and C21

= (_1)2+1M

21=

_M

21= (a

32a

13

_a

12a

33)

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Example 4.9 Find the minors and cofactors of the elements of the second row in the

determinant

|A| =

1 6 3

5 2 4

7 0 8

Solution : The elements of the second row are a21

=5; a22

=2; a23

=4.

Minor of a21

(i.e., 5) =6 3

0 8= 48

_0 = 48

Minor of a22

(i.e., 2) =1 3

7 8= 8

_21 =

_13

and Minor of a23

(i.e., 4) =1 6

7 0= 0

_42 =

_42

The corresponding cofactors will be

C21

= (_1)2+1M

21=

_(48) =

_48

C22

= (_1)2+2M

22= +(

_13) =

_13

and C23

= (_1)2+3M

23=

_(_42) = 42

4.7 VALUE OF A DETERMINANT USING COFACTOR

With the cofactor notation, we can give expansion of a determinant in terms of the elements

of any one row or column and their corresponding cofactors.

|A| =

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

= a11

C11

+a12

C12

+a13

C13

= a21

C21

+a22

C22

+a23

C23

= a31

C31

+a32

C32

+a33

C33

where Cijis the cofactor of the element a

ij

Example 4.10 Evaluate =

3 1 7

6 2 3

8 4 5

in terms of the elements of first row and their

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corresponding cofactors, and then in terms of the elements of first column and their

corresponding cofactors. Verify that both the results are same.

Solution :

Expanding the determinant in terms of the elements of first row and their

corresponding cofactors, we have the value of, if we expand byR1

,

i.e. first row is given by

= a11

C11

+ a12

C12

+ a13

C13

First we need to evaluate C11

, C12

and C13

C11

= (_1)1+1M

11= (

_1)2

2 3

4 5

= 10+12 = 22

C12

= (_1)1+2M

12= (

_1)

6 38 5

=_(_30+24) = 6

C13

= (_

1)1+3M13

= (_1)1+3

6 2

8 4

=_24

_16 =

_40

= 3 22 + 1 6 + 7 (_40) = 66 + 6 _ 280 = _208

Expanding the deteminant in terms of the elements of first column and their corresponding

cofactors, we have

= a11

C11

+ a21

C21

+ a23

C23

Now, C11

= (_1)1+1M

11= (

_1)2

2 3

4 5

= 10 + 12 = 22

C21

= (_1)2+1M

21= (

_1)3

1 7

4 5=

_(5

_28) = 23

C31

= (_1)3+1M

31= (

_1)4

1 7

2 3= (

_3_14) =

_17

= 3 22 + (_6) 23 + 8 (_17) = 66_138_136 = _208

The value of the determinant by both the methods of expansion is_

208. Thus, we verify

that both the results are same.

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Example 4.11 Show that the sum of the products of the elements of one row (or column)

by the cofactors of another row (or column) is always zero.

Solution :

Consider, =

2 3

4 5

Here, C11

= 5 and C12

=_

4

Let us find out a21 C11 + a22 C12

= 4 5 + 5 (_4)

= 20_

20 = 0

Similarly, if we consider

=

1 2 3

2 3 1

3 1 2

Here C21

= (_

1)2+12 3

1 2=

_1

C22

= (_

1)2+21 3

3 2=

_7

C23

= (_

1)2+31 2

3 1 =_(1

_6) = 5

Let us find out

a11

C21

+a12

C22

+a13

C23

= 1 (_1) + 2 (_7) + 3 (5)

=_

1_14+15 = 0

We can verify this property for any other row or column.

4.8 SARRUS DIAGRAM FOR THE EXPANSION OF

DETERMINANT OF ORDER 3

Another simple method for evaluating a determinant of order 3 is Sarrus diagram.

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Consider the determinant :

=

a a a

a a a

a a a

11 12 13

21 22 23

31 32 33

First of all we write down the three columns of the determinant, and then write the first row

columns again as shown below

11 12 13 11 12

21 22 23 21 22

31 32 33 31 32

a a a a a

a a a a a

a a a a a

From the sum of terms connected with downward arrows, we subtract the terms connected

with upwards arrows. Thus we get

a11a 22a33 + a12a23a31+a13a21a32_a

31a

22a

13

_a

32a

23a

11

_a

33a

21a

12

= a11

(a22

a33

_a

32a

23)

_a

12(a

21a

33

_a

23a

31)+a

13(a

21a

32

_a

31a

22)

= a11

a a

a a

22 23

32 33

_a

12

a a

a a

21 23

31 33

+a13

a a

a a

21 22

31 32

which is equal to the value of given determinant.

Example 4.12 Using Sarrus diagram, find the value of the determinant

=

1 2 3

7 5 0

3 2 4

Solution :

1 2 3 1 2

7 5 0 7 5

3 2 4 3 2The value of the determinant

= (_

1) (5)(_4) + 2.0.3 + 3. 7 (

_2)

_3.5.3

_(_2).0. (

_1)

_(_4). 7.2

= 20 + 0_

42_

45_

0 + 56

=_

11

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CHECK YOUR PROGRESS 4.2

1. Find the minors and cofactors of the elements of the second

row of the determinant

1 2 3

4 3 6

2 7 9

2. Find the minors and cofactors of the elements of the

third column of the determinat

2 3 2

1 2 1

3 1 2

3. Evaluate each of the following determinants using cofactors:

(a)

2 1 0

1 0 2

3 4 3(b)

1 0 1

0 1 1

1 1 0(c)

3 4 5

6 2 3

8 1 7

(d)

1

1

1

a bc

b ca

c ab(e)

b c a a

b c a b

c c a b

+

+

+(f)

1

1

1

a b c

b c a

c a b

+

+

+

4. Using Sarrus diagram, evaluate each of the following determinants:

(a)

2 1 3

1 1 2

2 2 4

(b)

0 2 7

1 2 5

1 2 1

(c)

1

1

1

a b

a c

b c

5. Using Sarrus diagram, show that

a h g

h b f

g f c= abc + 2fgh

_af2

_bg2

_ch2

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6. Solve for x, the following equations:

(a)

x 0 0

1 2 3

1 0 2= 0 (b)

x

x

3 3

3 3

2 3 3= 0 (c)

x x2 1

0 2 1

3 1 4= 28

4.9 PROPERTIES OF DETERMINANTS

We shall now discuss some of the properties of determinants. These properties will help us

in expanding the determinants.

Property 1: The value of a determinant remains unchanged if its rows and columns are

interchanged.

Let =

2 1 3

0 3 0

4 2 1

Expanding the determinant by first column, we have

= 2

3 0

2 1_

0

1 3

2 1+ 4

1 3

3 0

= 2 (3_

0)_

0 (1_

6) + 4 (0+9)

= 6 + 36 = 42

Let be the determinant obtained by interchanging rows and columns of. Then

=

2 0 4

1 3 2

3 0 1

Expanding the determinant by second column, we have (Recall that a determinantcan be expanded by any of its rows or columns)

( ) ( ) ( )

+

+

0

1 2

3 13

2 4

3 10

2 4

1 2

= 0 + (_3) (

_2_

12) + 0

= 42

Thus, we see that =

Property 2: If two rows ( or columns) of a determinant are interchanged, then the value of

the determinant changes in sign only.

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Let =

2 3 1

1 2 3

3 1 2

Expanding the determinant by first row, we have

= 22 3

1 2_

31 3

3 2+ 1

1 2

3 1

= 2 (4_3)

_3 (2

_9) + 1 (1

_6)

= 2 + 21_

5 = 18

Let be the determinant obtained by interchangingC1andC

2

Then =

3 2 12 1 3

1 3 2


31 3

3 2_

22 3

1 2+ 1

2 1

1 3

= 3 (2_

9)_

2(4_

3) + 1(6_1)

= _21_2+5 = _18

Thus we see that = _

Corollary

If any row (or a column) of a determinant is passed over n rows (or columns), then the

resulting determinant is = (1)n

For example ,

2 3 5

1 5 6

0 4 2

= (_

1)2

1 5 6

0 4 2

2 3 5

= 2 (10_24)

_3 (2

_0) + 5(4)

=_28

_6 +20=

_14

Property 3: If any two rows (or columns) of a determinant are identical then the value of

the determinant is zero.

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Proof : Let =

1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

be a determinant with identical columns C1 andC2 and let determinant obtained from

by interchanging C1andC

2

Then,

=

1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

which is the same as , but by property 2, the value of the determinant changes in sign, if its

any two adjacent rows (or columns) are interchnaged

Therefore =

Thus, we find that

or 2 = 0 = 0

Hence the value of a determinant is zero, if it has two identical rows (or columns).

Property 4: If each element of a row (or column) of a determinant is multiplied by the same

constant say, k 0, then the value of the determinat is multiplied by that constantk.

Let =

2 1 5

0 3 0

4 2 1


= 2(3_0)

_1 (0

_0) + (

_5) (0+12 )

= 6_60 =

_54

Let us multiply column 3 ofby 4. Then, the new determinantis :

=

2 1 20

0 3 04 2 4


= 2 (12_

0)_1 (0

_0) + (

_20) (0+12)

= 24_

240 =_216

= 4

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Corollary :

If any two rows (or columns) of a determinant are proportional, then its value is zero.

Proof : Let =

a b ka

a b ka

a b ka

1 1 1

2 2 2

3 3 3

Note that elements of column 3 are ktimes the corresponding elements of column 1

By Property 4, =k

a b a

a b a

a b a

1 1 1

2 2 2

3 3 3

= k 0 (by Property 2)

= 0

Property 5: If each element of a row (or of a column) of a determinant is expressed as the

sum (or difference) of two or more terms, then the determinant can be expressed as the

sum (or difference) of two or more determinants of the same order whose remaining rows

(or columns) do not change.

Proof : Let =

a + b c

a b c

a b c

1 1 1

2 2 2

3 3 3

+ +

Then, on expanding the determinant by the first row, we have

= ( )( ) ( )( ) ( )( )a b c b c b a c a c c a b a b1 2 3 3 2 1 2 3 3 2 1 2 3 3 2+ + + +

= a b c b c b a c a c c a b a b b c b c1 2 3 3 2 1 2 3 3 2 1 2 3 3 2 2 3 3 2( ) ( ) ( ) ( ) + +

+ ( ) ( )a c a c a b a b2 3 3 2 2 3 3 2

=

a b c

a b c

a b c

+ a b c

a b c

1 1 1

2 2 2

3 3 3

2 2 2

3 3 3

Thus, the determinant can be expressed as the sum of the determinants of the same

order.

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Example 4.13 Show that

1

1

1

2

2

2

w w

w w

w w= 0

where w is a non-real cube root of unity.

Solution : =

1

1

1

2

2

2

w w

w w

w w

Add the sum of the 2nd and 3rd column to the 1st column. We write this operation as

C C C C 1 1 2 3

+ +( )

=

1

1 1

1 1

2 2

2 2

2

+ +

+ +

+ +

w w w w

w w w

w w w=

0

0 1

0 1

2

2

w w

w

w= 0 (on expanding by C

1)

(since w is a non-real cube root of unity, therefore, 1+w+w2 = 0)


1

1

1

a bc

b ca

c ab= (a

_b) (b

_c) (c

_a)

Solution : =

1

1

1

a bc

b ca

c ab

=

0

0

1

a c bc ab

b c ca ab

c ab

[ R R R

1 1 3 and R R R

2 2 3 ]

=

0

0

1

a c b c a

b c a c b

c ab

( )

( ) = ( )( )a c b c

b

a

c ab

0 1

0 1

1

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Expanding byC1, we have

= ( )( )a c b cb

a

1

1

= ( )( )( )a c b c b a

= ( )( )( )a b b c c a

Example 4.15

Prove that

b c a a

b c a b

c c a b

+

+

+= 4 abc

Solution : =

b c a a

b c a b

c c a b

+

+

+

=

0 2 2

1 1 2 3

+

+

+

c b

b c a b

c c a b

R R R R[ ( )]

Expanding byR1, we get

= 0 2 2c a b

c a bc

b b

c a bb

b c a

c c

+

+

+

+( )

= 2 2c b a b bc b bc c c a[ ( ) ] [ ( )]+ +

= 2 2bc a b c bc b c a[ ] [ ]+

= 2bc a b c b c a[( ) ( )]+

= 2bc a b c b c a[ ]+ + +

= 4abc

Example 4.16 Evaluate:

=

a b b c c a

b c c a a b

c a a b b c

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Solution : =

a b b c c a

b c c a a b

c a a b b c

=

0

0

0

b c c a

c a a b

a b b c

[ ]C C C C

1 1 2 3 + +

= 0, on expanding by C1

Example 4.17 Prove that

1

1

1

bc a b c

ca b c a

ab c a b

( )

( )

( )

+

+

+

= 0

Solution : =

1

1

1

bc a b c

ca b c a

ab c a b

( )

( )

( )

+

+

+

=

1

1

1

bc bc ab ac

ca ca bc ba

ab ab ca cb

+ +

+ +

+ +[ ]C C C

3 2 3 +

= ( )ab bc ca

bc

ca

ab

+ +

1 1

1 1

1 1

= ( )ab bc ca+ + 0 (by Property 3)

= 0


=

a ab ac

ab b bc

ac bc c

2

2

2= 4a2b2c2

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Solution : =

a ab ac

ab b bc

ac bc c

2

2

2

= abc

a b c

a b c

a b c

= abc

+

+

L

NM

O

QP

a b c

c

b

R R R

R R R0 0 2

0 2 0

2 2 1

3 3 1

= abc ac

b( )

0 2

2 0(on expanding by C

1)

= abc a bc( )( ) 4

= 4a2b2c2


1 1 1

1 1 1

1 1 1

+

+

+

a

a

a= a2 (a+3)

Solution : =

1 1 1

1 1 1

1 1 1

+

+

+

a

a

a

=

aa a

a a

C C C C ++ +

+ +

+ +3 1 13 1 1

3 1 1

1 1 2 3[ ]

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=( )a a

a

+ +

+

3

1 1 1

1 1 1

1 1 1

=( )a a

a

C C C

C C C+

L

NM

O

QP3

1 0 0

1 0

1 0

2 2 1

3 3 1

= ( 3) (1)0

0a

a

a+

= ( )( )a a+ 3 2

= a a2 3( )+


1. Show that

x x x

x x x

x x x

+

+

+

3

3

3= 27 (x+1)

2. Show that

a b c a a

b b c a b

c c c a b

a b c

= + +

2 2

2 2

2 2

3( )

3. Show that

1 1 1

1 1 1

1 1 1

+

+

+

= + + +

a

b

c

bc ca ab abc

4. Show that

a a b a b

a b a a b

a b a b a

b a b

+ +

+ +

+ +

2

2

2

92

= +( )

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5. Show that

( )( )

( )( )

( )( )

a a a

a a a

a a a

+ + +

+ + +

+ + +

=

1 2 2 1

2 3 3 1

3 4 4 1

2

6. Show that

a b b c c a

b c c a b

c a a b b

a b c

b c a

c a b

+

=

+ +

+ + a +

+ + + c

2

7. Evaluate

(a)

a a b a b c

a a b a b c

a a b a b c

+ + +

+ + +

+ + +

2 3 2 4 3 2

3 6 3 10 6 3

(b)

( )

( )

( )

b c a a

b c a b

c c a b

+

+

+

2 2 2

2 2 2

2 2 2

8. Solve for x :

3 8 3

3 3 8 3

3 3 3 8

x x

x

x

= 0

4.11 SOLUTION OF A SYSTEM OF LINEAR EQUATIONS

BY CRAMERS RULE

Consider the system of linear equations

a1x+b

1y = c

1

a2x+b

2y = c

2

In this section, we shall learn to solve simultaneous linear equations in two unknowns and three

unknowns with the help of determinants.

A pair of values of x and y which satisfies both the equations simultaneously is called a solution

of the given system of linear equations and then the system is said to be consistent.

When the constants c1andc

2on the R.H.S. of the equations are both zero, the system is said

to be homogeneous system: otherwise, the system is said to be non-homogeneous. When the

system is homogeneous, it has always a solution x = y = 0. It is called a trival solution.

If the determinant =a b

a b

1 1

2 2

0, then the system of homogeneous equations has only one

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solution, i.e., the trival solution x = y = 0. However, if the determinant D =a b

a b

1 1

2 2

=0, then

the system forms a pair of dependent equations and in that case the system has infinite

number of solutions, i.e., the system has also non-trival solutions. Determinants are used

mainly to solve non-homogeneous linear equations in two or more variables.In this section,

we shall learn to solve simultaneous linear equations in two/three unknowns with the help of

determinants.The method of solving simultaneous linear equations by determinants is

commonly known as Cramers Rule.

4.11.1 Solving a system of Linear Equation in two variables

Consider the following system of linear equations :

2 3 5 0x y+ = ...... (i)

3 5 7 0x y+ = ...... (ii)

Usually, to find solutions of such a system, we apply the method of elimination.

Thus, first of all solving fory, we get

3 2 3 5 0

2 3 5 7 0

( )

( )

x y

x y

+ =

+ =

+ =

+ =

3 2 3 3 3 5

2 3 2 5 2 7

( ) ( ) ( ) ......( )

( ) ( ) ( ) .....( )

x y iii

x y iv

Subtracting (iii) from (iv), we get{2(5)

_3(3)} y = 2(7) _ 3(5)

i.e.,2 3

3 5y =

2 5

3 7

=y

2 5

3 7

2 3

3 5

......(A)

Observe that in the determinant of the numerator, the 1st column consists of the coefficeints

ofx and the 2nd columns consist of the constant terms.

And, in the determinant of the denominator of y, the 1st column and the 2nd column consist

of the coefficients ofx and y respectively.

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Similarly, solving (i) and (ii) forx, we get

5(2x + 3y 5) = 0

3(3x + 5y 7) = 0

+ =

+ =

RST

5 2 5 3 5 5 0

3 3 3 5 3 7 0

( ) ( ) ( ) ......( )

( ) ( ) ( ) .....( )

x y v

x y vi

Subtracting (vi) from (v), we get

{5(2) 3(3) }x = {5(5) 3(7)}

i.e.23

35x =

5 3

7 5

=x

5 3

7 52 3

3 5

......(B)

Again observe that in the determinant of the numerator ofx, the 1st column consists of the

constant terms and the 2nd column consists of the coefficients ofy.

And, in the determinant of the denominator ofx, the 1st column and the 2nd column consist

of the coefficients ofx andy respectively .

Thus, x =

5 37 5

2 3

3 5

=D

D

1(say)

and yD

D= =

2 5

3 7

2 3

3 5

2(say)

are the solutions of the given system of equations in the determinant form.

Consider another system of equations:

x y+ = 3 ...(i)

2 3 1x y = ...(ii)

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Again, solving (i) and (ii) by elimination method, forx, we get

+ =

=

3 3 3

1 2 3 1

( ) ( )

( )

x y

x y

=

=RST

3 3 3 3

1 2 1 3 1 1

( ) ( ) ( )

( ) ( ) ( )

x y

x y

Subtracting (iv) from (iii), we get

=

=

3 1 2 3 3 1 1

3 1 1 2 3 3 1 1

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

x x

xk p

i.e.

1 1

2 3

3 1

1 3 = x

=

=xD

D

3 1

1 3

1 1

2 3

1

(say)

Again, we observe that in the determinant D1, the 1st column consists of the

constant terms and the 2nd column consists of the coefficient ofy. Also, inD, the 1st and

the 2nd columns consist of the coefficients ofx andy respectively.

Similarly, solving (i) and (ii) fory, we get

2 2 2 3( ) ( ) ( )x y+ = ...(v)

1 2 2 3 1 1( ) ( ) ( )x y+ = ...(vi)

Subtracting (v) from (vi), we get

1 3 2 1 1 1 2 3( ) ( ) ( ) ( ) = k py

i.e.,

1 1

2 3

1 3

2 1 =y

=

=yD

D

1 3

2 1

1 1

2 3

2

(say)

...(iii)

...(iv)

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Again, we observe that the determinantD is nothing but the determinant of thecoefficient ofx andyand the determinantD

2is obtained by replacing the coefficient ofyby

the constant terms.

Thus, x =

3 1

1 31 1

2 3

and y =

1 3

2 11 1

2 3

are the solutions of the given equation

Therefore, if a x b y c a x b y c1 1 1 2 2 2

+ = + =and

is a given system of linear equations in two variables, then its solution will be

x

c b

c b

a b

a b

D

D= =

1 1

2 2

1 1

2 2

1

and y

a c

a c

a b

a b

D

D= =

1 1

2 2

1 1

2 2

2

provided Da b

a b=

1 1

2 2

0

4.11.2 Solution of a System of Linear Equations in Three Variables

Consider the following system of equations in three variables

x y z

x y z

x y z

+ + =

+ + =

+ + =

2 3 4

2 1

3 3 5 3...(C)

Let D =

1 2 3

2 1 1

3 3 5

i.e.,D is the determinant of the coefficients ofx,y andz.

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Then, xD

x

x

x

=

2 3

2 1 1

3 3 5

Multiplying the 2nd column byy and the 3rd column byz and adding

these to the 1st column, we get

xD

x y z

x y z

x y z

=

+ +

+ +

+ +

2 3 2 3

2 1 1

3 3 5 3 5

From (C), we get

xD D= =

4 2 3

1 1 13 3 5

1 (say)

i.e., xD D xD

D= =

1

1

Similarly, we will get

D2

1 4 3

2 1 1

3 3 5

= and D3

1 2 4

2 1 1

3 3 3

=

Then, as before, we can see that

yD D=2

and zD D=3

Thus, xD

Dy

D

Dz

D

D= = =1 2 3, , , where D 0

are the solutions of the given system of linear equations in three variables.

Therefore, if

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1 1

2 2 2 2

3 3 3 3

+ + =

+ + =

+ + =

is a given system of linear equations in three variables, then

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Solution: Now,

Da b

a b= = = =

1 1

2 2

2 3

3 510 9 1

Dc b

c b1

1 1

2 2

5 3

7 525 21 4= = = =

and Da c

a c2

1 1

2 2

2 5

3 714 15 1= = = =

Thus, by Cramers Rule xD

D= = =1

4

14

and yD

D= =

= 2

1

11

are the solutions of the given system of equations.Example 4.21 Solve the following system of equations by Cramers Rule:

2 3 3

2 5

3 5 2 1

x y z

x y z

x y z

+ =

+ + =

+ =

Solution: We have,

D =

= + = + + =

2 1 3

1 2 1

3 5 2

2 4 5 1 2 3 3 5 6 18 1 33 52 0( ) ( ) ( )

Also, to findD1, the 1st column will be replaced by constants.

D1

3 1 3

5 2 1

1 5 2

3 4 5 1 10 1 3 25 2 27 9 81 99=

= + = + =( ) ( ) ( )

To find,D2, 2nd column will be replaced by constants.

D2

2 3 3

1 5 1

3 1 2

2 10 1 3 2 3 3 1 15 18 3 42 63=

= = + + =( ) ( ) ( )

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Now, consider the system of equations

x y

x y

+ =

+ =

2 0

2 3 0

Here, D =

= + = 1 22 3

3 4 7 0

Also, D10 2

0 30 0 0= = =

and D21 0

2 00 0 0=

= =

Hence, x DD

= = =1 07

0

and xD

D= = =2

0

70

Thus, we find that for D 0 andD1= D2 = 0, the given system of equations will have onlythe trivial solutionx =y = 0

We already know that Cramers rule does not apply ifD = 0. The two cases arise namely

system of equations having (i) no solution, (ii) infinitely many solutions.

Consider the system of equations

2 4 5

2 3

x y

x y

+ =

+ =

Here, D = = =2 4

1 24 4 0

SinceD = 0, Cramers rule does not apply here.

Now, D15

310 12 2= = =

4

2

and D2

2 5

1 36 5 1= = =

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Hence, if D = 0 and D1

0, D2

0, the equations will have no solution.

Similar is the case for a system of three equations in three variables.For that, consider the

system of equations

x y z

x y z

x y z

+ + =

+ + =

+ + =

2

2 3 3

3 5 5...(i)

Here, D = = + = + =

1

1 1 10 9 1 5 3 1 3 2 1 2 1 0

1 1

2 3

1 3 5

( ) ( ) ( )

Now, xD

x

x

x

=

1 1

2 3

3 5

Multiplying the 2nd column byy and the 3rd column byz and adding them to the 1st column

we get

xD

x y z

x y z

x y z

=

+ +

+ +

+ +

1 1

2 3 2 3

3 5 3 5

= =

2 1 1

3 2 3

5 3 5

1D

[using(1)]

Thus, D1 2 10 9 1 15 15 1 9 10 2 0 1 1= + = =( ) ( ) ( )

So, xD D xD= =1

1

= =xD

1 1

0which is undefined

Similarly, we get yD D=2

and zD D= 3

where D2

1 2 1

1 3 3

1 5 5

1 15 15 2 5 3 1 5 3 0 4 2 2= = + = + = ( ) ( ) ( )

and D3

1 1 2

1 2 3

1 3 5

1 10 9 1 5 3 2 3 2 1 2 2 1= = + = + =( ) ( ) ( )

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yD D yD

Dy= = =

2

22

0which is undefined.

and zD D zD

Dz= = =

3

31

0which is undefined.

Thus, ifD = 0 andD1 0,D2 0 andD3 0, then the system of equations will have no

solution. In this case, we say that the system of equations is inconsistent.

Now, consider the system of equations

x y z

x y z

x y z

+ =

+ =

+ + =

3 6

3 3 4

5 3 3 10

Here, D =

= + + + + = + =

1 1 3

1 3 3

5 3 3

1 9 9 1 3 15 3 3 15 18 18 36 0( ) ( ) ( )

Also D1

6 1 3

4 3 3

10 3 3

6 9 9 1 12 30 3 12 30 108 18 126 0=

= + + + + = + =( ) ( ) ( )

D2

1 6 3

1 4 3

5 10 3

1 12 30 6 3 15 3 10 20 18 108 90 0= = + + + + = + =( ) ( ) ( )

D3

1 1 6

1 3 4

5 3 10

1 30 12 1 10 20 6 3 15 42 30 72 0=

= + + + + = + = =( ) ( ) ( )

Consider the first two equations,

i.e., x y z

x y z

+ =

+ =

3 6

3 3 4

Thus, forD = 0 and D D D1 2 3 0= = = for the given system of equations will haveinfinitely many solutions.

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These can be written as

x y z

x y z

=

+ = +

6 3

3 4 3

Solving these equations by determinants, we get

x

z

z z z z z z=

+

=

+ +

+=

+=

6 3 1

4 3 3

1 1

1 3

3 6 3 1 4 3

3 1

18 9 4 3

4

14 6

4

( ) ( )

xz

=7 3

2

and y

z

z z z z z z=

+

=

+

+=

+ +=

+

1 6 3

1 4 3

1 1

1 3

1 4 3 1 6 3

3 1

4 3 6 3

4

10 6

4

( ) ( )

yz

= +5 3

2

Letz = k, where kis any number, then we get

xk

=7 3

2, y

k=

+5 3

2and z k=

Thus, the system of equations has many solutions.

So, we conclude that for a given system of equations

(i) If D 0 and atleast one of D D Dn1 2

, , ..., is not equal to zero, then the system will

have non-zero, unique solution.(ii) If D 0 and each D

1= 0, then the system has only the trivial solution

x x xn1 2

0= = = =...

(iii) If D = 0 and some D1

0 , then the system has no solution.

(iv) If D = 0 and eachD1= 0, then the system has infinitely many solutions.

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Example 4.22 Solve the following system of equations:

x y z

x y z

x y z

+ + =

+ =

+ =

2

2 7 3 5

3 5 4

Solution : Here,

D =

= + + + = =

1 1 1

2 7 3

3 5 1

1 7 15 1 2 9 1 10 21 8 7 11 10 0( ) ( ) ( )

Now D1

2 1 1

5 7 3

4 5 1

2 7 15 1 5 12 1 25 28 16 7 3 6=

= + + + = =( ) ( ) ( )

D2

1 2 1

2 5 3

3 4 1

1 5 12 2 2 9 1 8 15 7 14 7 14=

= + + + = = ( ) ( ) ( )

D3

1 1 2

2 7 5

3 5 4

1 28 25 1 8 15 2 10 21 3 7 22 12= = + = + = ( ) ( ) ( )

Since D 0 and D D D1 2 3

0 0 0 , , , therefore, the system of equations will have

non-zero, unique solution.

Thus, by Cramers Rule

x

D

D= = =

1

6

10

3

5

yD

D= =

=2

14

10

7

5

zD

D= =

=3

12

10

6

5

are the solutions of the given system of equations.

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Example 4.23 Determine which of the following systems of equations will have a unique

solution; and also find the solutions in each case:

(i) 2 3 4 9

3 4 2 12

4 2 3 3

x y z

x y z

x y z

+ =

+ + =

=

(ii) x y z

x y z

x y z

+ =

+ + =

+ =

2 0

2 2 0

3 0

(iii) x y z

x y z

x y z

+ + =

+ + =

+ =

2 2

2 2 3

3 4

(iv) x y z

x y z

x y z

+ + =

+ =

+ + =

2 3 1

3 2 1

4 5 2

Solution:

(i) 2 3 4 9

3 4 2 12

4 2 3 3

x y z

x y z

x y z

+ =

+ + =

=

Here,

D =

= + + + = + =

2 3 4

3 4 2

4 2 3

2 12 4 3 9 8 4 6 16 16 3 40 53 0( ) ( ) ( )

Also

D1

9 3 4

12 4 2

3 2 3

9 12 4 336 6 4 24 12 72 126 144 342=

= + + + + + = + + =( ) ( ) ( )

D2

2

2 36 6 9 9 8 4 9 48 84 9 228 321=

= + + + + = + + =

9 4

3 12 2

4 3 2

( ) ( ) ( )

and

D3

2 3 9

3 4 12

4 2 3

2 12 24 3 9 48 9 6 16 72 171 90 189=

= + + = + + =( ) ( ) ( )

Since D 0 and D1

0 , D2

0 , D3

0

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Algebra


The system of equations will have a non-zero unique solution which is

xD

D= =

=

1

342

53

342

53

yD

D= = =

2

321

53

321

53

and zD

D= =

=

3

189

53

189

53

(ii) x y z

x y z

x y z

+ =

+ + =

+ =

2 0

2 2 0

3 0

Here,

D =

= + = + =

1 2 1

2 1 2

1 3 1

1 1 6 2 2 2 1 6 1 7 7 14 0( ) ( ) ( )

Also,

D1

0 2 1

0 1 2

0 3 1

0=

= (expanding by the 1st column)

D2

1 0 1

2 0 2

1 0 1

0=

=

and

D3

1 2 0

2 1 01 3 0

0=

=

Thus, we find that D 0 and D D D1 2 3 0= = =

The system of linear equations will not have a unique solution. In fact, it has atrivial solution x = 0, y = 0, z = 0

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Algebra


(iii) x y z

x y z

x y z

+ + =

+ + =

+ =

2 2

2 2 3

3 4

Here D =

=

1 2 1

2 1 2

1 3 1

0

Also D1

2 2 1

3 1 2

4 3 1

2 1 6 2 3 8 1 9 4 14 10 13 11=

= + + = + =( ) ( ) ( )

D C C 2 1 3

1 2 1

2 3 2

1 4 1

0= = =( )

and D3

1 2 2

2 1 3

1 3 4

1 4 9 2 8 3 2 6 1 13 10 14 11=

= + + = = ( ) ( ) ( )

Since D = 0 and D D1 2

0 0 =, and D3

0

The system of equations has no solution.

(iv) x y z

x y z

x y z

+ + =

+ =

+ + =

2 3 1

3 2 1

4 5 2

Here, D = = + + = + =

1 2 3

3 1 2

4 1 5

1 5 2 2 15 8 3 3 4 7 14 21 0( ) ( ) ( )

Also, D1

1 2 3

1 1 2

2 1 5

1 5 2 2 5 4 3 1 2 7 2 9 0= = + + = + =( ) ( ) ( )

D2

1 1 3

3 1 2

4 2 5

1 5 4 1 15 8 3 6 4 1 7 6 0= = + = + =( ) ( ) ( )

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Algebra


and D3

1 2 1

3 1 1

4 1 2

1 2 1 2 6 4 1 3 4 3 4 7 0= = + + = + =( ) ( ) ( )

Therefore, the given system of equations will have infinitely many solutions.


1. Solve the following systems of equations by Cramers Rule

(a) 2 4 3

3 5

x y

x y

=

+ =

(b) x y

x y

+ =

+ =

2 1

2 5 3

2. Obtain the solutions of the systems of equations using Cramers Rule

(a) 2 3 1

4 6 9

4 3 9 5

x y z

x y z

x y z

+ + =

+ + =

+ + =

(b)2 3 2 1

3 2

3 3

x y z

x y z

x y z

+ =

+ =

+ =

(c)3 4 5 6

2 1

2 3 5

x y z

x y z

x y z

+ =

+ =

+ + =

3. Solve the following systems of equations:

(a) 3 2 4

2 3

x y

x y

+ =

+ =

(b) 6 3 1

2 2 3

x y

x y

=

+ =

(c) 2 3 4 8

3 2

4 5 9

x y z

x y z

x y z

+ + =

+ =

=

(d) x y z

x y z

x y z

+ =

+ =

=

3 4

3 2 5 4

5 4 9

4. Determine which of the following systems of equations will have a unique solution.Also, find the solution in such a case.

( ) 2 4 (b) 2 0

3 5 7 2 0

3 2 0

a x y x y z

x y x y z

x y z

= + =

+ = + =

+ =

The expression a1b2 a2b1 is denoted by 1 12 2a b

a b

With each square matrix, a determinant of the matrix can be associated.Theminorof any element in a determinant is obtained from the given determinant by

deleting the row and column in which the element lies.

Thecofactorof an element aij in a determinant is the minor ofaij multiplied by( 1)i+j

LET US SUM UP

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MODULE - I

Algebra


A determinant can be expanded using any row or column. The value of thedeterminant will be the same.

A square matrix whose determinant has the value zero, is called a singular matrix. The value of a determinant remains unchanged, if its rows and columns are

interchanged.

If two rows (or columns) of a determinant are interchanged, then the value of thedeterminant changes in sign only.

If any two rows (or columns) of a determinant are identical, then the value of thedeterminant is zero.

If each element of a row (or column) of a determinant is multiplied by the sameconstant, then the value of the determinant is multiplied by the constant.

If any two rows (or columns) of a determinant are proportional, then its value iszero.

If each element of a row or column from of a determinant is expressed as the sum(or differenence) of two or more terms, then the determinant can be expressed as

the sum (or difference) of two or more determinants of the same order. The value of a determinant does not change if to each element of a row (or column)

be added to (or subtracted from) some multiples of the corresponding elements of

one or more rows (or columns).

The solution of a system of linear equations

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1 1

2 2 2 2

3 3 3 3

+ + =

+ + =

+ + =

is given by31 2,y and z DD Dx

D D D= = =

D

a b c

a b c

a b c

D

d b c

d b c

d b c

D

a d c

a d c

a d c

= = =

1 1 1

2 2 2

3 3 3

1

1 1 1

2 2 2

3 3 3

2

1 1 1

2 2 2

3 3 3

, , and

D

a b d

a b d

a b d

3

1 1 1

2 2 2

3 3 3

= , provided D 0

The system of linear equations

a x b y c z d

a x b y c z d

a x b y c z d

1 1 1 1

2 2 2 2

3 3 3 3

+ + =

+ + =

+ + =

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MODULE - I

Algebra


TERMINAL EXERCISE

(a) is consistent and has unique solution, when D 0

(b) is inconsistent and has no solution, whenD = 0, and D D D1 2 3, , are not all zero.

(c) is consistent and has infinitely many solutions, when D D D D= = = =1 2 3 0.

SUPPORTIVE WEB SITES

http://www.wikipedia.org

http://mathworld.wolfram.com

1. Find all the minors and cofactors of

1 2 3

3 4 2

2 3 1

2. Evaluate

43 1 6

35 7 4

17 3 2

by expanding it using the first column.

3. Using Sarrus diagram, evaluate

2 1 2

1 2 3

3 1 4

4. Solve forx, if

0 1 0

2

1 3

0x x

x

=

5. Using properties of determinants, show that

(a) 1

1

1

2

2

2

a a

b b

c c

b c c a a b= ( )( )( )
http://mathworld.wolfram.com/http://mathworld.wolfram.com/http://www.wikipedia.org/

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MODULE - I

Algebra


(b) 1

1

1

2 2

2 2

2 2

x y x y

y z y z

z x z x

x y y z z x

+ +

+ +

+ +

= ( )( )( )

6. Evaluate:

(a)

1 2 3

2 3 4

3 4 5

2 2 2

2 2 2

2 2 2

(b)

1

1

1

3 5

3 4

5 5

w w

w w

w w

,w being an imaginary cube-root of unity

7. Using Cramers rule, solve the following system of linear equations:

(a)x y

x y

=

+ =

2 3

2 3 4(b) 2 3 3

3 2 11x yx y

=+ =

(c)x y

x y+ =

=

2

4 3 3

8. Using Cramers rule, solve the following systems of equations:

(a) 2 3 8

3 2 3

3 1 2

x y z

x y z

x z y

+ + =

+ + =

+ = +

(b) x y z

x z y

x z y

= +

+ =

=

2 3

2

3 2 3

9. Determine which of the following systems of equations will have a unique solution:

(a) 2 6 1 03 2 0

x yx y

+ = + =

(b) 2 3 52 6

x y

x y =

=

(c) 2 3 1

4 6 3

6 3 2 5

x y z

x y z

x y z

+ + =

+ =

+ =

(d) 3 2 1

2 2

2 3 1

x y z

x y z

x y z

+ + =

+ =

+ =

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MODULE - I

Algebra


4. (a) Yes; x y= = 6 5, (b) Yes; x y z= = =0 0 0, ,

(c) x y z= = =0 3 2, ,

TERMINAL EXERCISE

1. M M M M M M

M M M

11 12 13 21 22 23

31 32 33

2 1 1 7 5 1

8 7 2

= = = = = =

= = =

, , , , , ,

, ,

C C C C C C

C C C

11 12 13 21 22 23

31 32 33

2 1 1 7 5 1

8 7 2

= = = = = =

= = =

, , , , , ,

, ,

2. 0 3. 31 4. x x= =0 1,

6. (a) 8 (b) 0

7. (a) x y= = 17

7

2

7, (b) x y= =3 1, (c)x y= =

9

7

5

7,

8. (a) x y z= = =2 3 3, , (b) x y z= = = 1

2

3

2

1

2, ,

9. (b) only

L.NO.04 MATHS 311

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