Chapter: 1
Matrices & Determinants
Matrices & Determinants
Session Objectives
Meaning of matrix
Type of matrices
Transpose of Matrix
Meaning of symmetric and skew symmetric
matrices
• Minor & co-factors
• Computation of adjoint and inverse of a
matrix
•
•
•
•
Matrices & Determinants
TYPES OF MATRICES
NAME
DESCRIPTION
EXAMPLE
Rectangular
matrix
No. of rows is not equal to
no. of columns
 6
 2

2
0
Square matrix
No. of rows is equal to no. of
columns
2

2


1
1
0
2
Diagonal
matrix
Non-zero element in principal
diagonal and zero 
in all other
positions
2
0


0
0
Scalar matrix
Diagonal matrix in which all
the elements on principal
diagonal and same
4
0


0
0
4
0
4
0
 1
5 
3

1
4

0
0

7

0
0

4

Matrices & Determinants
TYPES OF MATRICES
NAME
DESCRIPTION
Row matrix
A matrix with only 1
row
Column matrix
A matrix with only I
column
Identity matrix
Diagonal matrix

having each
diagonal element
equal to one (I)
Zero matrix
EXAMPLE
3
2 1 4
2
 
3

A matrix with all zero
entries
1 0
0 1 


0 0


0 0
Matrices & Determinants
TYPES OF MATRICES
NAME
DESCRIPTION
Upper Triangular
matrix
Square matrix
having all the entries
zero below the
principal diagonal
2
0


0
5
Square matrix
having all the entries
zero above the
principal diagonal
2
5


6
0
Lower Triangular
matrix
EXAMPLE
4
0
4
3
3
6

7

0
0

7

Matrices & Determinants
Determinants
If A = aij  is a square matrix of order 1,
then |A| = | a11 | = a11
a11 a12 
A
=
If
a

 21 a22 
|A| =
a11
a12
a21
a22
is a square matrix of order 2, then
= a11a22 – a21a12
Matrices & Determinants
Example
4 -3
Evaluate the determinant :
2
5
4 -3
Solution :
= 4 × 5 - 2 × -3  = 20 + 6 = 26
2
5
Matrices & Determinants
Solution
a11

If A = a21
a31
a12
a22
a32
a11 a12
| A |= a21 a22
a31 a32
a13 
a23  is a square matrix of order 3, then
a33 
a13
a
a23 = a11 22
a32
a33
a23
a33
- a12
a21
a23
a31
a33
+ a13
a21
a22
a31
a32
[Expanding along first row]
= a11 a22a33 - a32a23  - a12 a21a33 - a31a23  + a13 a21a32 - a31a22 
  a11a22 a33  a12 a31a23  a13 a21a32   a11a23 a32  a12 a21a33  a13 a31a22 
Matrices & Determinants
Example
2
Evaluate the determinant : 7
-3
3
1
4
-5
-2
1
Solution :
2
7
-3
3
1
4
-5
1
-2 =2
4
1
-2
7
-3
1
-3
-2
7
+ -5
1
-3
1
4
[Expanding along first row]
= 2 1 + 8  - 3 7 - 6  - 5 28 + 3 
= 18 - 3 - 155
= -140
Matrices & Determinants
Minors
-1
If A =
2
4
, then

3
M11 = Minor of a11 = 3, M12 = Minor of a12 = 2
M21 = Minor of a21 = 4, M22 = Minor of a22 = -1
Matrices & Determinants
Minors
 4 7 8
If A = -9 0 0  , then
 2 3 4
M11 = Minor of a11 = determinant of the order 2 × 2 square
sub-matrix is obtained by leaving first
row and first column of A
=
0 0
=0
3 4
Similarly, M23 = Minor of a23
M32 = Minor of a32 =
4 7
=
=12 -14 = -2
2 3
4 8
= 0+72 = 72 etc.
-9 0
Matrices & Determinants
Cofactors
Cij = Cofactor of aij in A = -1
i+ j
Mij ,
where Mij is minor of aij in A
Matrices & Determinants
Cofactors (Con.)
 4 7 8
A = -9 0 0 
 2 3 4
C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1
C23 = Cofactor of a23 =
(–1)2 + 3
0 0
=0
3 4
4 7
2
M23 = 
2 3
C32 = Cofactor of a32 = (–1)3 + 2M32 = -
4 8
= - 72 etc.
-9 0
Matrices & Determinants
Value of Determinant in Terms
of Minors and Cofactors
a11
If A = a21
a31
3
A 
a12
a22
a32
  1
j1
i j
a13 
a23  , then
a33 
3
aijMij 
 aijCij
j1
= ai1Ci1 + ai2Ci2 + ai3Ci3 , for i =1 or i = 2 or i = 3
Matrices & Determinants
Properties of Determinants
1.
The value of a determinant remains unchanged, if its
rows and columns are interchanged.
a1 b1
a2 b2
a3 b3
2.
c1
a1 a2 a3
c2 = b1 b2 b3
c3
c1 c2 c3
i.e. A  A '
If any two rows (or columns) of a determinant are interchanged,
then the value of the determinant is changed by minus sign.
a1 b1
a2 b2
a3 b3
c1
a2 b2
c2 = - a1 b1
c3
a3 b3
c2
c1
c3
 Applying R2
 R1 
Matrices & Determinants
Properties (Con.)
3.
If all the elements of a row (or column) is multiplied by a
non-zero number k, then the value of the new determinant
is k times the value of the original determinant.
ka1 kb1 kc1
a1 b1
a2 b2 c2 = k a2 b2
a3 b3 c3
a3 b3
c1
c2
c3
which also implies
a1 b1 c1
ma1 mb1 mc1
1
a2 b2 c2 =
a2
b2
c2
m
a3 b3 c3
a3
b3
c3
Matrices & Determinants
Properties (Con.)
4.
If each element of any row (or column) consists of
two or more terms, then the determinant can be
expressed as the sum of two or more determinants.
a1 + x b1 c1
a1 b1 c1
x b1 c1
a2 + y b2 c2 = a2 b2 c2 + y b2 c2
a3 + z b3 c3
a3 b3 c3
z b3 c3
5.
The value of a determinant is unchanged, if any row
(or column) is multiplied by a number and then added
to any other row (or column).
a1 b1 c1
a1 + mb1 - nc1 b1 c1
a2 b2 c2 = a2 + mb2 - nc2 b2 c2
a3 b3 c3
a3 + mb3 - nc3 b3 c3
 Applying C1  C1 + mC2 - nC3 
Matrices & Determinants
Properties (Con.)
6.
If any two rows (or columns) of a determinant are
identical, then its value is zero.
a1 b1
a2 b2
a1 b1
7.
c1
c2 = 0
c1
If each element of a row (or column) of a determinant is zero,
then its value is zero.
0 0 0
a2 b2 c2 = 0
a3 b3 c3
Matrices & Determinants
Properties (Con.)
8
a 0

Let A = 0 b
0 0
a 0
A =0 b
0 0
0

0 be a diagonal matrix, then
c 
0
0  abc
c
Matrices & Determinants
Row(Column) Operations
Following are the notations to evaluate a determinant:
(i) Ri to denote ith row
(ii) Ri  Rj to denote the interchange of ith and jth
rows.
(iii) Ri  Ri + lRj to denote the addition of l times the
elements of jth row to the corresponding elements
of ith row.
(iv) lRi to denote the multiplication of all elements of
ith row by l.
Similar notations can be used to denote column
operations by replacing R with C.
Matrices & Determinants
Evaluation of Determinants
If a determinant becomes zero on putting
x =  , then  x -   is the factor of the determinant.
x
5
2
For example, if Δ = x2
9
4 , then at x = 2
x3 16 8
  0 , because C1 and C2 are identical at x = 2
Hence, (x – 2) is a factor of determinant .
Matrices & Determinants
Sign System for Expansion of
Determinant
Sign System for order 2 and order 3 are
given by
+ – +
+ –
, – + –
– +
+ – +
Matrices & Determinants
Example-1
Find the value of the following determinants
42
(i) 28
1
7
6
4
14
3
2
6
2
(ii)
-10
-3 2
-1 2
5
2
Solution :
i 
42
1
6
6×7
1
6
28
14
7
3
4 = 4×7
2
2×7
7
3
4
2
6 1 6
=7 4 7 4
2 3 2
= 7×0
=0
 Taking out 7 common
from C1 
 C1 and C3 are identical
Matrices & Determinants
Example –1 (ii)
(ii)
6
2
-10
-3 2
-1 2
5
2
3   2  3 2
 1  2 
5   2 
1 2
5
2
3 3 2
 ( 2) 1 1 2
5
 ( 2)  0
5
Taking out  2 common from C1 
2
 C1 and C2 are identical
0
Matrices & Determinants
Example - 2
1
Evaluate the determinant 1
1
Solution :
1 a b+c
a
b
c
b+c
c+a
a+b
1 a a+b+c
1 b c+a = 1 b a+b+c
1 c a+b
1 c a+b+c
 Applying
c3  c2 +c3 
1 a 1
=  a+b+c  1 b 1
1 c 1
 Taking  a+b+c  common from C3 
= a + b + c × 0
 C1 and C3 are identical
=0
Matrices & Determinants
Example - 3
a b
Evaluate the determinant: a2 b2
c
c2
bc ca ab
Solution:
a
b
c
We have a2 b2 c2
bc ca ab
(a-b)
b- c
c
= (a-b)(a+b) (b- c)(b+c) c2
-c(a-b)
-a(b- c)
ab
1
1
c
=(a-b)(b - c) a+b b+c c2
-c
-a ab
Applying C1  C1 -C2
and C2  C2 - C3 
 Taking  a-b  and b - c  common 


from
C
and
C
respectively

1
2

Matrices & Determinants
Solution Cont.
0
1
c
=(a-b)(b - c) -(c - a) b+c c2
-(c - a) -a ab
0
1
 Applying c1  c1 - c2 
c
= -(a-b)(b - c)(c - a) 1 b+c c2
1 -a ab
0
1
c
= -(a-b)(b - c)(c - a) 0 a+b+c c2 - ab
1
-a
ab
Applying
R 2  R 2 -R 3 
Now expanding along C1 , we get
(a-b) (b-c) (c-a) [- (c2 – ab – ac – bc – c2)]
= (a-b) (b-c) (c-a) (ab + bc + ac)
Matrices & Determinants
Example-4
Without expanding the determinant,
3x+y 2x x
3
prove that 4x+3y 3x 3x = x
5x+6y 4x 6x
Solution :
3x+y
2x
L.H.S = 4x+3y
5x+6y
x
3x
4x
3x
2x
3x = 4x
6x
5x
3x
4x
3
2
1
1
2
1
= x3 4
5
3
4
3 + x2 y 3
6
6
3
4
3
6
x
y
3x + 3y
6x
6y
2x
x
3x
4x
3x
6x
3 2 1
= x3 4 3 3 + x2 y×0
5 4 6

C1 and C2 are identical in II determinant 
Matrices & Determinants
Solution Cont.
3 2 1
= x3 4 3 3
5 4 6
1 2 1
 Applying C1  C1 - C2 
= x3 1 3 3
1 4 6
1 2 1
= x3 0 1 2
0 1 3
 Applying R2 R2 -R1
= x3 ×(3-2)
and R 3 R 3 -R 2 
Expanding along C1 
=x3 = R.H.S.
Matrices & Determinants
Example -5
ω3
1
ω5
1
Prove that : ω3
ω5
ω5
ω4 = 0 , where wis cube root of unity.
1
Solution :
1
L.H.S = ω3
ω5
ω3
1
ω5
1
1
ω2
= 1
1
ω
ω2
ω2
1
= 0 = R.H.S.
ω5
1
ω4 = ω3
1
ω3 .ω2
ω3
1
ω3 .ω2
ω3 .ω2
ω3 .ω
1
 ω3 =1

C1 and C2 are identical
Matrices & Determinants
Example-6
x+a
b
c
x+b
c = x2 (x+a+b+c)
Prove that : a
a
b
x+C
Solution :
L.H.S=
x+a
b
c
a
x+b
c
a
b
x+C
x+a+b+c
b
c
= x+a+b+c x+b
x+a+b+c
c
b
x+c
 Applying C1  C1 +C2 +C3 
1
b
=  x+a+b+c  1 x+b
1
b
c
c
x+c
 Taking  x+a+b+c  common from C1 
Matrices & Determinants
Solution cont.
1
b
c
=(x+a+b+c) 0
x
0
0
0
x
 Applying R 2  R 2 -R1 and R 3  R 3 -R1 
Expanding along C1 , we get
(x + a + b + c) [1(x2)] = x2 (x + a + b + c)
= R.H.S
Matrices & Determinants
Example -7
Using properties of determinants, prove that
b+ c c+ a a+b
c+ a a+b b+ c =2(a+b+ c)(ab+bc+ ca- a2 -b2 - c2 ).
a+b b+ c c+ a
Solution :
b+c c+a a+b
L.H.S= c+a a+b b+c
a+b b+c c+a
2(a+b+c) 2(a+b+c) 2(a+b+c)
=
c+a
a+b
a+b
b+c
1
b+c
c+a
1
Applying R1  R1 +R 2 +R 3 
1
=2(a+b+c) c+a a+b b+c
a+b b+c c+a
Matrices & Determinants
Solution Cont.
0
0
1
=2(a+b+c) (c -b) (a- c) b+c
(a- c) (b - a) c+a
Applying C1  C1 - C2 and C2  C2 - C3 
Now expanding along R1 , we get
2(a+b+c) (c -b)(b - a)-(a- c)2 
= 2(a+b + c) bc - b2 - ac + ab - (a2 + c2 - 2ac)
=2(a+b+c) ab+bc+ac-a2 -b2 -c2 
=R.H.S
Matrices & Determinants
Example - 8
Using properties of determinants prove that
x+4 2x
2x x+4
2x
2x
2x
2x =(5x+4)(4- x)2
x+4
Solution :
x+ 4
L.H.S = 2x
2x
1
2x
2x
5x+ 4
2x
2x
x+ 4 2x = 5x+ 4 x+ 4 2x
2x x+ 4 5x+ 4 2x x+ 4
2x
Applying C1  C1 +C2 +C3 
2x
=(5x+ 4) 1 x + 4 2x
1 2x x + 4
Matrices & Determinants
Solution Cont.
1
2x
2x
=(5x+ 4) 0 -(x - 4)
0
0
x - 4 -(x - 4)
Applying R 2  R 2 -R1
and R 3  R 3 -R 2 
Now expanding along C1 , we get
(5x + 4) 1(x - 4)2 - 0 
=(5x+4)(4- x)2
=R.H.S
Matrices & Determinants
Example -9
Using properties of determinants, prove that
x+9 x
x x+9
x
x
x
x =243 (x+3)
x+9
Solution :
x+9
L.H.S=
3x+9
x
x
x
x
x+9
x
x
x+9
x
x
= 3x+9 x+9
x
3x+9
x
x+9
Applying C1  C1 +C2 +C3 
Matrices & Determinants
Solution Cont.
1
x
=(3x+9) 1 x+9
1
x
1
x
x
x+9
x
=3  x+3 0 9 0
0 -9 9
= 3(x +3) 81
x
Applying R 2 R 2 -R1 and R 3  R 3 -R 2 
Expanding along C1 
=243(x +3)
=R.H.S.
Matrices & Determinants
Example -10
(b+ c)2 a2
Show that (c + a)2 b2
(a+b)2 c2
bc
ca = (a2 +b2 + c2 )(a-b)(b - c)(c - a)(a+b+ c)
ab
Solution :
(b+ c)2 a2
L.H.S.= (c+ a)2 b2
(a+b)2 c2
a2 +b2 + c2 a2
 a2 +b2 + c2 b2
a2 +b2 + c2 c2
bc
b2 + c2 a2
ca = c2 + a2 b2
ab
a2 +b2 c2
bc
ca
ab
1 a2
=(a2 +b2 +c2 ) 1 b2
1 c2
 Applying
bc
ca
ab
bc
ca Applying C1  C1 - 2C3 
ab
C1  C1 +C2 
Matrices & Determinants
Solution Cont.
1
a2
bc
=(a2 +b2 +c2) 0 (b- a)(b+a) c(a-b)
0 (c-b)(c+b) a(b- c)
 Applying R 2 R 2 -R 1 and
R 3  R 3 -R 2 
1
a2
bc
=(a2 +b2 +c2 )(a-b)(b- c) 0 -(b+a) c
0 -(b+c) a
=(a2 +b2 +c2 )(a-b)(b- c)(-ab- a2 +bc+c2 )
Expanding along C1 
=(a2 +b2 +c2 )(a-b)(b - c) b  c - a +  c - ac +a 
=(a2 +b2 +c2 )(a-b)(b - c)(c - a)(a+b+c)=R.H.S.
Matrices & Determinants
Applications of Determinants
(Area of a Triangle)
The area of a triangle whose vertices are
(x1, y1), (x2 , y2 ) and (x3, y3 ) is given by the expression
x1
1
Δ = x2
2
x3
y1 1
y2 1
y3 1
1
= [x1 (y2 - y3 ) + x2 (y3 - y1 ) + x3 (y1 - y2 )]
2
Matrices & Determinants
Example
Find the area of a triangle whose
vertices are (-1, 8), (-2, -3) and (3, 2).
Solution :
x1
Area of triangle =
1
x2
2
x3
=
=
y1 1
-1
8
1
1
y2 1 = -2 -3 1
2
y3 1
3 2 1
1
-1(-3 - 2)- 8(-2 - 3)+1(-4+ 9)
2
1
5+ 40+5 = 25 sq.units
2
Matrices & Determinants
Condition of Collinearity of
Three Points
If A (x1 , y1 ), B (x2 , y2 ) and C (x3 , y3 ) are three points,
then A, B, C are collinear
 Area of triangle ABC = 0
x1
1
 x2
2
x3
y1 1
x1
y2 1 = 0  x2
y3 1
x3
y1 1
y2 1 = 0
y3 1
Matrices & Determinants
Example
If the points (x, -2) , (5, 2), (8, 8) are collinear,
find x , using determinants.
Solution :
Since the given points are collinear.
x -2 1
5
8
2
8
1 =0
1
 x 2-8 - -25-8 +1 40-16 =0
 -6x-6+24=0
 6x =18  x =3
Matrices & Determinants
Solution of System of 2 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
a1x +b1y = c1
... i
a2 x +b2 y = c2
...ii
Then x =
D1
D
where D =
, y=
D2
D
a1
b1
a2
b2
provided D  0,
, D1 =
c1
b1
c2
b2
and D2 =
a1
c1
a2
c2
Matrices & Determinants
Cramer’s Rule (Con.)
Note :
1
If D  0,
then the system is consistent and has unique solution.
2 
If D = 0 and D1 = D2 = 0,
then the system is consistent and has infinitely many
solutions.
3
If D = 0 and one of D1, D2  0,
then the system is inconsistent and has no solution.
Matrices & Determinants
Example
Using Cramer's rule , solve the following
system of equations 2x-3y=7, 3x+y=5
Solution :
D=
D1 =
2 -3
= 2+9 =11  0
3 1
7 -3
=7+15=22
5 1
2 7
D2 =
=10-21=-11
3 5
D0
By Cramer's Rule x =
D1 22
D
-11
=
= 2 and y = 2 =
= -1
D 11
D
11
Matrices & Determinants
Solution of System of 3 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
... i
a1x +b1y +c1z = d1
a2 x +b2 y +c2z = d2
... ii
a3x +b3y +c3z = d3
... iii
D
D
D
Then x = 1 , y = 2 , z = 3 provided D  0,
D
D
D
a1
b1
where D = a2
a3
c1
d1
b1
b2
c2 , D1 = d2
b2
b3
c3
b3
a1
b1
and D3 = a2
a3
b2
d2
b3
d3
d3
c1
a1
d1
c1
c2 , D2 = a2
c3
a3
d2
c2
d3
c3
d1
Matrices & Determinants
Cramer’s Rule (Con.)
Note:
(1) If D  0, then the system is consistent and has a unique
solution.
(2) If D=0 and D1 = D2 = D3 = 0, then the system has infinite
solutions or no solution.
(3) If D = 0 and one of D1, D2, D3  0, then the system
is inconsistent and has no solution.
(4) If d1 = d2 = d3 = 0, then the system is called the system of
homogeneous linear equations.
(i) If D  0, then the system has only trivial solution x = y = z = 0.
(ii) If D = 0, then the system has infinite solutions.
Matrices & Determinants
Example
Using Cramer's rule , solve the following
system of equations
5x - y+ 4z = 5
2x + 3y+ 5z = 2
5x - 2y + 6z = -1
Solution :
5 -1 4
D= 2 3 5
5 -2 6
5
-1 4
D1 = 2 3 5
-1 -2 6
= 5(18+10) + 1(12-25)+4(-4 -15)
= 140 –13 –76 =140 - 89
= 51  0
= 5(18+10)+1(12+5)+4(-4 +3)
= 140 +17 –4
= 153
Matrices & Determinants
Solution Cont.
5
5
4
D2 = 2 2 5
5 -1 6
5 -1
5
D3 = 2 3 2
5 -2 -1
= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)
= 85 + 65 – 48 = 150 - 48
= 102
= 5(-3 +4)+1(-2 - 10)+5(-4-15)
= 5 – 12 – 95 = 5 - 107
= - 102
D0
 By Cramer's Rule x =
and z =
D1 153
D
102
=
= 3, y = 2 =
=2
D
51
D
51
D3 -102
=
= -2
D
51
Matrices & Determinants
Example
Solve the following system of homogeneous linear equations:
x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0
Solution:
1

We have D = 1
3
1
-2
6
- 1

1  = 1 10 - 6  - 1 -5 - 3  - 1 6 + 6 
- 5 
= 4 + 8 - 12 = 0
 The system has infinitely many solutions.
Putting z = k, in first two equations, we get
x + y = k, x – 2y = -k
Matrices & Determinants
Solution (Con.)
 By Cramer's rule x =
y=
D2
D
=
1
1
k
-k
1
1
1
-2
=
D1
D
=
k
-k
1
-2
1
1
1
-2
=
-2k + k
k
=
-2 - 1
3
-k - k
2k
=
-2 - 1
3
These values of x, y and z = k satisfy (iii) equation.
x=
k
2k
, y=
, z = k , where k  R
3
3
Matrices & Determinants
Thank you
Matrices & Determinants