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x y z
z
y
x
z
y
x
Exercise:
- Solve the equations using matrix method x + 3y + 2z = 6, 3x 2y + 5z = 5, 2x 3y + 6z = 7
- Using matrix method, solve the following equations x + 3y + 3z = 12; x + 4y + 4z = 15; x + 3y + 4z = 13
- Using matrix inversion method solves the equations. x + y + z = 3; x + 2y + 3z = 4; x + 4y + 9z = 6
- Using matrix method, solve the simultaneous equation. x + y + z = 6; x y + 2z = 5 ; 2x + y z = 1
- Solve by matrix method the set of equations. x + y + z = 2 ; y + z = 1 ; z + x = 3
- Solve the following equations by matrix inversion method. 3x + y + 2z = 3 , 2x – 3y – z = – 3, x + 2y + z = 4
- Solve the equations by inversion matrix method. 3x + y + 2z = 3, 2x 3y z = 3, x + 2y + z = 4
Partial fraction
Significance : Partial fraction plays a very important role in separation of given expression.
Rational fraction : An expression of the type ( )
Q x
Px , where P(x) and Q(x) are polynomials in
x , is called rational fraction.
e.g 𝑥
(^2) +4𝑥+ 𝑥+2 ,^ is a rational fraction. There are two types of fractions proper fraction and improper fraction.
Proper fraction : In the fraction ( )
Q x
P x , if the degree of the polynomial P(x) is smaller than
the degree of the polynomial Q(x) then the fraction is said to be proper fraction.
e.g (^) 𝑥 (^3) +5𝑥+6𝑥+2 is a proper fraction.
Improper fraction : In the fraction ( )
Q x
P x , if the degree of the polynomial P(x) is greater than
or equal to the degree of the polynomial Q(x) then the fraction is said to be improper fraction
3
x
x eg , 𝑥
(^2) + 𝑥^2 −1 are improper fractions. Improper fraction to Proper fraction: Any improper fraction can be expressed as sum of a polynomial and a proper fraction by division method.
i.e. Improper fraction = Quotient +
Remainder Divisor
= Q +
R
D
Partial Fraction: Every proper fraction
P(x) Q(x) can be expressed as sum or difference of^ simple
fractions. These simple fractions are called as partial fractions of
P(x) Q(x)
e.g.
(x + 2) (x + 1) =^
(x + 2) +^
(x + 1)
Here
(x + 2) ,^
(x + 1) is proper fraction of^
(x + 2) (x + 1)
Methods to find partial fractions:
Depending upon the nature of factors of the denominator there are three cases CASE I: When denominator contains non repeated linear factors: If the denominator contains non-repeated linear factor of the type (ax + b) then for each such
factor there is partial fraction of the type
A
(ax + b)
In general ( )( )( ) ( ) ( ) ( )
x
C
x
B
x
A
x x x
ax b
Where A, B, C etc. are constants to be determined.
SOLVED EXMPLES
- Resolve into partial fractions
x^2 x
Solution:
x^2 x =^
x(x 1)
Let
x(x 1) =^
A
x +^
B
(x 1) … (1)
1 + 1 = C(1) (2)
2 = C(2)
C = 1
Putting the value of A, B, C in equation (1) x^2 + 1 x(x^2 1) =^
x +^
(x + 1) +^
(x 1)
3) Resolve into partial fractions
x + 3 (x – 1) (x + 1) (x + 5)
Solution : Let, x + 3 (x – 1) (x + 1) (x + 5) =^
A
x – 1 +^
B
x + 1 +^
C
x + 5 x + 3 = A(x + 1)(x + 5) + B(x – 1)(x + 5) + C(x – 1)(x + 1) Put x = 1 4 = A(2)(6) 4 = 12 A
A =
Put x = – 1
- 1 + 3 = B(–2)(4) 2 = – 8B
B = –^
Put x = – 5
- 5 + 3 = C(– 6)(– 4) – 2 = 24C
C =
x + 3 (x – 1) (x + 1) (x + 5) =
x – 1 +
x + 1 +
x + 5
EXERCISE:
Resolve into partial fractions
x^2 1 2)^
x^2 x 3)^
x^2 + 3x + 2
2x + 3 x^2 2x 3 5)^
x^2 + 4x + 1 (x 1) (x + 1) (x + 3) 6)^
2x 3 (x^2 1) (2x + 3)
3x 1 (x 4) (2x + 1) (x 1) 8)^
x^2 + 5x + 7 (x 1) (x + 2) (x + 4) 9)^
x + 4 x(x + 1) (x + 2)
Problem Reducible to Case I after Suitable Substitution
- Resolve into partial fractions
tan + 1 (tan + 2) (tan + 3)
Solution: Given
tan + 1 (tan + 2) (tan + 3)
Put tan = t then
t + 1 (t + 2) (t + 3)
Now
t + 1 (t + 2) (t + 3) is a proper fraction. Here factors of denominator are linear and unequal.
Let,
t + 1 (t + 2) (t + 3) =^
A
(t + 2) +^
B
(t + 3) … (1)
t + 1 (t + 2) (t + 3) =
A(t + 3) + B(t + 2) (t + 2) (t + 3) Comparing numerators, t + 1 = A(t + 3) + B(t + 2) … (2) To find A, Put t = 2 in equation (2), we get 2 + 1 = A( 2 + 3) + B(0) 1 = A(1) 1 = A A = 1 To find B, Put t = 3 in equation (2), we get 3 + 1 = A(0) + B( 3 + 2) 2 = B( 1) 2 = B B = 2 Putting the value of A and B in equation (1) t + 1 (t + 2) (t + 3) =^
(t + 2) +^
(t + 3) But t = tan tan + 1 (tan + 2) (tan + 3) =^
(tan + 2) +^
(tan + 3)
- Resolve into partial fractions
ex e2x^ + 4ex^ + 3
ex^ + 1 (ex^ + 2) (ex^ + 3) 4.^
tan (tan + 2) (tan + 3) 5.^
log x (log x 2) (log x 3)
CASE II: When denominator contains repeated linear factors:
( )( )^2 ( )( )( )^2
x
C
x
B
x
A
x x
ax b
Here ( x )is linear non- repeated factor and ( x )^2 is repeated linear factor.
SOLVED EXAMPLES:
- Resolve into partial fractions
(x 1) (x + 2)^2
Solution:
(x 1) (x + 2)^2 =^
A
(x 1) +^
B
(x + 2) +^
C
(x + 2)^2 … (1) 9 (x 1) (x + 2)^2 =
A(x + 2)^2 + B(x 1) (x + 2) + C(x 1) (x 1) (x + 2)^2 Comparing numerators
9 = A(x + 2)^2 + B(x 1) (x + 2) + C(x 1) … (2) To find A, Put x = 1 in equation (2), we get 9 = A(1 + 2)^2 + B(0) + C(0) 9 = A(3)^2 9 = A(9)
A =
9 ^ A = 1
To find C, Put x = 2 in equation (2), we get 9 = A(0) + B(0) + C( 2 1) 9 = C( 3)
C =
3 ^ C =^ ^3
To find B, Put x = 0 in equation (2), we get 9 = A(0 + 2)^2 + B(0 1) (0 + 2) + C(0 1) 9 = A(2)^2 + B( 1) (2) + C( 1) 9 = A(4) + B( 2) + C( 1) 9 = 1(4) + B( 2) 3( 1) 9 = 4 + B( 2) + 3 9 = 7 + B( 2) 9 7 = B( 2) 2 = B( 2)
B =
2 ^ B =^ ^1
Putting the value of A, B, C in equation (1), we get 9 (x 1) (x + 2)^2 =^
(x 1) –^
(x + 2) –^
(x + 2)^2
2 Resolve into partial fractions ( 1 )
x x
x
Solution: 1 ( 1 ) ( 1 )
x
C
x
B
x
A
x x
x
2
2
x x
xx A x B xC
x x
x
2 x 1 x ( x 1 ) A ( x 1 ) B x^2 C 2 Put x = 0 in equation 2 , we get 2 ( 0 ) 1 0 ( 0 1 ) A ( 0 1 ) B ( 0 )^2 C 1 0 ( 1 ) A ( 1 ) B ( 0 )^2 C 1 0 A ( 1 ) B 0 C 1 B Put x = 1 in equation 2 , we get 2 ( 1 ) 1 1 ( 1 1 ) A ( 1 1 ) B ( 1 )^2 C 2 1 1 ( 0 ) A ( 0 ) B 1 C
1 0 A ( 0 ) B 1 C
1 C
Put x= 0 in equation 2 , we get 2 ( 1 ) 1 1 ( 1 1 ) A ( 1 1 ) B ( 1 )^2 C
2 ( 1 ) 1 1 ( 2 ) A ( 2 ) B ( 1 )^2 C
3 2 A 2 B C
3 2 A 2 ( 1 )( 1 )
3 2 A 1
2 A 2
A 1
Put the values of A , B , and C equation 1 , we get
x x x x x
x
- Resolve in to partial fractions ( 1 )( 1 )
^2
x x
x
x x x x x
x
EXERCISE:
Resolve in to partial fraction
(x + 1)^2 (x + 2) 2.^
x^2 2x + 7 (x + 1) (x 1)^2 3.^
2x + 3 x^2 (x 1)
x^2 (x + 1) (x 2)^2 5.^
2x 3 (x + 1) (x^2 1) 6.^
x^2 + x + 1 (x 2) (x^2 4)
3x^2 + 5x (x^2 1) (x + 1) 8.^
x^2 (x + 1) (x + 2)^2 9.^
2x^2 + 5 (x 1)^2 (x 3)
CASE III : When denominator contains non repeated irreducible quadratic factor :
( )(^2 ) ( ) (^2 )
x
Bx C x
A
x x
ax b
Here ( x )is linear non- repeated factor and ( x^2 ) is an irreducible quadratic factor
SOLVED EXAMPLES:
1 Resolve in to partial fraction ( 2 )( 4 )
^2
x x
x
Solution :Let^32 (^2 2)
x
Bx C x
A
x x
x
2
2
x x
x A Bx C x
x x
x
3 x 2 ( x^2 4 ) A ( Bx C )( x 2 ) 2 Put x = - 2 in equation 2 , we get 3 ( 2 ) 2 (( 2 )^2 4 ) A ( B ( 2 ) C )( 2 2 ) 6 2 ( 4 4 ) A ( 2 B C )( 0 )
8 ( 8 ) A
A
Put x = 0 in equation 2 , we get
3 ( 0 ) 2 (( 0 )^2 4 ) A ( B ( 0 ) C )(( 0 ) 2 )
8 ( 8 ) A ( 0 )
0 2 ( 0 4 ) A ( 0 ) C )(( 0 ) 2 )
2 ( 4 ) A C ( 2 )
2 ( 4 )( 1 ) C ( 2 )
2 4 2 C
2 4 2 C
2 2 C
Put x = 1 in equation 2 3 ( 1 ) 2 ( 12 4 ) A ( B ( 1 ) C )( 1 2 ) 3 2 ( 1 4 ) A ( B C )( 1 2 ) 1 ( 5 ) A ( B C )( 3 ) 1 5 ( 1 ) 3 B 3 ( 1 ) 1 5 3 3 B
3 3 B
B 1
Put the values of A , B , C in equation 1
x
x x x x
x
- Resolve into partial fractions
x^2 + 23x (x + 3) (x^2 + 1) Given fraction x^2 + 23x (x + 3) (x^2 + 1) =^
A
(x + 3) +
(Bx + C) (x^2 + 1) … (1)
x^2 + 23x (x + 3) (x^2 + 1) =
A (x^2 + 1) + (Bx + C) (x + 3) (x + 3) (x^2 + 1) Comparing numerator’s x^2 + 23x = A (x^2 + 1) + (Bx + C) (x + 3) …(2) Put x = – 3 in equation (2) (– 3)^2 + 23 (– 3) = A ((– 3)^2 + 1) 9 – 69 = A (9 + 1)
- 60 = A (10) A = – 6 Put x = 0, A = – 6 in equation (2), we get 02 + 23(0) = – 6 (0^2 + 1) + (B (0) + C) (0 + 3) 0 = – 6 (1) + (C) (3)
C 1
Consider
5x x^2 4 =^
5x (x + 2) (x 2)
Let,
5x (x + 2) (x 2) =^
A
(x + 2) +^
B
(x 2) … (2)
5x (x + 2) (x 2) =
A(x 2) + B(x + 2) (x + 2) (x 2) Comparing numerators 5x = A(x 2) + B (x + 2) … (3) To find A, Put x = 2 in equation (3), we get 5( 2) = A( 2 2) + B(0) 10 = A( 4) 10 = A(4)
A =
4 ^ A =
To find B, Put x = 2 in equation (3), we get 5(2) = A(0) + B(2 + 2) 10 = B(4)
B =
4 ^ B =
Putting the value of A, B in equation (2), we get 5x (x^2 4) =^
(x + 2) +^
(x 2) Equation (1) becomes x^3 + x x^2 4 = x +^
2(x + 2) +^
2(x 2)
- Resolve into partial fractions :
x^2 + 1 x^2 1
Solution : Given fraction
x^2 + 1 x^2 1 is a improper fraction. We first convert this improper fraction into proper fraction by division method. Divide numerator by denominator
x + 1 x – 1
2 2 = Q +
R D
x + 1 x – 1
2 2 = 1 +^
2 x – 1
2
Proper fraction
Divisor (D)
x 2 – 1 x + 1 x – 1
2 2
2
1 Quotient (Q)
Remainder (R) .... (1)
Consider,
x^2 1 =^
(x + 1) (x 1)
Let,
(x + 1) (x 1) =^
A
(x + 1) +^
B
(x 1) … (2)
(x + 1) (x 1) =
A(x 1) + B(x + 1) (x + 1) (x 1) Comparing numerators 2 = A(x 1) + B(x + 1) … (3) To find A, Put x = 1 in equation (3), we get 2 = A( 1 1) + B(0) 2 = A( 2)
A =
2 ^ A =^ ^1
To find B, Put x = 1 in equation (3), we get 2 = A(0) + B(1 + 1) 2 = B(2)
B =
B = 1
Putting the value of A, B in equation (2), we get 2 x^2 1 =^
(x + 1) +^
(x 1) Equation (1) becomes x^2 + 1 x^2 1 =^1 ^
(x + 1) +^
(x 1)
