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Mythila Publishers, Puduvayal 41 X std Mathematics Made Easy

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1-=y6-5=y

5= y+6

5= 4+ y+2

(1)in 4=zand2xapplying By

2=x6=3x

8-14=3x

12=8+3x

14=2(4)+3x

(4),in 4=zapplying By

)()(

).....()()(

)(

).....()()(

Þ

Þ

=Þ-=--

=+

=-

=--´

=+-

=+-´

=++

=+-

=++

412345

1423

23

523322

1632

1822422

4142321

92

5

zz

zx

zx

zx

zyx

zyx

zx

zyx

zyx

a) 1432

0111

0421

=+=+-=+-xzzyyx

;;

4

1z4=

z

14=cIf

3

1=y3=

13=bIf

2

12=

12,=aIf

3=b6-=2b-

4-=2b-2

(1),in2=aapplying By

4=c8-=2c-

6-=2c-2

(4),in2=aapplying By

2=a8=4a

14=2c+3a

6-=2c-a(3)+(4)

-(4)---62

2220

402,2(2)+(1)

-(3)---14=2c+3a

-(2)---1-=c-b

-(1)---4-=2b-a

1;

1,

1

=Þ

Þ

=Þ

Þ

Þ

Þ

===

y

xx

c = -a -

c = -b - a +

c = -b + a -

cz

by

ax

b)15

22

4

5

11

3

11

4

1

3

1

4

1

2

1=+-==-+

zyxyxzyx;;

., z , y x

ba

aa

c a

c a

c a

c b a

cb

a

ba

c b a

c

ba

cz

by

ax

213is,That

1(2)in3

1Subs

3

13215,

,(4)in2

1cSubs

2

1c77 – 154c– (5),,(4)Solve

......(5)163032)(

32606

......(4)3415

(3)and(1)inSub.(2)

.......(3)3260315.

15

224

5

......(2);3

.......(1)3436

;4

1

3

1

42

1,

1,

1

===

=Þ=

=Þ=-

=

=Þ=

=+¸

=+

=-

=+-

=+-

=

=-+

=-+

===

c) 1323

27

16

12

=++=+=-zyzyx

;

.14z,18y,10x Therefore,

1442270

(2),in10x ng Substituti

181230

(1),in10x ng.Substituti

1033033)(

484214,(1)

282421)(

156642,(3)

126621,3(2)

)3.....(.783z2y

1323

)2.......(4227

372

27

12

)1.......(123

262

16

12

1323

;27

16

12

===

=Þ=-

=

=Þ=-

=

=Þ=+

=-´

=++

=+´

=-´

=+

=+

=-

=-Þ+=-

=-

=-Þ+=-

=++=+=-

z z

y y

x x

y x

y x

z y

z x

zy

z x

zxzx

y x

yxyx

zyzyx



d) )( zyzy

x +-=+=+=+ 11052102

320

25.=z30,=y35,=x Solutions,

30=y105=y+75

105=y+3(25)

(3),in25=zApplying

35=x 30-=100-2x

30-=4(25)-2x

(4),in25=zApplying

25=z325-=13z-

295=9z+2x

30-=4z-2x (5),-(4)

-(5)---295=9z+2x

315=9z+3y+0x 3,(3)

20-=0z+3y-2x (1),

(4)---30-=4z-2x

10-=4z-3y+0x

20-=0z+3y-2x (2).+(1)

-(3)---105=y+3z

z)+(y-110=5+2z

(2)-----10-=4z-3y

52102

3

(1)-----20-=3y-2x

102

320

Þ

Þ

Þ

´

+=+

+=+

zy

yx

2. Discuss the nature of solutions of the following system of equations a) 1145,2,52 -=+---=+-=-+ zyxzyxzyx

00(4)(5)

)4.......(32

)5......(323),(

936(2)(1)

)3......(1145

)2........(2

)4.......(32(2)(1)

)2.......(2

)1......(.52,

)3......(1145

)2........(2

)1.........(,52

=-

=+

=+¸

=+-

-=+--

-=+-

=++

=+-

=-+

-=+--

-=+-

=-+

y x

y x

y x

zyx

zyx

y x

. z y x

z y x

zyx

zyx

zyx

system has an infinite number of solutions. b) 2122133 =+--=+--=-+ zyxzyxzyx ;;

10),(

4222)4(

)5.......(322),3()1(

)3.......(2

)1.......(133

)4.......(2),2()1(

)2.......(122

)1.......(133

)3.......(2

)2.......(122

)1.......(133

-=-

=-´

=-+

=+--

=-+

=-+

=+--

=-+

=+--

=+--

=-+

zx

zx

zyx

zyx

zx

zyx

zyx

zyx

zyx

zyx

system is inconsistent and has no solution. c) 321252362 =--=+--=-+ zxzyxzyx ;;

solution.manyyIinfinitel

0=0

6-=4z+4z-6-

6-=4z+2z)+2(3-

(4)inx ofvalueApply

2z+3=x (3),From

(4)-----6-=4z+2x -

12=5z+2y3x-

6=z2y+x (2),+(1)

(3)-----3=2zx

(2)-----12=5z+2y3x -

(1)-----6=z2y+x

--

-

-

--

-

d)2

12343132

-=++-=-+-+-=+ zyxzyxxzy ;);(

solution.nohassystem

7=0

9-=10y+4x

2-=10y+4x (5).-2(4)

(5)---9-=10y+4x

1-=2z+4y+6x

8=2z6y+2x (3),+2(2)

(4)---1-=5y+2x

4=z3y+x

3=z+2y+3x (2),+(1)

-(3)-----1-=2z+4y+6x

2

1=z+2y+3x

-(2)-----4=z3y+x

-(1)-----3=z+2y+3x

1)+x 3(=z+2y

\

---

---

-

---

-

e) 27234

=+++

=+

=+

zyxyxxzzy

;



solution.unique3=x

81=27x

81=4z+7x

0=4z-20x (5),+4(4)

(5)---81=4z+7x

81=3z+3y+3x

0=z+3y-4x 3(3),-(1)

-(4)---0=z-5x

0=2z-3y+x

0=z+3y-4x (2),+(1)

(3)---27=z+y+x

-(2)---0=2z-3y+x

3y+3x =2x +2z

23

(1)---0=z+3y-4x

4x+4z=3z+3y

x)+4(z=z)+3(y

34

\

+=

+

+=

+

yxxz

xzzy

3. In an interschool atheletic meet, with 24 individual events, securing a total of 56 points, a first place secures 5 points, second place secures 3 points, and a third place secures 1 point. Having as many third place finishers as first and second place finishers, find how many athletes finished in each place

No. of I, II and III place finishers = x , y, z

12.finishersplacethirdofNo.

8finishersplacesecondofNo.

4finishersplacefirstofNo.

8y12y8

(3)in12z,4x Sub.

4x 82x )(

363y3x ,(3)3

443y5x

12yx

(3),in12zSub.

12z24zz,(1)in(3)Sub.

(3).........zyx

(2).......56z3y5x

(1).......24zyx

56pointsofnumber Total

24eventsofnumber Total

=

=

=

=Þ=+

==

=Þ=-

=+´

=+

=+

=

=Þ=+

=+

=++

=++

=

=

4. Sum of thrice the first number, second number and twice the third number is 5. If thrice the second number is subtracted from sum of first

number and thrice the third we get 2. If the third number is subtracted from the sum of twice the first, thrice the second, we get 1. Find numbers.

3.z2,y.1,x Therefore,

.1x 5623x

(1),in3zand2ySub.

3z77z14

(5),.in2ySub.

2y63y)(

17z10y

(5)7......7z7y)(

33z9y6x ,3(3)

104z2y6x ,2(1)

(4)......17z.10y),(

69z9y3x ,3(2)

52zy3x ,1(1)

(3).......1z3y2x

(2).......23y3zx

......(1).52zy3x

===

=Þ=++

==

=Þ=+-

=

=Þ=+

-=-

=+--

=-+´

=++´

-=--

=+-´

=++´

=-+

=-+

=++

5. Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now ?

511598424

(1)in84,24sub.

841296

(3)in24Sub

241687x )(

(4).........14423

24282,(3)

(4).........14423)(

)2........(780643

6364444,(1)

)3....(.124

)4(4)4(

)2........(780643

654

1

3

1

2

1

)1........(159

533

zfather grand

fatherher

, Vani,ofagePresent

y y

z x

z z

x

x

. z x

zx

. z x

zyx

z yx

zx

xz

zyx

x yz

z yx

z yx

y

x

=Þ=++

==

=Þ=-

=

=Þ=-

=-

=-´

=--

=++

=++´

=-

-=-

=++

=++

=++

=++

=

=

=



∴ Vani' present age = 24 years Father's present age = 51 years Grand father's age = 84 years

6. Sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than fve times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, Find original 3 digit number ?

731.numberdigitReversed

137

7(3)10100(1)

z10y100x nodigit3

7z11z31

11zyx

(1)in3y1,Subx

1x 1192x

(4)in3ySub

38428y,)(

375y22x ,1(5)

12133y22x ,11(4)

(5)...........375y22x ,27by

999135y94x )(

4695z40y499x ,1(2)

104595z95y95x ,95(1)

(4).........113y2x (3),(1)

(3)...........0z-2yx

z2yx Given(3)

(2)...........46-95z-40y499x

46z)y10x (1005x y10z100Given(2)

(1)...........11zyx Given(1),

x10y100znumber Reversed

z10y100x number Rqd

digitsunit'

digits10'

digits100'

=

=

++=

++=

=Þ=++

=++

==

=Þ=+

=

=Þ=-

=+´

=+´

=+¸

=+-

-=-+´

=++´

=++

=+

=+

=+

+++=++

=++

++=

++=

=

=

=

y

z

y

x

7. There are 12 pieces of five, ten and twenty rupee

currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its

value will be increased by ₹20. Find the number of currencies in each sort.

(3).......12520z5y10x

(2).......10520z10y5x

(1).........12zyx

currenciesrupee5ofNo.



=++

=++

=++

=

=

=

z

y

x

2notesrupee20ofNo.

3notesrupee10ofNo.

7notesrupee5ofNo.

7x 1223x

(1)in3y2,zSub

3y96y

(4)in2zSub

2z105z.,)(

279z3y3,(4)

174z3y,1(5)

(5).....174z3y

8520z15y.,)(

12520z5y10x ,1(3)

21040z20y10x 2,(2)

(4)..........93zy

4515z5y.,)(

10520z10y5x ,1(2)

605z5y5x 5,(1)

=

=

=

=Þ=++

==

=Þ=+

=

=Þ=-

=+´

=+´

=+

=+-

=++´

=++´

=+

-=---

=++´

=++´

3.2 GCD and LCM of Polynomials

1. Find the GCD of the polynomials

a) 3552and2 2323 -+-+-+ xxxxxx

3552g(x)

2f(x)23

23

-+-=

+-+=

xxx

xxx

2

223 +-+ xxx 3552 23 -+- xxx

4222 23 +-+ xxx

777 2 x x -+-

−7 ( 12 x + x - ) x

12 x + x - 223 +-+ xxx

xxx +- 23

222 2 +- xx

222 2 +- xx

0

GCD= 12 x + x -

b) 1261263333 2323 ++++++ xxxxxx and

)1(3

3333)(

)22(6

126126)(

23

23

23

23

+++=

+++=

+++=

+++=

xxx

xxxxg

xxx

xxxxf

1

123 +++ xxx 22 23 +++ xxx

123 +++ xxx

12 +x



1+x

12 +x 123 +++ xxx

003 +++ xx

12 +x

12 +x

0

G.C.D = 3( 12 +x )

c) 1821123and4860306 2323 -+--+- xxxxxx

)8105(6

4860306)(

)674(3

1821123)(

23

23

23

23

-+-=

-+-=

-+-=

-+-=

xxx

xxxxg

xxx

xxxxf

1

8105 23 -+- xxx 674 23 -+- xxx

8105 23 -+- xxx

232 +- xx

2-x

232 +- xx 8105 23 -+- xxx

xxx 23 23 +-

882 2 -+- xx

462 2 -+- xx42 -x

= )2(2 -x

1-x

2-x 232 +- xx

xx 22 -

2+- x2+- x

0

GCD coefficients = 3, )2(3GCD -= x

2. Find the GCD of the polynomials

a) 3523 2334 +-+--+ xxxxxx and

)32(325)(

23)(223

34

-+=+-+=

--+=

xxxxxxg

xxxxf

2+x

3523 +-+ xxx 230 324 --++ xxxx

xxxx 35 234 +-+

3452 23 --+ xxx

61022 23 +-+ xxx

963 2 -+ xx )32(3 2 -+= xx

1-x

322 -+ xx 3523 +-+ xxx

xxx 32 23 -+

322 +-- xx

322 +-- xx

0

GCD = 322 -+ xx

b) 11111 234 -+-- xxxx and

1111)(

1)(23

4

-+-=

-=

xxxxg

xxf

11+x

1111 23 -+- xxx 1000 234 -+++ xxxx

xxxx 1111 234 -+-

11111 23 -+- xxx

1211112111 23 -+- xxx

120120 2 +x

= )1(120 2 +x

11-x

12 +x 1111 23 -+- xxx

xxx +- 23 0

1111 2 -- x

1111 2 -- x

0

GCD= 12 +x

c) xxxxxxxx 88144241263 234234 -++--+ and

)842(3

241263)(

)4472(2

88144)(

23

234

23

234

--+=

--+=

-++=

-++=

xxxx

xxxxxg

xxxx

xxxxxf

2

842 23 --+ xxx 4472 23 -++ xxx

16842 23 --+ xxx

12123 2 ++ xx

)44(3 2 ++ xx

2-x

442 ++ xx 842 23 --+ xxx

xxx 44 23 ++

882 2 --- xx

882 2 --- xx

0

GCD = 442 ++ xx3. Find the LCM of the following

a) 4224 48,8 yxyx

23

444

424

4242

243

2424

yx48=

y32=L.C.M

yx32=

yx32222=48

2=

222=8

´´´

´´´

´´´´´´

´´

´´´´

x

yx

yx

yxyx



b) 232 84 yxyx ,

23

3

233

223

22

22

yx8=

y2x32=L.C.M

yx2=

yx3222=y8x

yx2=

yx22=y4x

´´

´´

´´´´

´´

´´´

c) cbaba 2223 129 ,-

cb12a=

c..b.a3.2=L.C.M

c.b?.3.a2=

.cb.a3..2=cb12a

b.a.2=

b.a.3.3-=b9a

22

222

222

22222

232

2323

d) 222 81216 nnmm ,,-

22

224

232

22222

4

n48m=

n.3m2=L.C.M

n.2=8n

nm3.2-=n12m-

m.2=16m

4. Find the LCM of the following

a) 205105 2 -- xx ,

)4(5

)2)(2(5LCM

)2)(2(5205

)2(5105

2

2

-=

+-=

+-=-

-=-

x

xx

xxx

xx

b) 121 24 +-- xxx ,

)1)(1(

)1)(1)(1(LCM

)1)(1(12

)1)(1)(1(1

4

2

2

24

--=

+-+=

--=+-

+-+=-

xx

xxx

xxxx

xxxx

c) 423 22 -+- ppp ,

1)2)(p2)(p+(p=L.C.M

2)-(p2)+(p=4-p

2)1)(p(p232

2

--

--- = p + p

d) 364352 22 --- xxx ,

3)3)(x +1)(x +4(2x =L.C.M

3)+3)(x 4(x =

9)4(x=36-4x

3)1)(x +(2x =35x 2x22

2

-

-

-

---

5. Find the LCM of the following

a) 9327 223 --- xxx ,)(,

)27)(3(

)93)(3)(3(LCM

)3()3(9

)3()3()3(

)93)(3(27

3

2

2

2

23

-+=

+++-=

+-=-

--=-

++-=-

xx

xxxx

xxx

xxx

xxxx

b) 33323 2786432 yxyxxyx --- ,)(,)(

)9y+6xy-(4x3y)-(2x8x=L.C.M

)9y+6xy-(4x3y)-(2x=

)(3y)+(2x)(3y)-((2x)3y)-(2x=

(3y)-(2x)=27y-8x

3y)-8(2x=

3y)]-[2(2x=6y)-(4x

3y)-(2xx=

3y)]-[x(2x=3xy)-(2x

2232

22

22

3333

3

33

22

222

3.3 Relationship between LCM and GCD

1. Find LCM ,GCD. Verify GCD LCMx gxf .)().( =

a) 22 3521 xyyx ,

verified.

735735

)7)(105(=)(35xy21x2y

GCD×LCM=g(x)×f(x)

7xy=

y·x ·7

105

537=L.C.M

7535=g(x)

7321=f(x)

3333

222

22

22

22

22

yx = yx

xy y x

GCD =

y x =

· y · x · ·

· x y · = xy

y · x · y = x

b) )(),( 3322 812 yxyx --

verified.

))((96

y)-4(x)()(24=GCD×LCM

))((96

)(8)(12=g(x)×f(x)

)()(24

))(()(24=L.C.M

y)-4(x=GCD

)y)(-2.2.2(x

)(8=g(x)

))(.(3.2.2

),(12=f(x)

3322

2222

3322

3322

2222

22

22

33

22

yxyx

yxyxyx

yxyx

yxyx

yxyxyx

yxyxyxyx

yxyx

yx

yxyx

yx

--=

+--

--=

--

+--=

++--

+-=

-

-+=

-

c) )(),)(( 111 33 ++- xxx



verified.Hence

1)+(x )y(x

1)+(x )y(x=GCD×LCM

1)+(x )y(x

1)+(x1)+1)(x (x=g(x)×f(x)

yx=

)(y)(x=

)y+)(xy(x=

1)+x-1)(x+x+1)(x+1)(x-(x=L.C.M

1+x =GCD

1)+x -1)(x+(x =

1)+(x=g(x)

1)+1)(x +x +1)(x-(x =

1)+1)(x -(x=f(x)

66

66

66

33

66

2323

3333

22

2

3

2

3

-=

-

-=

-

-

-

-

d) )(),( xyxxyyx ++ 222

verified.iprelationshtheHence

y)+y(xx= y)+y(xx

y)+x(x· y)+xy(x= y)+x(x· y)+(xxy

GCD×LCM=g(x)×f(x)

y)+xy(x=L.C.M

y)+x(x=GCD

y)+x(x=xy)+(x=g(x)

y)+(xxy=)xy+y(x=(f(x)

2222

2

22

2. Find the LCM

a) axaxxax 3327 234 --- isGCDwhose)(,

222

222

2

22

22

33

3a)-(x)9a+3ax-x(x=

3a)-(x

3a)-(x)9a+3ax-3a)(x-(xx=

GCD

g(x)×f(x)=LCM

GCD×LCM=g(x)(x).f

3a)-(xGCD

3a)-(x=g(x)

)9a+3ax-3a)(x-(xx=

)(3a)+x(3a)-3a)(x-(xx=

)(3a)-x(x=f(x)

=

b) 265124 22 -+--+ aaaaa isGCDwhose,

GCD×LCM=g(x)×(x)f

2-aGCD

2)-3)(a-(a=65=g(x)

2)-6)(a+(a=124=f(x)2

2

=

+-

-+

aa

aa

3)-(a2)-6)(a+(a=LCM

2-a

2)-3)(a-(a2)-6)(a+(a=

GCD

g(x)×f(x)=LCM

3. Find the GCD

)2)(1(24x:LCM)23(8),(12 323434 --+-- xxxxxxx

1)-(x 4x=GCD

2)-1)(x-(x24x

2)+3x-(x8x1)-(x12x=

LCM

g(x)×f(x)=GCD

GCD×LCM=g(x)×(x)f

2)-1)(x-(x24x=GCD

2)+3x-(x8x=

)2x+3x-8(x=g(x)

1)-(x12x=

)x-12(x=f(x)

2

3

223

3

22

234

3

34

2. ))((LCM),(),( 2233422433 yxyxyxyyxxyx ++++++

) y+xy-(x=

) y+xy+)(x y+(x

) y+xy+)(x y2+xy-)(x y+(x=

LCM

g(x)×f(x)=GCD

GCD×LCM=g(x)×(x)f

) y+xy+)(x y+(x=LCM

) y+xy+)(x y+xy-(x=

(xy)-) y+(x=

) y+x2y+(x=g(x)

) y+(x=f(x)

22

2233

22233

2233

2222

2222

424

33

4. Given the LCM and GCD of two polynomials p(x) and q(x) find the unknown polynomial

i) LCM : 701110 23 ++- aaa GCD : 7-a

p(x) : 35122 +- aa q(x) = ?

7)-(a2)+(a=

35)+12a-(a

7)-a70)(+11a+10a-(a=

(x)p

GCD×LCM=q(x)

GCD×LCM=q(x)×(x)p

2

23

ii) LCM: )()( 422422 yyxxyx +++

GCD : )( 22 yx -



q(x) = ))(( 2244 yxyxyx +--

` p(x) :?

) y+xy+(x=

xy)-y+(x

) y+xy+)(x y+xy-(x=

xy)-y+(x

(xy)-) y+(x=

xy)-y+)(xy-(x

)y+yx+(x)y-(x=

xy)-y+)(xy-(x

)y-)(xy+yx+(x)y+(x=

q(x)

GCD×LCM=p(x)

22

22

2222

22

2222

2244

422444

2244

22422422

3.4 Rational Expressions

1. Reduce the rational expressions to its lowest form

i) 9

32 -

-

x

x

3

1

33

3

9

32

+=

-+

-=

-

-

x

xx

x

x

x

))((

ii)168

162

2

++

-

xx

x

4

4

44

44

168

162

2

+=

++

-+=

++

-

x

x-

xx

xx

xx

x

))((

))((

iii)xx

x

+

-2

2 1

x

1)-(x

1)+x(x

1)-(x1)+(x

=

=+

-

xx

x2

2 1

iv)44

18112

2

+-

+-

xx

xx

2)-(x

9)-(x

2)-2)(x-(x

2)-9)(x-(x

=

=+-

+-

44

18112

2

xx

xx

v)xxx

xx

98

81923

2

-+

+

1)-9)(x +x(x

9)+9x(x

)98x(

9)+9x(x

98

819223

2

=

-+=

-+

+

xxxxx

xx

1-x

9=

vi)ppp

pp

64242

40323

2

+-

--

4)-(p2p

5+p

4)-(p8)-(p2p

5)+8)(p-(p

32)+12p-2p(p

5)+8)(p-(p2

=

=

=+-

--

ppp

pp

64242

40323

2

2. Find the excluded values, if any

i) x

x

8

10+

0 valueexcludedThus,

when,undefinedis

=

=

=+

0

088

10

x

xx

x

ii)5138

272 ++

+

pp

p

1,8

5

0)1)(58(ifundefinedis)1)(58(

27

--=

=++++

+

p

pppp

p

iii)12 +x

x

valuesexcludedrealnobecan there

01

101

0

2

2

2

¹+

+³+

³

x

x

x

iv)252 -y

y

5and-5 valuesexcluded

or

))((

0rDenominato

))((

=

-==

=+-

=

+-=

-

55

055

55252

yy

yy

yy

y

y

y

v) 652 +- tt

t

3.and2 valuesexcluded

3tor2t

03)-2)(t-(t

=

==

=

=+- 0652 tt

vi)2

862

2

-+

++

xx

xx



1 valueexcluded

0rdenominato

1)-2)(x+(x

4)+2)(x+(x

=

=-

=-

+=

=-+

++

01

1

4

2

862

2

x

x

x

xx

xx

vii)xxx

x

6

2723

3

-+

-

3and20, valueexcluded

,,

2)-3)(x+x(x

0rdenominato

2)-3)(x+x(x

9)+3x+3)(x-(x

6)-x+x(x

)3+x(3)+3)(x-(x

2

2

22

=

=-==

=

=

=

=-+

-

230

0

6

2723

3

xxx

xxx

x

3.5. Operations of Rational Expressions

1. Simplify

i) 52

3 27

9 x

y

y

x´

2

3

yx=

ii) 34

224 1

1 ba

x

x

bx -´

-

ba

xx

ba

xx

x

bx

4

4

34

24

1

11

1

)(

))((

+=

+-´

-=

iii)4

2

2

2

20

6

2

4

y

xz

z

yx´

3

3

5

3

y

x=

iv) t

t

t

t

10

126

84

5 3 -´

-

4

3

10

26

24

5

2

3

t

t

t

t

t

=

-´

-=

)(

)(

v) 3

22

3

12

7

2110

)( -

-+´

-

+-

p

pp

p

pp

3

4

333

34

7

37

-

+=

---

-+´

-

--=

p

p

ppp

pp

p

pp

))()((

))(())((

vi)2032

169

43

42

22

-+

-´

+

+

xx

yx

yx

x

5)-(2x

)(

5)- 4)(2x+(x

))((

yx

yxyx

yx

x

43

4343

43

4

+=

-+´

+

+=

vii)22

22

22

33 2

693 yx

yxyx

yxyx

yx

-

++´

++

-

)(

))((

))((

))((

))((

yx

yxyx

yxyx

yxyx

yxyx

yxyxyx

23

2322

22

+

++=

-+

++´

++

++-=

2. Simplify

i) 4

4

3

714

y

x

y

x¸

33

44

6

7

314

yx

x

y

y

x

=

´=

ii)4

4

4

162

+

-¸

+

-

x

x

x

x

44

4

4

44

+=-

+´

+

+-=

xx

x

x

xx ))((

iii) 145

49

44

2832

2

2

2

--

-¸

++

-+

bb

b

bb

bb

2+b

4-b

7)+7)(b-(b

2)+(b7)-(b

2)+2)(b+(b

))((

=

´+-

=74 bb

iv)2

2

12

6

4

2

y

xx

y

x --¸

+

3

3

23

12

4

2 2

-=

+-´

+=

x

y

xx

y

y

x

))((

v)4113

3118

123

32162

2

2

2

--

++¸

--

--

xx

xx

xx

xx

1

492

11

412

138

413

113

1238

3118

4113

123

3216

2

2

2

2

2

2

-

+-=

+-

--=

++

-+´

-+

-+=

++

--´

--

--=

x

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

xx

))((

))((

))((

))((

))((

))((



vi)50355

56

672

3522

2

2

2

---

++¸

++

++

aa

aa

aa

aa

55)+1)(a+(a

5)+2)(a+5(a-

2)+3)(a+(2a

3)+1)(2a+(a-=´=

vii)tt

tt

t

tt

42

823

3

822122

22

+

-+¸

+-

)(

))((

)())((

123

4

432

22

3

12432

-=

-+

+´

--=

t

tt

tt

t

tt

3. If 422

82

33

432

2

2

2

--

-+=

-

-+=

aa

aay

a

aax , find 22 -yx

9

4

3

2

3

2

2)-4)(a+(a

1)+2)(a-2(a

1)-1)(a+3(a

1)-4)(a+(a

422

82

33

43

22

22

2

2

2

2

=

úû

ùêë

é=ú

û

ùêë

é

=

´=

--

-+¸

-

-+=

-yx

y

x

y

x

aa

aa

a

aa

y

x

4. If a polynomial 1452 --= xxxp )( is divided by

another polynomial q(x) we get 2

7

+

-

x

x, find q(x)

44

222

7

2

7145

2

7

2

2

++=

++=

=+

-´

+

-=

--

+

-=

xx

xxx

xxq

x

x

xq

xx

x

x

xq

xp

))((q(x)

2)+7)(x-(x)(

)(

)(

)(

3.6. Addition and Subtraction

1. Simplify

a)2

1

2

1

-

-+

-

+

x

xx

x

xx )()(

2

2

2

2

)1()1(

22

-=

-

-++=

-

-++=

x

x

x

xxxx

x

xxxx

b)283

412

283

36202

2

2

2

--

+++

--

++

xx

xx

xx

xx

)4)(7(

328

283

41236202

22

+-

+=

--

---++=

xx

x

xx

xxxx

7

8

)4)(7(

)4(8

-=

+-

+=

x

xx

x

c)2

1

3

2

-

-+

+

+

x

x

x

x

)3)(2(

722

)3)(2(

324

)3)(2(

)1)(3()2)(2(

2

22

+-

-+=

+-

-++-=

+-

-+++-=

xx

xx

xx

xxx

xx

xxxx

d)xy

y

yx

x

-+

-

33

22

22

33

33

yxyx

yx

yxyxyx

yx

yx

yx

y

yx

x

++=

-

++-=

-

-=

--

-=

))((

e)4

252

4

212 2

-

+--

-

-+

x

xx

x

xx )())((

4

42

4

)252()232(

4

)252(

4

)232(

22

22

-

-=

-

+----=

-

+--

-

--=

x

x

x

xxxx

x

xx

x

xx

f)1

1

1

42 -

+-

- x

x

x

x

1

1

)1)(1(

)1)(1(

)1)(1(

)12(

)1)(1(

)12(4

)1)(1(

)1)(1(4

1

1

)1)(1(

4

2

2

+

-=

+-

---=

+-

+--=

+-

++-=

+-

++-=

-

+-

+-=

x

x

xx

xx

xx

xx

xx

xxx

xx

xxx

x

x

xx

x

g)2

1

2

32222

23

+-

+

++

xx

xx

)(

22

3

22

223

)2(

12

)2(

)2()32(

+

+=

+

+-++=

x

x

x

xxx

h)158

1

23

1

65

1222 +-

-+-

++- xxxxxx



))()((

)(

))()()((

))((

))()()((

))()()((

))()()((

))(())(())((

))(())(())((

513

9

5123

29

5123

1811

5123

2315856

5123

125351

35

1

12

1

23

1

2

222

---

-=

----

--=

----

+-=

----

-+-+-++-=

----

-----+--=

---

--+

--=

xxx

x

xxxx

xx

xxxx

xx

xxxx

xxxxxx

xxxx

xxxxxx

xxxxxx

2. Which rational expression should be subtracted

from 8

868

2

+

++

x

xx to get

42

32 +- xx

42

1

42

3

42

4

42

3

)42)(2(

)2)(4(

42

3

8

86)(

42

3)(

8

86

2

22

22

28

2

23

2

+-

+=

+--

+-

+=

+--

+-+

++=

+--

+

++=

+-=-

+

++

xx

x

xxxx

x

xxxxx

xx

xxx

xxxp

xxxp

x

xx

3. 22

21

12

12

12

12

BA

B

BAx

xB

x

xA

--

-+

-=

-

+= find,If

14

28

14

144144

12

12

12

12

B)+(A

121

B)-(AB)+(A

B-A

B)-(AB)+(A

2B-B)+(A

B-(A)(

2121

2

2

2

22

22

22

+

+=

-

+-+++=

+

-+

-

+=+

=-

--

=

=

+-

-=

--

-

x

x

x

xxxx

x

x

x

xBA

BA

B

BA

BA

B

BABA

B

BA

28

1421

28

14

B)+(A

1

2

2

22

2

2

+

+=

--

-

+

+=

x

x

BA

B

BA

x

x

4. If ,1

1,

1 A

+=

+=

xB

x

x then prove that

2

222

)1(

)1(2)()(

+

+=

¸

--+

xx

x

BA

BABA

2

2

2

222

2

2

22

22

222222

)1(

)1(2

1

)1(

)1(2)()(

1

1

1

)1(

)1(2

1

1

12

)B+2(A=

)B+2AB-(A+)B+2AB+(A)()(

+

+=

´+

+=

¸

--+

=+

´+

=

+

+=

úúû

ù

êêë

é÷ø

öçè

æ

++÷

ø

öçè

æ

+=

=--+

xx

x

xx

x

BA

BABA

xx

x

x

B

A

x

x

xx

x

BABA

5. Pari needs 4 hours to complete a work. His

friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?

minutes.24hours2=5

12=both bytaken Hrs

12

56

1

4

1=togetherbydonework

6

1=hr1in doneWork

hours6=Yuvan bytaken Hrs4

1=hr1in doneWork

hours 4=paribytaken Hrs

=

+

6. Iniya bought 50 kg of of apples and bananas. She

paid twice as much per kg for the apple as she did for the banana. If Iniya bought ₹ 1800 worth

of apples and ₹ 600 worth bananas, then how many kgs of each fruit did she buy?

20,30503

2xin(1),.sub.

3

2600

900in(3),Sub.

900

getwe(2),in2sub.By

)3......(600

)2....(1800

)1(..........50

,banana&appleofprice

,banana&appleofweight

==Þ=+

=Þ=

=

=

=

=

=+

=

=

yxx

xyy

xq

xq

qp

qy

px

y x

qp

yx



3.7 Square Root of Polynomials

1. Find the square root of the following expressions

a) 201648256 )()()()( dxcxbxax ----

10824

201648

16

256

)()()()(

)()()()(

dxcxbxax

dxcxbxax

----=

----=

b)4124

888

81

121

)()()(

)()()(

cbbacb

cbyxba

---

-++

262

444

4124

888

)()()(9

)()()(11

)()()(81

)()()(121

cbbacb

cbyxba

cbbacb

cbyxba

---

-++=

---

-++

c) 14412

16128

81

144

hgf

cba

726

864

726

864

14412

16128

3

4

9

12

81

144

hgf

cba

hgf

cba

hgf

cba

=

=

d)448

16124

100

400

zyx

zyx

2

64

224

862

448

16124

2

10

20

100

400

x

zy

zyx

zyx

zyx

zyx

=

=


a) 25204 2 ++ xx

|5+2x|=

5)+(2x=

)5+2(2x)(5)+((2x)

2

22=

++= 25204 2 xx

b)

16

1

2

1

21427

2

2

+-

++

xx

xx

÷ø

öçè

æ-

+=

÷ø

öçè

æ-

+=

÷ø

öçè

æ+-

+++=

+-

++

4

1

147

4

1

147

4

1

4

12

214147

16

1

2

1

21427

2

2

22

2

2

2

x

x

x

x

xx

xxx

xx

xx

)(

)(

.

c)36

211

xx++

3

2

3

323

2

36

11

11

112

11

211

x

x

xx

xx

+=

÷ø

öçè

æ+=

++=

++=

..

d) ))()(( 13282137294 222 ----+- xxxxxx

|1)+2)(7x-1)(x-(4x|

1)+(7x1)-2)(4x-1)(x+2)(7x-1)(x-(4x

))()((

=

=

----+-= 13282137294 222 xxxxxx

e) ))()(( 13212316 222 ++-+-+ xxxxxx

|)1)(12)(13(|

)1)(12)(1)(13)(12)(13(

)132)(123)(16( 222

++-=

+++-+-=

++-+-+=

xxx

xxxxxx

xxxxxx

f) ))()(( 23

11

3

424

2

31

6

172 222 ++++++ xxxxxx

6

|2)+3)(x+2)(4x+(3x|

3

3)+2)(4x+(x

2

2)+2)(3x+(x

6

3)+2)(4x+(3x

)()()(

))()((

=

=

++++++=

++++++=

3

6114

2

483

6

61612

23

11

3

424

2

31

6

172

222

222

xxxxxx

xxxxxx

g) ]2)152(5][2)103(15[ 22 ++++++ xxxx

]22)322(3[ 2 +++ xx

)2)(23(

)23(2)23(

22322322)322(3

)2)(15(

)2()2(5

25252)152(5

)15)(23(

)15(2)15(3

2103152)103(15

22

22

22

++=

+++=

+++=+++

++=

+++=

+++=+++

++=

+++=

+++=+++

xx

xxx

xxxxx

xx

xxx

xxxxx

xx

xxx

xxxxx

)2)(15)(23(

)2)(23)(2)(15)(15)(23(

+++

++++++

xxx

xxxxxx


a) 9182424916 22 +-+-+ yxxyyx



334

)334(

3.3.24.3.23.4.23)3()4(

2

222

+-=

+-=

-+-+-+=

yx

yx

yxyxyx

b) 222 16254030249 yzyzxzxyx ++-+-

|5z)+ 4y-(3x|=

5z)+ 4y-(3x=

2.4y.5z2.3x.5z+2.3x.4y(-4y)+(5z)+(3x)=

2

222 ++

++-+-= 222 16254030249 yzyzxzxyx

3.8. Square Root by Division Method

1. Find the square root of

a) 12171664 234 +-+- xxxx

18 2 +- xx28x

4

234

64

12171664

x

xxxx +-+-

xx -21623

23

16

1716

xx

xx

+-

+-

1216 2 +- xx

1216

12162

2

+-

+-

xx

xx

0

12171664 234 +-+- xxxx = 18 2 +- xx

b) 9364212 234 +-+- xxxx

362 +- xx2x

4

234 9364212

x

xxxx +-+-

xx 62 2 -23

23

3612

4212

xx

xx

+-

+-

3122 2 +- xx

9366

93662

2

+-

+-

xx

xx

0

369364212 2234 +-=+-+- xxxxxx

c) 94242837 432 +++- xxxx

372 2 -- xx22x

4

234

4

94237284

x

xxxx +++-

xx 74 2 -23

23

4928

3728

xx

xx

+-

+-

3144 2 -- xx

94212

942122

2

++-

++-

xx

xx

37294237284 2234 --=+++- xxxxxx

d) 1816 24 ++ xx

104 2 ++ xx24x

4

234

16

108016

x

xxxx ++++

xx 08 2 +23

23

00

80

xx

xx

+

+

108 2 ++ xx

108

1082

2

++

++

xx

xx

0

141816 224 +=++ xxx

e) 144216183198121 234 ++-- xxxx

12911 2 ++ xx211x

4

234

121

144216183198121

x

xxxx ++--

xx 922 2 +23

23

81198

183198

xx

xx

+-

--

121811 2 ++ xx

144216264

1442162642

2

++-

++-

xx

xx

0

12911144216183198121 2234 ++=++-- xxxxxx

2. Find the square root of

a)2

2

2

2 93013

204

x

y

x

y

y

x

y

x+-++

Multiply Nr. & Dr by x2y2

x

y

y

x

xy

yxyx

xy

yyxyxx

35

2

352913204 2242234

--=

--=

+++

22 352 yxyx --22x

4

42234

4

913204

x

yyxyxx +++

xyx 54 2 -

223

223

2520

1320

yxyx

yxyx

+

+

22 3104 yxyx --422

422

912

912

yyx

yyx

+

+

b)2

2

2

2 1027

10

x

y

x

y

y

x

y

x+-+-

Multiply Nr. & Dr by x2y2

xy

+ yxyyxy + xx 432234 102710 --

x

y

y

x

xy

yxyx+-=

+-5

5 22



22 5 yxyx +-2x

4

432234 102710

x

+ yxyyxy + xx --

xyx 52 2 -223

223

2510

2710

yxy + x

yxy + x

-

-

22 102 yxyx +-4322

4322

102

102

+ yxyyx

+ yxyyx

-

-

3. Find a and b if the following are perfect squares

a) baxxxx ++++ 234 28129

423 2 ++ xx23x

4

234

9

28129

x

baxxxx ++++

xx 26 2 +23

23

412

2812

xx

xx

+

+

446 2 ++ xx

161624

242

2

++

++

xx

baxx

0

16

016

16

016

=

=-

=

=-

b

b

a

xax

b) abxxxx +++- 234 37124

732 2 +- xx22x

4

234

4

37124

x

abxxxx +++-

xx 34 2 -23

23

912

3712

xx

xx

+-

+-

764 2 +- xx

494228

282

2

+-

++

xx

abxx

0

49

049

42

042

=

=-

-=

=+

a

a

b

xbx

c) 100220361 234 ++++ xxbxax2121110 xx ++

10

100

361220100 432 axbxxx ++++

x1120 +

2

2

121220

361220

xx

xx

+

+

2122220 xx ++432

432

144264240

240

xxx

axbxx

++

++

0

144

0144

264

0264 4433

=

=-

=

=-

a

xax

b

xbx

d) 168 234 +++- bxaxxx

442 +- xx2x

4

234 168

x

bxaxxx +++-

xx 42 2 -23

23

168

8

xx

axx

+-

+-

482 2 +- xx

16328

16)16(2

2

+-

++-

xx

bxxm

0

32

032

032

2

0816

08)16( 22

-=

=+

=+

=

=--

=--

b

b

xbx

m

m

xxm

e) bx

a

xxx+++-

234

1361

Multiply Nr. & Dr by x2

2

4321361

x

bxaxxx +++-

2231 xx +-

1

1

1361 432 bxaxxx +++-

x32 -

2

2

96

136

xx

xx

+-

+-

2262 xx +-432

432

4124

4

xxx

bxaxx

+-

++

0

4

04

12

012 4433

=

=-

-=

=+

b

xbx

a

xax

3.9 Quadratic Equations

1. Find zeroes of the expression 1282 ++ xx

zerosare6and2

01248366)(p

0121642)(p

2,6

02

,06

0)2)(6(

01282

--

=+-=-

=+-=-

-=-=

=+

=+

=++

=++ xx

x

x

xx

xx

2. Determine the quadratic equations, whose sum and product of roots are General form of the quadratic equation

0rootsofproductroots)theof(sum2 =+- xx

a) 149, 01492 =+- xx

b) 209,- 02092 =++ xx

c) 43

5,

01253

043

5

2

2

=+-

=+-

xx

xx

d) 12

3-

-,



022

0)1(2

3

2

2

=-+

=-+÷ø

öçè

æ --

xx

xx

e)2

5

2

7,

-

0572

02

5

2

7

2

2

=++

=+÷ø

öçè

æ --

xx

xx

f)2

1

5

3 --,

05610

02

1

5

3

2

2

=-+

=÷ø

öçè

æ -+÷

ø

öçè

æ --

xx

xx

g) 22 )5(,)2( +-- aa

0)5()2(

0)5()2(222

222

=++-+

=++---

axax

axax

3. Find the sum and product of the roots

a

c

a

b

cbxax

=

-=+

=++

ab

ba

ba

roots,ofproduct

roots,ofsum

,0ofrootsbeandLet 2

a) 06582 =-+ xx

651

65,8

1

8-=

-=-=

-=+ abba

b) 0752 2 =++ xx

2

7,

2

5=

-=+ abba

c) 02832 =-+ xx

281

28,3

1

3-=

-=-=

-=+ abba

d) 032 =+ xx

01

0,3

1

3==-=

-=+ abba

e)2

1013

aa=+

3

10,

3

1

0103

1013

1013

2

2

2

-=

-=+

=-+

=+

=+

abba

aa

aa

a

aa

f) 043 2 =-- yy

3

4,

3

1

3

)1( -==

--=+ abba

g) 02 322 =-- kxkkx

23

22

,1)(

kk

k

k

k-=

-==

--=+ abba

3.10 Solving Quadratic Equations

1. Solve by factorization method

a) 0274 2 =-- xx

2,4

1

02,014

0)2)(14(

-=

=-=+

=-+

x

xx

xx

b) 030192 2 =++ mm

2

5,2

2

5,2

0152,02

0)152)(2(

0)2(15)2(2

--=

-=-=

=+=+

=++

=+++

Roots

mm

mm

mm

mmm

c) 03622 2 =+- xx

2

3

032or032

0)32)((32(

0)32(3)32(2

03662 2

=

=-=-

=--

=---

=+--

x

xx

xx

xxx

xxx

d) 02572 2 =++ xx

2,2

5:Roots

052,02

0)52)(2(

0)2(5)2(2

025522 2

-

=+=+

=++

=+++

=+++

xx

xx

xxx

xxx

e) 04213 24 =+- xx

6,,7areroots

6or7Thus

6or7a

06)()7(

04213

.Let

22

2

2

±±=

==

=

=--

=+-

=

x

xx

aa

a a

ax

f) )5()6(3 2 +=- ppp

2,2

9

0)2)(92(

01852

51832

22

-=

=+-

=--

+=-

p

pp

pp

ppp



g) 23)7( =-aa

9,2are,Roots

0)2)(9(

0187

18)7(

23)7(

2

22

-=

=+-

=--

=-

=-

a

aa

aa

aa

aa

h) 08

12 2 =+- xx

4

1,

4

1are,Roots

0)14)(14(

01816 2

=

=--

=+-

x

xx

xx

i)2

12

1

1=

-+

- x

x

x

x

2,1are,roots

0)1)(2(

01

][5]12[2

2

5)1(

2

222

2

22

-=

=+-

=--

-=+-+

=-

-+

x

xx

xx

xxxxx

xx

xx

2. Solve by completing the square method

a) 012 2 =-- xx

16

1

4

1

2

12

2

1,1

4

2,

4

4

4

3

4

1

16

9

4

1

16

9

4

1

16

1

2

1

16

1

2

2

1

2

02

1

2

2

2

2

2

2

=

=

=

-=Þ

-=

±=

±=-

=÷ø

öçè

æ-

+=+-

=-

=--

b

b

b

xx

x

x

x

xx

xx

xx

b) 04129 2 =+- xx

9

4

3

2

9

122

3

20

3

2

03

2

9

4

9

4

9

4

9

12

9

4

9

12

2

2

2

2

=

=

=

=Þ=-

=÷ø

öçè

æ-

+-=+-

-=-

b

b

b

xx

x

xx

xx

c) 0232 =-- xx

2

9

2

3

32

2

133

2

13

2

3

2

13

2

3

2

13

2

3

2

92

2

93

23

2

2

2

2

=

=

=

±=

±=

±=-

=÷ø

öçè

æ-

+=+-

=-

b

b

b

x

x

x

x

xx

xx

d) 231

75+=

-

+x

x

x

1

13

21

21

4

1

12375

=

=

=

-==

±=

±=-

=

+=+

=

-+=+

22

2

2

2

2

b

1b

22b

,

1)-(x

132-

32-

0=3-2-

0=9-6-3

))((

xx

x

x

xx

xx

xx

xx

xxx

3. Solve by Formula method

a) 0222 =-+ xx

1313

2

322

2

322

2

322

2

842

12

21422

2

4

221

0

022

2

2

2

2

---=

--+-=

±-=

+±-=

--±-=

-±-=

-===

=++

=-+

or

or

)(

))((

,,

with compare

x

x

a

acbbx

cba

cbxax

xx

b) 0252 2 =+- xx

0withcompare

02522

2

=++

=+-

cbxax

xx

a

acbbx

cba

2

4

2,5,2

2 -±-=

=-==



2

1or2

4

35or

4

35

4

35

4

16255

)2(2

)2)(2(4)5()5( 2

=

-+=

±=

-±=

--±--=

x

x

x

c) 0332 2 =-- xx

4

333or

4

333

4

333

4

2493

)2(2

)3)(2(4)3()3(

2

4

3,3,2

0withcompare

0332

2

2

2

2

-+=

±=

+±=

---±--=

-±-=

-=-==

=++

=--

x

x

a

acbbx

cba

cbxax

xx

d) 023203 2 =-- yy

13

236

2620

6

26206

67620

6

27640020

32

23342020

2

4

23203

0

023203

2

2

2

2

-=

-+=

±=

+±=

---±--=

-±-=

-=-==

=++

=--

or

or

)(

))(()()(

,,

with compare

x

x

x

a

acbbx

cba

cbxax

yy

e) 02362 2 =+- ff

0withcompare

023622

2

=++

=+-

cbxax

ff

2

1or2

2

33or

2

33

22

326

22

24366

)2(2

)23)(2(4)6()6(

2

4

23,6,2

2

2

=

-+=

±=

-±=

--±--=

-±-=

=-==

x

x

x

a

acbbx

cba

f) 05523 2 =-+ pp

5or3

5

6

5452or

6

5452

6

5452

6

602052

)3(2

)5)(3(4)52()52(

2

4

5,52,3

0withcompare

2

2

2

-=

--+-=

±-=

+±-=

--±-=

-±-=

-===

=++

p

p

p

a

acbbp

cba

cbxax

g) 01236 222 =-+- )( baayy

6or

6

72

1212or

72

1212

72

14412

72

14414414412

)36(2

))(36(4)12()12(

)(,12,36

0withcompare

2

222

222

22

2

babax

babax

ba

baaa

baaa

bacaba

cbxax

-+=

-+=

±=

+-±=

---±--=

-=-==

=++

h) 0222 =+++- )()( qpxqppqx



p

qp

q

qpx

pq

qpqpqp

pq

qpqpqp

pq

qpqpqp

pq

qpqpqp

pq

pqqpqpqp

pq

qppqqpqpx

qpcqpbpqa

cbxax

++=

-±++=

-±++=

-+±+=

-+±+=

-++±+=

+-+-±+=

+=+-==

=++

,

2

)]())[((

2

)]())[((

2

))(()(

2

][][)(

2

}]4)[(][)(

)(2

))((4])([)(

)(,)(,

0withcompare

2

222

222

2222

22

2

3.11Problems Involving Quadratic Equations

1. Product of Kumaran’s age two years ago and his age four years from now is one more than twice his present age. What is his present age?

years3agepresents'Kumaran

3negative,betcan'Age

3

03)3)((

2182

214)(2)(Given,

years.4)(agehisnow,fromyears4

years.2)(agehisago,yearsTwo

years.KumaranofagePresent

2

=

-¹

±=

=+-

+=-+

+=+-

+=

-=

=

x

x

xx

x xx

xxx

x

x

x

2. A girl is twice as old as her sister. Five years hence, the product of their ages (in years) will be 375. Find their present ages.

years.20=2(10)=girl ofAge

years10sister ofageHence

10,=0=10)-35)(x +(2x

0=350-15x +2x

0=375-25+10x +5x +2x

375=)5+5)(2x +(x

375=agestheir ofProduct

5+2x =girl ofAge

5+x =sisterofagehence,years5

2xgirlofAge,sisterofAge

235

2

2

=

=Þ

==

-xx

x

3. If the difference between a number and its

reciprocal is 5

24 , find the number.

1/5.and5numbersRequired

55

1

05015

0515

05245

5

241

5

241

1reciprocalits

xnumber required

2

2

=

--

-+

--

-

-

=

=

= or xx =

= or x = x

) = ) (xx (

= x x

= x

x

= x

x

x

4. A ladder 17 feet long is leaning against a wall. If the ladder, vertical wall and the floor from the bottom of the wall to the ladder form a right triangle, find the height of the wall where the top of the ladder meets if the distance between bottom of the wall to bottom of the ladder is 7 feet less than the height of the wall?

ft15wall ABofHeight

8ABve,betcan'Height

8(or)15

08)(15(

01207

4914289

7)((17)

BC ABAC

:theoremPythagorasBy

feet7)(BC

ft,17 ACABC,rt.In

feet7)(BC

feetwall ABofHeight

2

22

222

222

=

-¹-

-=

=+-

=--

+-+=

-+=

+=

-=

=D

-=

=

x

x)x

xx

xxx

xx

x

x

x

5. Hypotenuse of right angled triangle is 25 cm Perimeter 56 cm. Find length of smallest side.

cm.7measuresideSmallest

25and247,triangletheofSides

24=731=x 31

7=x and24=x

0=7)24)((

016831

62961625

)-(31+x=theorem,25PythogorasBy

312556=BC

56=ACBCAB

ABcm,25= ACHypotenuse

2

22

222

=

=

--

--

-

-

-=--

++

=

xx

= x + x

x+ x+ = x

x

xx

x



6. A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from two gates should pole be erected?

m.16andm12idistanceRequired

12.=xand16-=x

0=12-xand0=16+x

0=12)-16)(x+(x

0=192- 4x+x

0= 400-16+8x+2x

400= 4+2x(4)+x+x

20= 4)+(x+x

2

2

222

222

=

7. A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains.

0km/hr8=trainexpressofspeedAverage

km/hr60trainpassenger ofspeedAverage

80negative,betcan'speed

60.or 80

0=80)+(x )06(x

0480020

120

4800

1=20)+x(x

])20240[(

120

240240Given,

20

240=trainexpressbytakenTime

240=trainpassenger bytakenTime

Speed

Distance=Time

20+=trainexpressofspeedAverage

trainpassenger ofspeedAverage

km24=coveredDistance

2

2

=

-¹

-=

-

=-

-

-

=

x

x

x + x

= x + x

xx +

= x + x

x +

x

x

x

8. A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.

km/hr45bustheofspeedOriginal

45

0=60)+(x 45)(x

0270015

152700

180

1

15

15

2

1=

15)+x(x

])1590[(

2

1

15

9090

15

90=speedincreasedintakenTime

90=speedoriginal intakenTime

Speed

Distance=Time

15+x =speedIncreased

xbustheofspeedoriginal

km90=coveredDistance

2

2

2

=

-

-

-

-

=

x =

= x + x

x + = x

= x + x

xx +

= x + x

x +

x

9. A garden measuring 12m by 16m is to have a pedestrian pathway that is ‘w’ meters wide installed all the way around so that it increases the total area to 285 m2. What is the width of the pathway?

m.'1.5width,betcan'

2

3,

2

310)32)(312(

0=93-56w+4w

0=285-192+56w+4w

285=4w+32w+24w+192

285=2w)+(16·2w)+(12

m285=Area

2w.+16width

2w+12pathwayincluding

gardenoflength

wwidthpathway

2

2

2

2

=-

-=Þ=-+

=

=þýü

=

vew

www

10. There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. Total cost of laying the flower bed

and gravelling the path at ₹3

, ₹4 per square metre respectively is ₹364. Find the width of the gravel path.



6mbedflower ofLength

636

36444003

364)x(10043x.

./mRs.4pathgravel laying ofCost

./mRs.3bedflower laying ofcost

m)x(100pathgravel theofArea

mxbedflower of Area

mbedflower theofSide

m100fieldsquareofArea

10mfieldsquareoflength

2

22

22

2

2

22

22

2

=

±=Þ=

=+

=-+

=

=

-=

=

=

=

=

x x

x – x

ei

x

11. A flock of swans contained x2 members. As the clouds gathered, 10x went to a lake and one-eighth of the members flew away to a garden. The remaining three pairs played about in the water. How many swans were there in total?

14412=swansofnumber Total

)(7

412

0=)47)(12(

048807

x=6+8

+10.

623=swansofnumber Remaining

x8

1=gardentoflewswansofNo.

10=laketowentswansofNo.

=swansofnumber Total

2

2

22

2

2

=

´-

+-

--

=´

, x = x =

xx

= xx

xxei

x

x

12. From a group of 2x2 black bees , square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?

72=2(6)=beesofnumber Total

)(2

3-=x 6,=x

0=6)-3)(x +(2x

01892

229

16.

2=beesofnumber Remaining

9

16x=2x

9

8=treetoreturnedbeesofNo.

x=x

1=treetowentbeesofNo.

2x=beesofnumber Total

2

2

22

22

2

2

´

--

´

= xx

x = + x

x + ei

13. Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice?

m42=9ofgalleryfromDistance

m284ofgalleyfromDistance

28(or)140d

028)(d140)(d

03920112

0196005605

9456019600

9d)1404900(4

9=d)²4(70

)70(9

4

d-70=9ofgalleryfromDistance

d4ofgalleyfromDistance

ratiodistanceofsq.intensitysoundofRatio

2

2

22

22

=

-=

=-+

=-+

=-+

=+-

=+-

-

-

=

=

dd

dd

ddd

dd

d²

²d

d² =

x

x

14. Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If had your eggs, I would have

earned ₹15”, to which the second replied: “If I

had your eggs, I would have earned ₹3

26 . How

many eggs did each had in the beginning?

40,60=woman1,2ineggsofNo.

0)40)(200(

08000160

4800400009

)100(49(1),From

49

3

2015(2),in,Sub

3

20

3

26

1515Given,

)2.........(

:moneyequal for eggssold

=priceselling their

)1.........(100

100

woman1,2ineggsofNo.

2

22

22

22

=-+

=-+

+-=

-=

=

=

=Þ

=Þ=

=

-=

=+

=

xx

xx

xxx

xx

yx

yx

xy

qp

xqqx =

yppy

qypx

p,q.

xy

yx

x,y



15. A ball rolls down a slope and travels a distance d = t2 −0.75t feet in t seconds. Find the time when the distance traveled by ball is 11.25 feet.

)3(sec753

3,753

030753

03753

02511750

2511750

2511where

750

2

2

2

-¹=

-==

=+=-

=+-

=--

=-

=

-=

t . t

t .

, t .t

) ) (t.(t

. t.t

.t .t

ft. d

t . td

3.12. Nature of Roots

1. Determine the nature of roots

rootReal No

rootsEqual andReal

rootsUnequal andReal

RootsofNature

0

0

0

42

<D

=D

>D

-=D ac b

a) 0202 =-- xx

rootsunequal andReal

081

)20)(1(4)1(

20112

.

, c , b a

>=D

---=D

-=-==

b) 016249 2 =+- xx

rootsEqual andReal

0

)16)(9(4)24(

162492

.

, c , b a

=D

--=D

=-==

c) 0922 2 =+- xx

rootsReal No

068

)9)(2(4)2(

9222

.

, c , b a

<-=D

--=D

=-==

d) 021115 2 =++ xx

UnequalandReal

01

)2)(15(4)11(

211152

, c , b a

>=D

-=D

===

e) 012 =-- xx

UnequalandReal

05

)1)(1(4)1(

1112

, c , b a

>=D

---=D

-=-==

f) 02332 2 =+- tt

RootsReal No

015

)23)(2(4)3(

2332

2

, c , ba

<-=D

--=D

=-==

g) 02269 2 =+- yy

rootsEqual andReal

0

)2)(9(4)26(

2269

2

.

, c , b a

=D

--=D

=-==

h) 016249 22222 =+- dcabcdxxba

rootsEqual andReal

0

)16)(9(4)24(

1624922222

2222

.

dcbaabcd

dcabcd, c, b baa

=D

--=D

=-==

2. Find the value(s) of ‘k’ for which the roots of the following equations are real and equal.

a) 081)48(2 =++- kkx

4,16

1

0)4)(116(

046516

0324166464

0)81)((4)]48[(

04

0.

andareRoots

81)48(

2

2

2

2

=

=--

=+-

=-++

=-+

=-

=D

=+-==

k

kk

k k

k kk

kk

ac b

equalreal

, c k k, b a

b) 012)65( 2 =++- kxxk

2,30)2)(3(

065

024204

0)1)(65(4]2[

04

12)65(

2

2

2

2

=Þ=--

=+-

=+-

=--

=-

==-=

kkk

k k

k k

kk

ac b

k, c , b ka

c) 016)26(2 =+++ xkkx

16)26( , c kk, b a =+==

9

1,10)1)(19(

01109

044036

0)16)((4)]26[(

04

2

2

2

2

=Þ=--

=+-

=+-

=-+

=-

kkk

k k

k k

kk

ac b

3. Find the values of ‘k’ such that quadratic equation has no real roots?



0119 2 =++++ xkxk )()(

.75Thus,

0)7()5(

0352

0(1))94()1(

04

0roots,real No

119

2

2

2

k

kk

kk

k k

ac b

, c k , b k a

<<-

<-+

<--

<+-+

<-

<D

=+=+=

4. i) 02 22222 =+++++ srqsprxqpx )()( has no real

roots - Prove ii) If qrps = , show that the roots real and equal.

equal.andreal bewill rootso0,

])[(4

])[(4

thenqr psIf

equal.notareoots,0As

])[(4

]2[4

]

2[4

)()(4)][2(

4

)(2

0)(2)(

2

2

2

2222

2222

22222222

22222

2

2222

22222

S

qrps

qrps

r

qrps

r qpqrs sp

sqrq

sprpsqpqrsrp

srqpqspr

acb

s r, c qspr, b q pa

srqsprxqpx

=D

--=

--=D

=

<D

--=D

-+-=

--

--++=

++-+=

-=D

+=+=+=

=+++++

5. If the roots of 02 =-+-+- )()()( acxcbxba are

real and equal, prove that b, a, c are in A.P

acb

acb

cba

cba

ac bcabcba

ababcacbccb

ababcacbccb

acbacb

accbba

=+

=+

=--

=--

=-+-++

=-++--+

=+----+

=----

=-

-=-=-=

2

2

02

0)(2

04244

044442

0)(4)2(

0)()(4)(

04ACB

C,B, AHere

2

222

222

222

2

2

b, a, c are in A.P.6. If a, b are real, show that roots of the equation

0962 =--+-- )()()( baxbaxba are real and

unequal

4ACB

)(9c),(6B,A2 -=D

--=+-=-= bababa

unequal.&real areRoots

realareb&a0,)(72

)][2(36

])()[(36

)(36)(36

0b))(a9()(4b)(a36

22

22

22

22

2

>+=

+=

-++=

-++=

=----+=

ba

ba

baba

baba

ba

7. 0)(2)( 2222 =-+--- acbxbcaxabc roots are real

& equal prove: either abccbaa 30 333 =++= or

acbbcaabc

acbxbcaxabc

-=--=-=

=-+---222

2222

C)(2B,A

0)(2)(

abccba

abccba

a

abccbaa

ac abbc aa

bcaacabcbbcacba

acbabcbca

acbabcbca

3

03or

0

0)3(

03

0)()2(

0)()()(

0)()(4)(4

04ACB

333

333

333

3324

233222224

2222

2222

2

=++

=-++

=

=-++

=++-

=+----+

=----

=----

=-

3.13 Relation b/w Roots and Coefficients

1. Write each of the following expression in terms of abba and+

ab

baabba

ab

baba

ab

bbaa

a

b

b

a

baab

baabba

ba

ba

abba

ab

abba

ab

ba

a

b

b

a

)(32)(

)(3

3333iii)

1)(39

1339)13)(13(ii)

11

2)(

33i)

2

22

22

2222

2

22

++-+=

+++=

+++=

++

+

++-=

+--=--

+=+

-+=

+=+

2. The roots of the equation 0572 2 =+- xx are α and β. Without solving for the roots, find

i) ba

11+ ii)

a

b

b

a+ iii)

2

2

2

2

+

++

+

+

a

b

b

a

2

5,

2

7

2

)7(

then,,be0572ofRoots 2

==--

=+

=+-

abba

baxx



6

13

42

72

2

5

82

74

2

52

2

7

4)(2

8)(42)(

422

4444

2

2

2

2iii)

10

29

2

5

2

52

2

7

2)(

ii)

5

7

2

52

7

11i)

2

2

22

2

2

22

=

+÷ø

öçè

æ+

+÷ø

öçè

æ+÷

ø

öçè

æ-÷

ø

öçè

æ

=

+++

+++-+=

+++

+++++=

+

++

+

+

=

÷ø

öçè

æ-÷

ø

öçè

æ

=

-+=

+=+

=

=

+=+

baab

baabba

baab

bbaa

a

b

b

a

ab

abba

ab

ba

a

b

b

a

ab

ba

ba

3. If α,β are roots of 01072 =++ xx find

values of

i) ba - ii) 22 ba + iii) 33 ba -

iv) 44 ba + v)a

b

b

a+ vi)

a

b

b

a 22

+

10,7


=-=+

=++

abba

baxx

641)10(229

)(2)(iv)

117

)3)((10(3)3(

)(3)(iii)

29

)10(2)7(

2)(ii)

3)10(4)7(

4)(i)

22

222244

3

333

2

222

2

2

=-=

-+=+

=

+=

-+-=-

=

--=

-+=+

=--=

-+=-

abbaba

baabbaba

abbaba

abbaba

10

133

10

)7)(10(3343

)(3)(vi)

10

29

10

2049

2)(v)

322

2

-=

---=

+-+=+

=-

=

-+=+

ab

baabba

a

b

b

a

ab

abba

a

b

b

a

4. If α,β are roots of 0273 2 =-+ xx find values of

i)a

b

b

a+ ii)

a

b

b

a 22

+

3

2,

3

7


-=

-=+

=-+

abba

baxx

9

67

3

2

3

2

3

73

3

7

)(3)(

ii)

6

61

3

2

3

22

3

7

2)(

i)

3

3

3322

2

2

22

=-

÷ø

öçè

æ -÷ø

öçè

æ --÷

ø

öçè

æ -

=

+-+=

+=+

-=

-

÷ø

öçè

æ --÷

ø

öçè

æ -

=

-+=

+=+

ab

baabba

ab

ba

a

b

b

a

ab

abba

ab

ba

a

b

b

a

5. If one root of the equation 0642 2 =+- ayy is

twice the other then find the values of a.

322

64,

2


===+

=+-

abba

ba

a

ayy

242

84roots,ofSum

8,416

32)2(roots,ofProduct

(given)2

2

±=Þ=±±

±=±=Þ=

=

=

aa

abb

bb

ba

6. If one root of the equation 0813 2 =++ kxx(having real roots) is the square of other, find k.

273

81,

3


==-

=+

=++

abba

ba

k

kxx



363

93roots,ofSum

9,327

27)(roots,ofProduct

(given)

3

2

2

-=Þ-

=+

==Þ=

=

=

kk

abb

bb

ba

7. If the difference between the roots of the

equation 0132 =+- kxx is 17 find k.

k

kxx

==+

=+-

abba

ba

,13


30)2)(15(

2,15302)2()1(

.....(2),13

.(1)given)....(17

-=Þ=-

=

-==Þ=+

=+

=-

kk

kab

baa

ba

ba

8. If α, β are the roots of 027 2 =++ axx and if

7

13-=-ab . Find the values of a.

7

2,

7=

-=+ abba

a

15or1577

12

7roots,ofsum

2,7

1,

7

1,2

0)2)(17(

02137

7

2

7

137

7

2)

7

13(

7

13

7

13

2

2

-=Þ-

=±±

-=+

-=Þ-

=

=-+

=--

=-

=-

-=Þ-

=-

aa

aba

ba

aa

aa

aa

aa

abab

9. If α and β are roots of 012 2 =-- xx form the equation whose roots are

2

1,

2

1 -==+ abba

i) ba

1,

1 ii) abba 22 , iii) abba ++ 2,2

4

2

1

2

1

1

1rootsofProduct

1

2

12

1

rootsofSum

1,

1EqnrqdofRootsi)

=

÷ø

öçè

æ÷ø

öçè

æ -=

=

-=-

=

+=

=

abab

ba

ba

0rootsofProductroots)ofSum(

:EquationQuadraticofForm2 =+- xx

04EquationRequired 2 =++ xx

8

1

2

1

)(

Product

4

1

2

1

2

1

)(

Sum

,EqnrqdofRootsii)

3

3

2222

22

-=

÷ø

öçè

æ -=

=

´=

-=

-´=

+=

+=

=

ab

abba

baab

abba

abba

0128

08

1

4

1Equation,Required

2

2

=-+

=÷ø

öçè

æ -+÷

ø

öçè

æ --

xx

xx

02

1

2

1

)(2

4)(25

)(25

)2)(2(Product

2

3

2

13

)(3

22Sum

2,2EqnrqdofRootsii)

2

2

22

=-

+=

-+=

-++=

++=

++=

=

÷ø

öçè

æ=

+=

+++=

++=

abba

abbaab

baab

abbaba

abba

abba

032

002

3Equation,Required

2

2

=-

=+÷ø

öçè

æ-

xx

xx

10. The roots of the equation 0462 =-+ xx are α, β. Find the quadratic equation whose roots are

i) 2ba ,2 ii) abba 22 , iii) ba

2,

2

4,6 -=-=+ abba

01644Equation,Required

164

)(

Product

44)4(26

2)(

Sum

,EqnrqdofRootsi)

2

2

2

22

2

2

22

22

=+-

==

=

´=

=---=

-+=

+=

=

xx

ab

ba

abba

ba

ba

06424Eqn,Rqd

64)4(

)(

Product

24)6)(4(

)(

Sum

,EqnrqdofRootsii)

2

3

3

2222

22

=--

-=-=

=

´=

=--=

+=

+=

=

xx

ab

abba

baab

abba

abba

1)4(

4

4Product

34

)6(2

)(2rootsofiii)Sum

-=-

=

=

=-

-=

+=

abab

ba

013Eqn,Rqd 2 =-- xx



3.14. MATRICES

1. Following information regarding No. of men

and women workers in 3 factories I, II and III.

Factory Men Women I 23 18 II 47 36 III 15 16

i) Represent above information in matrix form

÷÷÷

ø

ö

ççç

è

æ

=

1615

3647

1823

A

ii) What does the entry in the second row and

first column represent?

There are 47 men workers in factory II.

2. If a matrix has 16 elements, what are the

possible orders it can have?

1 × 16, 16 × 1

2 × 8, 8 × 2

4 × 4,

3. i) If a matrix has 18 elements, what are the possible orders it can have?

1 × 18, 18 × 1,

2 × 9, 9 × 2,

3 × 6 6 × 3,

ii) What if it has 6 elements? 1 × 6, 6 × 1, 3 × 2, 2 × 3

4. In the matrix, write

úúúúú

û

ù

êêêêê

ë

é

-

-

11186

0341

52

371

3498

(i) The number of elements

A has 4 rows and 4 columns

Number of elements = 4 × 4= 16

(ii) The order of the matrix

4 × 4

(iii) Write the elements

1110

52

3,7

444334

242322

=-==

===

aaa

aaa

5. Construct a 33´ matrix whose elements are

÷÷÷

ø

ö

ççç

è

æ

=´

333231

232221

131211

AFormMatrix 33Rqd

aaa

aaa

aaa

i. 22 jiaij =

÷÷÷

ø

ö

ççç

è

æ

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

81369

36164

941

81

)3()3(

36

)2()3(

9

)1()3(

36

)3()2(

16

)2()2(

4

)1()2(

9

)3()1(

4

)2()1(

1

)1()1(

2233

2232

2231

2223

2222

2221

2213

2212

2211

A

aaa

aaa

aaa

ii. || jiaij 2-=

÷÷÷

ø

ö

ççç

è

æ

=

=

-=

=

-=

=

-==

-=

=

-=

=

-==

-=

=

-=

=

-=

311

420

531

3

)3(23

1

)2(23

1

)1(234

)3(22

2

)2(22

0

)1(225

)3(21

3

)2(21

1

)1(21

333231

232221

131211

A

aaa

aaa

aaa

iii.3

)( 3jiaij

+=

÷÷÷÷÷÷

ø

ö

çççççç

è

æ

=

=

+=

=

+=

=

+=

=

+=

=

+=

==

+=

=

+=

==

+=

=

+=

723

125

3

643

125

3

649

3

649

3

8

72

3

)33(

3

125

3

)23(

3

64

3

)13(3

125

3

)32(

3

64

3

)22(

93

27

3

)12(3

64

3

)31(

93

27

3

)21(

3

8

3

)11(

3

33

3

32

3

31

3

23

3

22

3

21

3

13

3

12

3

11

A

aaa

aaa

aaa

6. Find the transpose of A.

i.÷÷÷

ø

ö

ççç

è

æ

-=

283

971

345

A÷÷÷

ø

ö

ççç

è

æ

-=Þ

293

874

315TA

ii.÷÷÷

ø

ö

ççç

è

æ

-

-

=

574

982

135

A÷÷÷

ø

ö

ççç

è

æ

-

-

=Þ

591

783

425TA



7. Find the transpose of -A.

i. ÷÷ø

öççè

æ

--

-=

135

942A

÷÷÷

ø

ö

ççç

è

æ

-

--

=-Þ÷÷ø

öççè

æ

-

--=-

19

34

52

135

942 TAA

ii.

÷÷÷÷

ø

ö

çççç

è

æ

-

-

-

=

53

25

37

A

÷÷

ø

ö

çç

è

æ

-

--=-Þ

÷÷÷÷

ø

ö

çççç

è

æ

-

-

-

=-523

357

53

25

37T

AA

8. Verify AATT =

i.

÷÷÷÷

ø

ö

çççç

è

æ

-=

1382

57.017

225

A

AA

AA

TT

TTT

=

÷÷÷÷

ø

ö

çççç

è

æ

-=Þ

÷÷÷÷

ø

ö

çççç

è

æ-

=

1382

57.017

225

,

12

52

37.02

8175

ii.÷÷÷

ø

ö

ççç

è

æ

=

34

98

51

A

AA

AA

TT

TTT

=

÷÷÷

ø

ö

ççç

è

æ

=Þ÷÷ø

öççè

æ=

34

98

51

395

481

9. Find the values of x, y and z from equations

i. ÷÷ø

öççè

æ=

÷÷

ø

ö

çç

è

æ

532

3312 zy

x

equal.areelementsing Correspond

equal.arematricesgivenIf

3,3,12

Elements,Corres.Equating

=== xzy

ii. ÷÷ø

öççè

æ=÷÷

ø

öççè

æ

+

+

85

26

5

2

xyz

yx

Elements,Corres.Equating

)2.....(8

8

)1....(.6

0.....55

yxxy

yx

zz

=Þ=

=+

=Þ=+

4or 224

0)2)(4(

086

68

(1),in2)(Sub.

2

=Þ=

=--

=-+-

=+

xory

yy

yy

yy

iii.÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

+

+

++

7

5

9

zy

zx

zyx

474(1),in''ofvalueSub.

39z(1)5in2)(Sub.

297(1),in3)(Sub.

)3..(..........7

)2..(..........5

)1..(..........9

=Þ=+

=Þ=+

=Þ=+

=+

=+

=++

yyz

z

xx

zy

zx

zyx

10. Find the value of a, b, c, d from the equation

÷÷ø

öççè

æ=÷÷

ø

öççè

æ

+-

+-

20

51

32

2

dcba

caba

19getwe),4(in7Sub

7getwe),3(in1Sub

2getwe),2(in1Sub

,1gives)2()1(

)4.........(23

)3.........(52

)2.........(02

)1.........(1

-==

=-=

-=-=

-=-

=+

=+

=-

=-

dc

ca

ba

a

dc

ca

ba

ba

3.17. Operations on Matrices

1. Find A+B.

a)÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

=

042

131

071

987

654

321

BA ,

÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

+++

+++

+++

=

÷÷÷

ø

ö

ççç

è

æ

+÷÷÷

ø

ö

ççç

è

æ

=+

9129

785

392

094827

163514

037211

042

131

071

,

987

654

321

BA

b)÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

=

18524781

80171218

40153820

40328053

30216250

23141522

BA ,

÷÷÷

ø

ö

ççç

è

æ

=

÷÷÷

ø

ö

ççç

è

æ

+÷÷÷

ø

ö

ççç

è

æ

=+

5884127134

110387468

63295342

18524781

80171218

40153820

40328053

30216250

23141522

BA



c)÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

-

-

-

=

69

43

81

923

645

231

BA ,

It is not possible to add A and B Because they have different orders

2. If ÷÷÷

ø

ö

ççç

è

æ

-

=

38

43

91

A Verify that

0)()( =+-=-+ AAAA

0)()(

00

00

00

38

43

91

38

43

91

)(

00

00

00

38

43

91

38

43

91

)(

=+-=-+

÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

-

+÷÷÷

ø

ö

ççç

è

æ

-

--

--

=+-

÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

-

--

--

+÷÷÷

ø

ö

ççç

è

æ

-

=-+

AAAA

AA

AA

3. verify that ABBA +=+ if

÷÷÷

ø

ö

ççç

è

æ

=÷÷÷

ø

ö

ççç

è

æ

-

=

01

33

75

,

38

43

91

BA

ABBA

AB

BA

+=+

÷÷÷

ø

ö

ççç

è

æ

-

=÷÷÷

ø

ö

ççç

è

æ

-

+÷÷÷

ø

ö

ççç

è

æ

=+

÷÷÷

ø

ö

ççç

è

æ

-

=÷÷÷

ø

ö

ççç

è

æ

+÷÷÷

ø

ö

ççç

è

æ

-

=+

39

76

166

38

43

91

01

33

75

39

76

166

01

33

75

38

43

91

4. Verify that CBACBA ++=++ )()( if

÷÷÷

ø

ö

ççç

è

æ

-

-=÷÷÷

ø

ö

ççç

è

æ

--

=÷÷÷

ø

ö

ççç

è

æ

-

-=

142

321

438

117

291

432

401

132

134

CBA ,,

÷÷÷

ø

ö

ççç

è

æ

--

=++

÷÷÷

ø

ö

ççç

è

æ

--

=+

654

4104

9914

)(

255

572

8610

CBA

CB

÷÷÷

ø

ö

ççç

è

æ

--

=+

516

1123

566

BA

CBACBA

CBA

++=++

÷÷÷

ø

ö

ççç

è

æ

--

=++

)()(

654

4104

9914

)(

5. Find 2A+B

if÷÷÷

ø

ö

ççç

è

æ

-

-

=÷÷÷

ø

ö

ççç

è

æ

--

=

057

421

3114

134

931

687

BA ,

÷÷÷

ø

ö

ççç

è

æ

--

=

÷÷÷

ø

ö

ççç

è

æ

-

-

+÷÷÷

ø

ö

ççç

è

æ

--

=

÷÷÷

ø

ö

ççç

è

æ

-

-

+÷÷÷

ø

ö

ççç

è

æ

--

=+

2111

2281

92718

057

421

3114

268

1862

121614

057

421

3114

134

931

687

22 BA

6. Find 4A - 3B

÷÷÷÷

ø

ö

çççç

è

æ

-

--

=

÷÷÷÷

ø

ö

çççç

è

æ -

=

965

32

7

4

1347

491

24

3

2

1245

BA ,

÷÷÷÷

ø

ö

çççç

è

æ

--

--

=

÷÷÷÷

ø

ö

çççç

è

æ

--

----

+÷÷÷

ø

ö

ççç

è

æ -

=

÷÷÷÷

ø

ö

çççç

è

æ

-

--

-

÷÷÷÷

ø

ö

çççç

è

æ -

=-

115411

9242

15

4

51441

271815

92

21

4

391221

16364

2432

8620

965

32

7

4

1347

3

491

24

3

2

1245

434 BA

7. Find the value of (i) B -5A (ii) 3A - 9B if

÷÷ø

öççè

æ=÷÷

ø

öççè

æ=

941

837

738

940BA ,

÷÷ø

öççè

æ

--

---=

÷÷ø

öççè

æ-÷÷

ø

öççè

æ=-

÷÷ø

öççè

æ

---

--=

÷÷ø

öççè

æ

---

--+÷÷

ø

öççè

æ=

÷÷ø

öççè

æ-÷÷

ø

öççè

æ=-

602715

456563

941

8379

738

94039Bii)3A

261139

37177

351540

45200

941

837

738

9405

941

8375i) AB



8. Compute BAiiCBAi2

3

2

123 --+ ))

÷÷÷

ø

ö

ççç

è

æ

--=÷÷÷

ø

ö

ççç

è

æ

-

--

=÷÷÷

ø

ö

ççç

è

æ

=

341

271

035

,

510

3112

468

,

678

053

381

CBA

CBA -+ 23i)

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ -

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

-=

---

--

+-

-

+=

----

--

+=

251923

84414

11314

341

271

035

1020

6224

8

182124

0159

928

341

271

035

510

3112

468

2

678

053

381

3

12163

÷÷÷÷÷÷

ø

ö

çççççç

è

æ

-

--

-

=

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ

--

--

-

+÷÷÷

ø

ö

ççç

è

æ

=

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ

-

--

-÷÷÷

ø

ö

ççç

è

æ

=

-=-

2

924

2

914

2

32

1513

2

23

1530

9336

121824

678

053

381

2

1

510

3112

468

3

678

053

381

2

1

]3[2

1

2

3

2

1ii) BABA

9. Find X and Y if ÷÷ø

öççè

æ=-÷÷

ø

öççè

æ=+

40

03

53

07YXYX ,

÷÷

ø

ö

çç

è

æ=Þ÷÷

ø

öççè

æ=+

÷÷

ø

ö

çç

è

æ=Þ÷÷

ø

öççè

æ=+

÷÷ø

öççè

æ=-

÷÷ø

öççè

æ=+

2

1

2

302

13

042,)2()1(

2

9

2

305

93

0102,)2()1(

)2......(40

03

)1.......(,53

07

YY

XX

YX

YX

10. Find the value of a, b, c, d, x, y from the following matrix equation

÷÷ø

öççè

æ

-+÷÷

ø

öççè

æ=÷÷

ø

öççè

æ

--+÷÷

ø

öççè

æ

05

10

4

22

42

3

3

8

cb

aa

ab

d

:ElementsCorres.Equate

45

122

423

83÷÷ø

öççè

æ

-

+=÷÷

ø

öççè

æ

--

++

cb

a

ab

ad

4

3

447

44

2

3

523

7

128

1

23

=

=-

=--

=

-=-

=

+=+

-=

=+

c

c

ca

b

bb

a

aa

d

d

11. Find the values of x, y, z if

÷÷ø

öççè

æ=÷÷

ø

öççè

æ

++++

--

61

01

7

33

zyxyx

zxx

10

64

6

17

12

012

03

4

13

-=

-=+

-=+

=++

=

=-

=-

=

=-

y

y

y x

y x

z

z

z x

x

x

12. Find the values of x, y, z if 1684343 =++- yzzyx

10

166

14y

410

4

84

10

414

4

z

z

y

z y

z y

x

x

y x

=

=+

=

=-

=-

=+-

-=

=+

=+

13. Find values of x, y if ÷÷ø

öççè

æ=÷÷

ø

öççè

æ-+÷÷

ø

öççè

æ

- 6

4

3

2

3

4yx

6

24

(2)in4Sub

4)(

23)2(

22,2)1(

)2.........(633

)1(..........424

6

4

33

24

6

4

3

2

3

4

y

y

x

x

yx

y x

yx

y x

y x

y x

y

y

x

x

=

=+-

=

=+

=+-¸

=-¸

=+-

=-

÷÷ø

öççè

æ=÷÷

ø

öççè

æ

+-

-

÷÷ø

öççè

æ=÷÷

ø

öççè

æ-+÷÷

ø

öççè

æ

-

14. Solve for x, y ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ

-

-+÷

÷ø

öççè

æ

8

522

2

2

y

x

y

x

2,4

0)2)(4(

082

1,5

0)1)(5(

054

8

5

2

4

22

2

2

-=

=+-

=--

-=

=+-

=--

÷ø

öçè

æ=÷÷

ø

öççè

æ

-

-

y

yy

yy

x

xx

xx

yy

xx

15. Find the non-zero values of x



÷÷ø

öççè

æ +=÷÷

ø

öççè

æ+÷÷

ø

öççè

æ

x

x

x

x

x

xx

610

2482

44

582

3

22 2

4of,valuezeroNon

42083

4,00)4(

128

44812

1220

48162

883

12162

610

2482

88

1016

3

22

2

2

2

2

2

2

2

=

=Þ=+

=Þ=-

=+

=Þ=

÷÷ø

öççè

æ +=÷

÷ø

öççè

æ

++

+

÷÷ø

öççè

æ +=÷÷

ø

öççè

æ+÷

÷ø

öççè

æ

x

xx

xxx

xxx

xx

x

x

xxx

xx

x

x

x

x

xx

xx

3.18.Multiplication Of Matrices

1. Find order of the product matrix AB if

(i) (ii) (iii) (iv) (v) Orders of A 3 × 3 4 × 3 4 × 2 4 × 5 1 × 1

Orders of B 3 × 3 3 × 2 2 × 2 5 × 1 1 × 3

i) A 3 × 3 B3 × 3

∴ Order of AB = 3 × 3

ii) A 4 × 3 B 3 × 2


iii) A4 × 2 B 2 × 2


iv) A 4 × 5 B 5 × 1


v) A 1 × 1 B 1 × 3


2. If A is of order qp ´ and B is of order

rq ´ what is the order of AB and BA?

A p × q B q × r

∴ Order of AB = p × r

B q × r A p × q

No. of columns in B ¹ No. of rows in A

∴ Order of BA is not defined

3. A has ‘a’ rows and 3a + columns. B has‘b’

rows and –b17 columns, and if both

products AB, BA exist, find a, b?

10b7,

10b7b17

(1)in7aSub

7a142a(2),(1)

(2)........17ba

(1)........3ba

(2).......ab17

AinrowsofNo.BincolumnsofNo.

(1)b........3a

inrowsofNo. AincolumnsofNo.

exists,BA,AB

b)(17bB

3)(aaA

==

=Þ=-

=

=Þ=+

=+

-=-

=-

=

=+

=

-´+´

a

B

4. Find AB ,If ÷÷÷

ø

ö

ççç

è

æ

=÷÷ø

öççè

æ=

135

142

138

,513

021BA

÷ø

öçè

æ=

÷ø

öçè

æ

++++++

++++++=

÷ø

öçè

æ=

úúúúúúú

û

ù

êêêêêêê

ë

é

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

÷÷÷

ø

ö

ççç

è

æ

=

÷÷÷

ø

ö

ççç

è

æ

÷ø

öçè

æ=

++++++

++++++

91851

31112

51354925224

021083048

1

1

1

513

1

4

3

513

5

2

8

513

1

1

1

021

1

4

3

021

5

2

8

021

135

142

138

513

021

)1(5)1(1)1(31)4(1)3(3)5(5)2(1)8(3

0)1(2)1(10)4(2)3(10)2(2)8(1

AB

5. Show that BA, satisfy commutative

property w.r. to matrix multiplication (or) Find AB, BA . Check if BAAB = ,if

i) ÷÷ø

öççè

æ=÷÷

ø

öççè

æ=

31

02,

31

12BA

BAAB

BA

BA

AB

AB

¹\

÷ø

öçè

æ=

÷ø

öçè

æ++

++=

÷ø

öçè

æ÷ø

öçè

æ=

÷ø

öçè

æ=

÷ø

öçè

æ++

++=

÷ø

öçè

æ÷ø

öçè

æ=

105

24

9132

0204

31

12

31

02

95

35

9032

3014

31

02

31

12



ii) ÷÷ø

ö

ççè

æ

-=

÷÷ø

ö

ççè

æ -=

22

222,

22

222BA

BAAB

BA

AB

=\

÷ø

öçè

æ=÷÷

ø

öççè

æ

++-

+-+=

÷÷ø

öççè

æ

-÷÷ø

öççè

æ -=

÷ø

öçè

æ=÷÷

ø

öççè

æ

+-

-+=

÷÷ø

öççè

æ

-÷÷ø

öççè

æ -=

80

08

442222

242444

22

222

22

222

80

08

442222

242444

22

222

22

222

iii) ÷÷ø

öççè

æ -=÷÷

ø

öççè

æ=

52

31,

34

52BA

BAAB

BAAB

¹\

÷ø

öçè

æ --=

÷ø

öçè

æ++

--=

÷ø

öçè

æ÷ø

öçè

æ -=

÷ø

öçè

æ=

÷ø

öçè

æ+-+

+-+=

÷ø

öçè

æ -÷ø

öçè

æ=

2524

410

1510204

95122

34

52

52

31

310

1912

151264

256102

52

31

34

52

iv) ÷÷ø

öççè

æ

-

-=÷÷

ø

öççè

æ=

13

21,

13

21BA

true.ispropertyeCommutativ

BA AB

50

05

1633

2261

13

21

13

21

50

05

1633

2261

13

21

13

21

=\

÷ø

öçè

æ-

-=

÷ø

öçè

æ+-+-

+--=

÷ø

öçè

æ÷ø

öçè

æ-

-=

÷ø

öçè

æ-

-=

÷ø

öçè

æ+--

+--=

÷ø

öçè

æ-

-÷ø

öçè

æ= BAAB

6. Solve ÷÷ø

öççè

æ=÷÷

ø

öççè

æ÷÷ø

öççè

æ

5

4

21

12

y

x

1x(1)in2Sub

263)(

522)2(

.421)1(

)2.....(52

)1......(42

5

4

2

2

=Þ=

=Þ-=--

=+´

=+´

=+

=+

÷ø

öçè

æ=÷

ø

öçè

æ

+

+

y

yy

yx

yx

yx

yx

yx

yx

7. Show that A(BC)(AB)C = , If

i) ÷÷ø

öççè

æ

-=

÷÷÷

ø

ö

ççç

è

æ -

=-=12

21,

31

12

11

,111 CBA

)()(

29

1

3

3

111

7

4

1

111

17

34

31

111)(

17

34

31

1

2)31(

2

1)31(

1

2)12(

2

1)12(

1

2)11(

2

1)11(

12

21

31

12

11

)29(

1

241

2

141

12

2141)(

41

611221

3

1

1

111

1

2

1

111

31

12

11

111

BCACAB

BCA

BC

CAB

AB

AB

=

-=

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ-

÷÷÷

ø

ö

ççç

è

æ -

-=

÷÷÷

ø

ö

ççç

è

æ

-

-

-=

÷÷÷

ø

ö

ççç

è

æ

-

-

=

úúúúúúúú

û

ù

êêêêêêêê

ë

é

÷ø

öçè

æ-

÷ø

öçè

æ

÷ø

öçè

æ

-÷ø

öçè

æ

÷ø

öçè

æ

--÷

ø

öçè

æ-

=

÷ø

öçè

æ-÷

÷÷

ø

ö

ççç

è

æ -

=

-=

úû

ùêë

é÷ø

öçè

æ

-÷ø

öçè

æ=

÷ø

öçè

æ

-=

=

+--+-=

úúú

û

ù

êêê

ë

é

÷÷÷

ø

ö

ççç

è

æ-

-÷÷÷

ø

ö

ççç

è

æ

-=

÷÷÷

ø

ö

ççç

è

æ -

-=

ii) ÷÷ø

öççè

æ=÷÷

ø

öççè

æ=÷÷

ø

öççè

æ=

21

02,

51

04,

31

21CBA

(AB)CA(BC)

3029

2022

107

08

31

21A(BC)

107

08

21

02

51

04BC

3029

2022

21

02

157

106(AB)C

157

106

51

04

31

21AB

=

÷ø

öçè

æ=

÷ø

öçè

æ÷ø

öçè

æ=

÷ø

öçè

æ=

÷ø

öçè

æ÷ø

öçè

æ=

÷ø

öçè

æ=

÷ø

öçè

æ÷ø

öçè

æ=

÷ø

öçè

æ=

÷ø

öçè

æ÷ø

öçè

æ=



8. Verify that AC. AB C) A (B +=+

i) ÷÷ø

öççè

æ-=÷÷

ø

öççè

æ

-=÷÷

ø

öççè

æ

-=

23

67,

24

21,

31

21CBA

AC. AB C) A (B

AC

AB

C) A (B

C B

+=+\

÷ø

öçè

æ-=÷

ø

öçè

æ-+÷

ø

öçè

æ

-

-=+

÷ø

öçè

æ-=÷

ø

öçè

æ-÷ø

öçè

æ

-=

÷ø

öçè

æ-

-=÷

ø

öçè

æ-

÷ø

öçè

æ-

=

÷ø

öçè

æ-=÷

ø

öçè

æ-

-÷ø

öçè

æ-

=+

÷ø

öçè

æ-

-=÷

ø

öçè

æ-+÷

ø

öçè

æ-

=+

.43

127

016

84

413

43 AC AB

016

84

23

67

31

21

413

43

24

21

31

21

43

127

41

86

31

21

41

86

23

67

24

21

ii) ÷÷ø

öççè

æ

-=÷÷

ø

öççè

æ -=÷÷

ø

öççè

æ

-=

314

231,

253

211,

15

31CBA

÷ø

öçè

æ-=

÷ø

öçè

æ-+÷÷

ø

öççè

æ

-=+

÷ø

öçè

æ-=

÷ø

öçè

æ-

÷ø

öçè

æ-

=

÷÷ø

öççè

æ

-=

úúúú

û

ù

êêêê

ë

é

÷ø

öçè

æ-÷

ø

öçè

æ --÷

ø

öçè

æ-

÷ø

öçè

æ÷ø

öçè

æ -÷ø

öçè

æ

=

÷ø

öçè

æ -÷ø

öçè

æ-

=

÷ø

öçè

æ-=

÷ø

öçè

æ-

÷ø

öçè

æ-

=+

÷ø

öçè

æ-

=

÷ø

öçè

æ-

+÷ø

öçè

æ -=+

15411

19201

7149

11611

8102

81410

7149

11611

314

231

15

31

8102

81410

2

2)15(

5

1)15(

3

1)15(

2

231(

5

131(

3

1)31(

253

211

15

31

15411

19201

561

422

15

31)(

561

422

314

231

253

211)(

ACAB

AC

AB

CBA

CB

9. Show that . A B(AB) TTT = If

i) ÷÷÷

ø

ö

ççç

è

æ

-

-

=÷÷ø

öççè

æ

-=

40

41

12

,112

121BA

÷ø

öçè

æ

-=

÷÷÷

ø

ö

ççç

è

æ

-÷ø

öçè

æ

-

-=

÷ø

öçè

æ

-=

÷ø

öçè

æ

-=

÷÷÷

ø

ö

ççç

è

æ

-=

÷ø

öçè

æ

-

-=

÷ø

öçè

æ

+--++

++-+-=

÷÷÷

ø

ö

ççç

è

æ

-

-

÷ø

öçè

æ

-=

49

50

11

12

21

441

012

49

50)(

45

90

11

12

21

441

012

242014

281022

40

41

12

112

121

TATB

TAB

TA

TBAB

ii) ÷÷÷

ø

ö

ççç

è

æ

-

=÷÷ø

öççè

æ=

15

21

71

,821

925BA

÷ø

öçè

æ=÷

ø

öçè

æ-+-+

++++=

÷÷÷

ø

ö

ççç

è

æ

÷ø

öçè

æ

-=

÷ø

öçè

æ=

÷ø

öçè

æ=÷

ø

öçè

æ

-+++

-+++=

÷÷÷

ø

ö

ççç

è

æ

-÷ø

öçè

æ=

330

4352

8479435

40214525

89

22

15

127

511

330

4352)(

343

3052

8474021

94354525

15

21

71

821

925

TT

T

AB

AB

AB

10.Prove that IA.AT = ,If ÷÷ø

öççè

æ

-=

qq

qq

cossin

sincosA

IAA

θθθθθθ

θθθθθθ

θθ

θθ

θθ

θθAA

θθ

θθA

θθ

θθA

T

T

T

=\÷ø

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æ=

÷÷ø

öççè

æ

+-

-+=

÷ø

öçè

æ -÷ø

öçè

æ-

=

÷ø

öçè

æ -=Þ÷

ø

öçè

æ-

=

10

01

sincossincoscossin

sincoscossinsincos

cossin

sincos

cossin

sincos

cossin

sincos

cossin

sincos

22

22

11.Verify that IA =2 when ÷÷ø

öççè

æ

-

-=

56

45A

IA

AA

=\÷ø

öçè

æ=

úúúú

û

ù

êêêê

ë

é

÷ø

öçè

æ-

--÷

ø

öçè

æ-

÷ø

öçè

æ-

--÷

ø

öçè

æ-

=

÷ø

öçè

æ-

-÷ø

öçè

æ-

-=´

2

10

01

5

4)56(

6

5)56(

5

4)45(

6

5)45(

56

45

56

45



12.Show that I.BA =+ 22 If

÷÷ø

öççè

æ=÷÷

ø

öççè

æ=

q

q

q

q

sin0

0sin,

cos0

0cosBA

IBA

BA

B

A

=+\÷ø

öçè

æ=

÷÷ø

öççè

æ

+

+=

÷÷ø

öççè

æ+÷÷

ø

öççè

æ=+

÷÷ø

öççè

æ=

÷ø

öçè

æ÷ø

öçè

æ=

÷÷ø

öççè

æ=

÷÷ø

öççè

æ

++

++=

÷ø

öçè

æ÷ø

öçè

æ=

22

22

22

2

2

2

222

2

2

2

2

2

2

2

2

10

01

sincos0

0sincos

sin0

0sin

cos0

0cos

sin0

0sin

sin0

0sin

sin0

0sin

cos0

0cos

cos000

000cos

cos0

0cos

cos0

0cos

qqqq

qq

qq

qq

qq

qq

qq

qq

qq

qq

13.Show that 22 I)( A)(A adbc da -=+- ,If

÷÷ø

öççè

æ=÷÷

ø

öççè

æ=

10

01, I

dc

baA

RHS adbc

adbc

adbc

adbc

daddbccdcacdac

bdabbdabadabca

dadcdca

bdabada

dbccdac

bdabbca

dc

ba da

dc

ba

dc

ba

da

=-=

÷ø

öçè

æ-=

÷ø

öçè

æ

-

-=

÷÷ø

öççè

æ

--+--+

--+--+=

÷÷ø

öççè

æ

++

++-÷÷

ø

öççè

æ

++

++=

÷ø

öçè

æ+-÷

ø

öçè

æ÷ø

öçè

æ=

+-=

2

22

22

2

2

2

2

2

I)(

10

01)(

0

0

)(

A)(ALHS

14. ÷ø

öçè

æ=÷

ø

öçè

æ=÷

ø

öçè

æ=

21

02,

51

04,

31

21CBA Show that

TT BA -=- TB)(Ai.

÷ø

öçè

æ-

-=-

÷ø

öçè

æ=

÷ø

öçè

æ=

÷ø

öçè

æ-

-=-

÷ø

öçè

æ-

-=

÷ø

öçè

æ--

--=-

22

03

50

14

32

11

22

03B)(A

20

23

5311

0241

T TT

T

T

BA

B

ABA

BC ACB) C (A -=-ii.

÷ø

öçè

æ--

-=

÷ø

öçè

æ-÷

ø

öçè

æ=

÷ø

öçè

æ÷ø

öçè

æ-÷

ø

öçè

æ÷ø

öçè

æ=

-=

÷ø

öçè

æ--

-=

÷ø

öçè

æ÷ø

öçè

æ

-

-=

÷ø

öçè

æ÷ø

öçè

æ

--

--=

-=

42

44

107

08

65

44

21

02

51

04

21

02

31

21

RHS

42

44

21

02

20

23

21

02

5311

0241

LHS

BC AC

B) C(A

15.Show that 075 22 IAA =+- if ÷

ø

öçè

æ-

=21

13A

RHS

IAA

=

÷ø

öçè

æ=

÷ø

öçè

æ

+-++-

+-+-=

÷ø

öçè

æ+÷

ø

öçè

æ

--÷

ø

öçè

æ

-=

÷ø

öçè

æ+÷

ø

öçè

æ-

-÷ø

öçè

æ+---

+-=

÷ø

öçè

æ+÷

ø

öçè

æ

--÷

ø

öçè

æ

-÷ø

öçè

æ

-=

+-=

00

00

7103055

0557158

70

07

105

515

35

58

70

07

105

515

4123

2319

10

017

21

135

21

13

21

13

75LHS 22

mythila publishers, puduvayal 41 x std mathematics …...mythila publishers, puduvayal 41 x std...

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