Post on 08-Apr-2018
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MATHEMATICS
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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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Determinants and Their Applications
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
Expanding the determinant by first row, we have
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
Expanding the determinant by first row, we have
= 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
Expanding the determinant by first row, we have
= 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)
Example 4.14 Show that
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
Example 4.18 Show that
=
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
Example 4.19 Show that
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( )+
CHECK YOUR PROGRESS 4.3
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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Notes
MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
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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MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
(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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Notes
MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
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.
CHECK YOUR PROGRESS 4.4
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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Notes
MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
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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Notes
MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
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= ( )( )( )
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Notes
MATHEMATICS
MODULE - I
Algebra
Determinants and Their Applications
(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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Notes
MODULE - I
Algebra
Determinants and Their Applications
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