chp-1-matrices-determinants1 (2).ppt

Matrices & Determinants
Chapter: 1

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

TYPES OF MATRICES
NAME DESCRIPTION EXAMPLE
Rectangular
matrix
No. of rows is not equal to
no. of columns
Square matrix No. of rows is equal to no. of
columns
Diagonal
matrix
Non-zero element in principal
diagonal and zero in all other
positions
Scalar matrix Diagonal matrix in which all
the elements on principal
diagonal and same








5
0
2
1
2
6

2 1 3
2 0 1
1 2 4




















7
0
0
0
4
0
0
0
2










4
0
0
0
4
0
0
0
4

TYPES OF MATRICES
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 A matrix with all zero
entries

3 2 1 4
 

2
3












1
0
0
1
0 0
0 0







TYPES OF MATRICES
Upper Triangular
matrix
Square matrix
having all the entries
zero below the
principal diagonal
Lower Triangular
matrix
Square matrix
having all the entries
zero above the
principal diagonal










7
0
0
6
4
0
3
5
2










7
3
6
0
4
5
0
0
2

Determinants
If is a square matrix of order 1,
then |A| = | a11 | = a11
ij
A= a
 
 
If is a square matrix of order 2, then
11 12
21 22
a a
A =
a a
 
 
 
|A| = = a11a22 – a21a12
a a
a a
1
1 1
2
2
1 2
2

Example
4 - 3
Evaluate the determinant :
2 5
 
4 - 3
Solution : = 4×5 - 2× -3 = 20 + 6 = 26
2 5

Solution
If A = is a square matrix of order 3, then
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
 
 
 
 
 
[Expanding along first row]
11 12 13
22 23 21 23 21 22
21 22 23 11 12 13
32 33 31 33 31 32
31 32 33
a a a
a a a a a a
| A |= a a a = a - a + a
a a a a a a
a a a
     
11 22 33 32 23 12 21 33 31 23 13 21 32 31 22
= a a a - a a - a a a - a a + a a a - a a
   
11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 22
a a a a a a a a a a a a a a a a a a
     

Example
2 3 - 5
Evaluate the determinant : 7 1 - 2
-3 4 1
 
2 3 - 5
1 - 2 7 - 2 7 1
7 1 - 2 = 2 - 3 + -5
4 1 -3 1 -3 4
-3 4 1
     
= 2 1+ 8 - 3 7 - 6 - 5 28 + 3
= 18 - 3 - 155
= -140
[Expanding along first row]
Solution :

Minors
-1 4
If A = , then
2 3
 
 
 
21 21 22 22
M = Minor of a = 4, M = Minor of a = -1
11 11 12 12
M = Minor of a = 3, M = Minor of a = 2

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
4 7
= =12-14=-2
2 3
M32 = Minor of a32 etc.
4 8
= =0+72=72
-9 0

Cofactors
 i+j
ij ij ij
C = Cofactor of a in A = -1 M ,
ij ij
where M is minor of a in A

Cofactors (Con.)
C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1
0 0
= 0
3 4
C23 = Cofactor of a23 = (–1)2 + 3 M23 =  
4 7
2
2 3
C32 = Cofactor of a32 = (–1)3 + 2M32 = etc.
4 8
- =-72
-9 0
4 7 8
A = -9 0 0
2 3 4
 
 
 
 
 

Value of Determinant in Terms
of Minors and Cofactors
11 12 13
21 22 23
31 32 33
a a a
If A = a a a , then
a a a
 
 
 
 
 
 
3 3
i j
ij ij ij ij
j 1 j 1
A 1 a M a C

 
  
 
i1 i1 i2 i2 i3 i3
= a C +a C +a C , for i=1 or i=2 or i=3

Properties of Determinants
1. The value of a determinant remains unchanged, if its
rows and columns are interchanged.
1 1 1 1 2 3
2 2 2 1 2 3
3 3 3 1 2 3
a b c a a a
a b c = b b b
a b c c c c
i e A A

. . '
2. If any two rows (or columns) of a determinant are interchanged,
then the value of the determinant is changed by minus sign.
 
1 1 1 2 2 2
2 2 2 1 1 1 2 1
3 3 3 3 3 3
a b c a b c
a b c = - a b c R R
a b c a b c
Applying 

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.
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
ka kb kc a b c
a b c = k a b c
a b c a b c
which also implies
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
a b c ma mb mc
1
a b c = a b c
m
a b c a b c

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.
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
a +x b c a b c x b c
a +y b c = a b c + y b c
a +z b c a b c z b c
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).
 
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 1 1 2 3
3 3 3 3 3 3 3 3
a b c a +mb - nc b c
a b c = a +mb - nc b c C C + mC -nC
a b c a +mb - nc b c
Applying 

Properties (Con.)
6. If any two rows (or columns) of a determinant are
identical, then its value is zero.
2 2 2
3 3 3
0 0 0
a b c = 0
a b c
7. If each element of a row (or column) of a determinant is zero,
then its value is zero.
1 1 1
2 2 2
1 1 1
a b c
a b c = 0
a b c

Properties (Con.)
 
a 0 0
8 Let A = 0 b 0 be a diagonal matrix, then
0 0 c
 
 
 
 
 
a 0 0
= 0 b 0
0 0 c
A abc


Row(Column) Operations
Following are the notations to evaluate a determinant:
Similar notations can be used to denote column
operations by replacing R with C.
(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.



Evaluation of Determinants
If a determinant becomes zero on putting
is the factor of the determinant.
 
x = , then x -
 
2
3
x 5 2
For example, if Δ = x 9 4 , then at x =2
x 16 8
, because C1 and C2 are identical at x = 2
Hence, (x – 2) is a factor of determinant .
  0


Sign System for Expansion of
Determinant
Sign System for order 2 and order 3 are
given by
+ – +
+ –
, – + –
– +
+ – +

 
42 1 6 6×7 1 6
i 28 7 4 = 4×7 7 4
14 3 2 2×7 3 2
 
1
6 1 6
=7 4 7 4 Taking out 7 common from C
2 3 2
Example-1
6 -3 2
2 -1 2
-10 5 2
42 1 6
28 7 4
14 3 2
Find the value of the following determinants
(i) (ii)
Solution :
1 3
= 7×0 C and C are identical
= 0
 
 

Example –1 (ii)
6 -3 2
2 -1 2
-10 5 2
(ii)
 
 
 
   
    
 
3 2 3 2
1 2 1 2
5 2 5 2
 
    
 
 
    
 

1
1 2
3 3 2
( 2) 1 1 2 Taking out 2 common from C
5 5 2
( 2) 0 C and C are identical
0

Evaluate the determinant
1 a b+c
1 b c+a
1 c a+b
Solution :
 
3 2 3
1 a b+c 1 a a+b+c
1 b c+a = 1 b a+b+c Applying c c +c
1 c a+b 1 c a+b+c

    3
1 a 1
= a+b+c 1 b 1 Taking a+b+c common from C
1 c 1
 
 
Example - 2
  1 3
= a + b + c × 0 C and C are identical
= 0
 
 

2 2 2
a b c
We have a b c
bc ca ab
 
2
1 1 2 2 2 3
(a-b) b-c c
= (a-b)(a+b) (b-c)(b+c) c Applying C C -C and C C -C
-c(a-b) -a(b-c) ab
 
   
2
1 2
1 1 c
Taking a-b and b-c common
=(a-b)(b-c) a+b b+c c
from C and C respectively
-c -a ab
 
 
 
Example - 3
bc
2 2 2
a b c
a b c
ca ab
Evaluate the determinant:
Solution:

 
2
1 1 2
0 1 c
=(a-b)(b-c) -(c-a) b+c c Applying c c -c
-(c-a) -a ab

2
0 1 c
=-(a-b)(b-c)(c-a) 1 b+c c
1 -a ab
 
2
2 2 3
0 1 c
= -(a-b)(b-c)(c-a) 0 a+b+c c -ab Applying R R -R
1 -a ab

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)
Solution Cont.

Without expanding the determinant,
prove that
3
3x+y 2x x
4x+3y 3x 3x =x
5x+6y 4x 6x
3x+y 2x x 3x 2x x y 2x x
L.H.S= 4x+3y 3x 3x = 4x 3x 3x + 3y 3x 3x
5x+6y 4x 6x 5x 4x 6x 6y 4x 6x
3 2
3 2 1 1 2 1
= x 4 3 3 +x y 3 3 3
5 4 6 6 4 6
Example-4
Solution :
 
3 2
1 2
3 2 1
= x 4 3 3 +x y×0 C and C are identical in II determinant
5 4 6

Solution Cont.
 
3
1 1 2
1 2 1
= x 1 3 3 Applying C C -C
1 4 6

 
3
2 2 1 3 3 2
1 2 1
= x 0 1 2 ApplyingR R -R and R R -R
0 1 3
 
 
3
1
3
= x ×(3-2) Expanding along C
=x = R.H.S.
3
3 2 1
= x 4 3 3
5 4 6

Prove that : = 0 , where wis cube root of unity.
3 5
3 4
5 5
1 ω ω
ω 1 ω
ω ω 1
3 5 3 3 2
3 4 3 3
5 5 3 2 3 2
1 ω ω 1 ω ω .ω
L.H.S = ω 1 ω = ω 1 ω .ω
ω ω 1 ω .ω ω .ω 1
 
2
3
2 2
1 2
1 1 ω
= 1 1 ω ω =1
ω ω 1
=0=R.H.S. C and C are identical
 
 
Example -5
Solution :

Example-6
2
x+a b c
a x+b c =x (x+a+b+c)
a b x+C
Prove that :
 
1 1 2 3
x+a b c x+a+b+c b c
L.H.S= a x+b c = x+a+b+c x+b c
a b x+C x+a+b+c b x+c
Applying C C +C +C

Solution :
 
  1
1 b c
= x+a+b+c 1 x+b c
1 b x+c
Taking x+a+b+c commonfrom C
 
 

Solution cont.
 
2 2 1 3 3 1
1 b c
=(x+a+b+c) 0 x 0
0 0 x
Applying R R -R and R R -R
 
Expanding along C1 , we get
(x + a + b + c) [1(x2)] = x2 (x + a + b + c)
= R.H.S

 
1 1 2 3
2(a+b+c) 2(a+b+c) 2(a+b+c)
= c+a a+b b+c Applying R R +R +R
a+b b+c c+a

1 1 1
=2(a+b+c) c+a a+b b+c
a+b b+c c+a
Example -7
Solution :
Using properties of determinants, prove that
2 2 2
b+c c+a a+b
c+a a+b b+c =2(a+b+c)(ab+bc+ca-a -b -c ).
a+b b+c c+a
b+c c+a a+b
L.H.S= c+a a+b b+c
a+b b+c c+a

 
1 1 2 2 2 3
0 0 1
=2(a+b+c) (c-b) (a-c) b+c Applying C C -C and C C - C
(a-c) (b-a) c+a
 
Now expanding along R1 , we get
2
2(a+b+c) (c-b)(b-a)-(a-c)
 
 
2 2 2
=2(a+b+c) bc-b -ac+ab-(a +c -2ac)
 
 
Solution Cont.
2 2 2
=2(a+b+c) ab+bc+ac-a -b -c
=R.H.S
 
 

Using properties of determinants prove that
2
x+4 2x 2x
2x x+4 2x =(5x+4)(4-x)
2x 2x x+4
Example - 8
1 2x 2x
=(5x+4)1 x+4 2x
1 2x x+4
Solution :
 
1 1 2 3
x+4 2x 2x 5x+4 2x 2x
L.H.S= 2x x+4 2x =5x+4 x+4 2x Applying C C +C +C
2x 2x x+4 5x+4 2x x+4


Solution Cont.
 
2 2 1 3 3 2
1 2x 2x
=(5x+4) 0 -(x- 4) 0 ApplyingR R -R and R R -R
0 x- 4 -(x- 4)
 
Now expanding along C1 , we get
2
(5x+4) 1(x-4) -0
 
 
2
=(5x+4)(4-x)
=R.H.S

Example -9
Using properties of determinants, prove that
x+9 x x
x x+9 x =243 (x+3)
x x x+9
x+9 x x
L.H.S= x x+9 x
x x x+9
 
1 1 2 3
3x+9 x x
= 3x+9 x+9 x Applying C C +C +C
3x+9 x x+9

Solution :

 
1
=3(x+3) 81 Expanding along C
=243(x+3)
=R.H.S.

1 x x
=(3x+9)1 x+9 x
1 x x+9
Solution Cont.
  2 2 1 3 3 2
1 x x
=3 x+3 0 9 0 Applying R R -R and R R -R
0 -9 9
 
 
 

Example -10
Solution :
2 2 2 2 2
2 2 2 2 2
1 1 3
2 2 2 2 2
(b+c) a bc b +c a bc
L.H.S.= (c+a) b ca = c +a b ca Applying C C -2C
(a+b) c ab a +b c ab
 

 
 
2 2 2 2
2 2 2 2
1 1 2
2 2 2 2
a +b +c a bc
a +b +c b ca Applying C C +C
a +b +c c ab
 
2
2 2 2 2
2
1 a bc
=(a +b +c )1 b ca
1 c ab
2 2
2 2 2 2 2
2 2
(b+c) a bc
(c+a) b ca =(a +b +c )(a-b)(b-c)(c-a)(a+b+c)
(a+b) c ab
Show that

Solution Cont.
 
2
2 2 2
2 2 1 3 3 2
1 a bc
=(a +b +c ) 0 (b-a)(b+a) c(a-b) Applying R R -R and R R -R
0 (c-b)(c+b) a(b-c)
 
 
2 2 2 2 2
1
=(a +b +c )(a-b)(b-c)(-ab-a +bc+c ) Expanding along C
2 2 2
=(a +b +c )(a-b)(b-c)(c-a)(a+b+c)=R.H.S.
2
2 2 2
1 a bc
=(a +b +c )(a-b)(b-c) 0 -(b+a) c
0 -(b+c) a
    
2 2 2
=(a +b +c )(a-b)(b-c) b c-a + c-a c+a
 
 

Applications of Determinants
(Area of a Triangle)
The area of a triangle whose vertices are
is given by the expression
1 1 2 2 3 3
(x , y ), (x , y ) and (x , y )
1 1
2 2
3 3
x y 1
1
Δ= x y 1
2
x y 1
1 2 3 2 3 1 3 1 2
1
= [x (y - y )+ x (y - y )+ x (y - y )]
2

Example
Find the area of a triangle whose
vertices are (-1, 8), (-2, -3) and (3, 2).
Solution :
1 1
2 2
3 3
x y 1 -1 8 1
1 1
Area of triangle= x y 1 = -2 -3 1
2 2
x y 1 3 2 1
 
1
= -1(-3-2)-8(-2-3)+1(-4+9)
2
 
1
= 5+40+5 =25 sq.units
2

Condition of Collinearity of
Three Points
If are three points,
then A, B, C are collinear
1 1 2 2 3 3
A (x , y ), B (x , y ) and C (x , y )
1 1 1 1
2 2 2 2
3 3 3 3
Area of triangle ABC = 0
x y 1 x y 1
1
x y 1 = 0 x y 1 = 0
2
x y 1 x y 1

 

If the points (x, -2) , (5, 2), (8, 8) are collinear,
find x , using determinants.
Example
Solution :
x -2 1
5 2 1 =0
8 8 1

      
x 2-8 - -2 5-8 +1 40-16 =0

-6x-6+24=0

6x=18 x=3
 
Since the given points are collinear.

Solution of System of 2 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
 
2 2 2
a x+b y = c ... ii
 
1 1 1
a x+b y = c ... i
1 2
D D
Then x = , y = provided D 0,
D D

1 1 1 1 1 1
1 2
2 2 2 2 2 2
a b c b a c
where D = , D = and D =
a b c b a c

Cramer’s Rule (Con.)
then the system is consistent and has infinitely many
solutions.
  1 2
2 If D = 0 and D = D = 0,
then the system is inconsistent and has no solution.
 
1 If D 0
Note :
,

then the system is consistent and has unique solution.
  1 2
3 If D = 0 and one of D , D 0,


Example
2 -3
D= =2+9=11 0
3 1

1
7 -3
D = =7+15=22
5 1
2
2 7
D = =10-21=-11
3 5
Solution :
1 2
D 0
D D
22 -11
By Cramer's Rule x = = =2 and y = = =-1
D 11 D 11


Using Cramer's rule , solve the following
system of equations 2x-3y=7, 3x+y=5

Solution of System of 3 Linear
Equations (Cramer’s Rule)
Let the system of linear equations be
 
2 2 2 2
a x+b y+c z = d ... ii
 
1 1 1 1
a x+b y+c z = d ... i
 
3 3 3 3
a x+b y+c z = d ... iii
3
1 2 D
D D
Then x = , y = z = provided D 0,
D D D
, 
1 1 1 1 1 1 1 1 1
2 2 2 1 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3
a b c d b c a d c
where D = a b c , D = d b c , D = a d c
a b c d b c a d c
1 1 1
3 2 2 2
3 3 3
a b d
and D = a b d
a b d

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.

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
1
5 -1 4
D = 2 3 5
-1 -2 6
= 5(18+10)+1(12+5)+4(-4 +3)
= 140 +17 –4
= 153
= 5(18+10) + 1(12-25)+4(-4 -15)
= 140 –13 –76 =140 - 89
= 51 0


3
5 -1 5
D = 2 3 2
5 -2 -1
= 5(-3 +4)+1(-2 - 10)+5(-4-15)
= 5 – 12 – 95 = 5 - 107
= - 102
Solution Cont.
1 2
3
D 0
D D
153 102
By Cramer's Rule x= = =3, y= = =2
D 51 D 51
D -102
and z= = =-2
D 51


2
5 5 4
D = 2 2 5
5 -1 6
= 5(12 +5)+5(12 - 25)+ 4(-2 - 10)
= 85 + 65 – 48 = 150 - 48
= 102

Example
Solve the following system of homogeneous linear equations:
x + y – z = 0, x – 2y + z = 0, 3x + 6y + -5z = 0
Solution:
     
1 1 - 1
We have D = 1 - 2 1 = 1 10 - 6 - 1 -5 - 3 - 1 6 + 6
3 6 - 5
= 4 + 8 - 12 = 0
 
 
 
 
 
The systemhas infinitely many solutions.

Putting z = k, in first two equations, we get
x + y = k, x – 2y = -k

Solution (Con.)
1
k 1
D -k - 2 -2k + k k
By Cramer's rule x = = = =
D -2 - 1 3
1 1
1 - 2

2
1 k
D 1 - k -k - k 2k
y = = = =
D -2 - 1 3
1 1
1 - 2
k 2k
x = , y = , z = k , where k R
3 3
 
These values of x, y and z = k satisfy (iii) equation.

Thank you

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