Math part 1

ii) sin2 ( π − x) + sin2 ( π + x) = 1 π EXERCISE 3.2
44 4
-y 1) Find the value of :
Solution : consider π − x = y ∴x =
4

L.H.S. = sin2 ( π − x) + sin2 ( π + x) i) sin690° ii) sin (495°)
44
iii) cos 315° iv) cos (600°)
= sin2 y + sin2 ( π + π − y )
44 v) tan 225° vi) tan (- 690°)

= sin2 y + cos2 y vii) sec 240° viii) sec (- 855°)

= 1= R.H.S. ix) cosec 780° x) cot (-1110°)

2) Prove the following:

iii) sin2 π + sin2 3π + sin2 5π + sin2 7π = 2 cos (π + x) cos (−x)
88 88
i)  π  = cot2 x
sin (π )  2 
Solution : L.H.S. − x cos + x

= sin2 π + sin2 3π + sin2 5π + sin2 7π ii) cos  3π + x  cos (2 π + x)[cot
8888  2 

= sin2 π + sin2 3π  3π − x  + cot(2 π + x)] = 1
88  2 

+ sin2  4π + π  + sin2  4π + 3π 
 8   8 

π π  π 3π  iii) sec 840°.cot (- 945°) + sin 600° tan (- 690°)
 2 8  8 
= sin2 π + sin2 3π + sin2 + + sin2  2 + = 3
8 8 2

= sin2 π + sin2 3π + cos2 π + cos2 3π iv) cos ec( 90° − x)sin (180° − x) cot (360° − x) = 1
88 8 8 sec(180° + x) tan (90° + x)sin (−x)

= 1+1

= 2 = R.H.S. v) sin3 (π + x) sec2 (π − x) tan (2π − x) = tan3 x

iv) cos2  π + x  sin (π − x) cosec2 − x
 2 
 π   2π   3π   9π 
cos 2  10  + cos 2  5  + cos 2  5  + cos 2  10  =2

Solution : L.H.S vi) cosq + sin (270° + q) - sin (270° − q)
+ cos (180° + q) = 0

= cos2  π  + cos 2  2π  + cos2  3π  + cos 2  9π 
 10   5   5   10 
Let's Learn
= cos2 π  + cos2  π − π  + cos2  π + π  + cos2  π − π  3.3 Trigonometric functions of multiple angles.
 10   2 10   2 10  10 

= cos 2  π  + sin2  π  + cos 2  π  + sin2  π  Angles of the form 2q, 3q, 4q etc. are integral
 10   10   10   10 
multiple of q these angles are called multiple
= 1+1 angles and angles of the form θ , 3θ etc. are
= 2 = R.H.S.
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called submultiple angles of q.

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3.3.1 Trigonometric functions of double angles = 2 (1 - sin2q) - 1
(2q) = 2 - 2sin2q - 1
= 1 - 2sin2q ..…… (3)
Theorem : For any angle q,
2tanθ = cos2q - sin2q

1) sin 2q = 2sinq cosq = 1+ tan2θ = cos2θ − sin2θ
2) cos2q = cos2q − sin2q = 2cos2q - 1 1

= 1 − 2 sin2q = 1− tan2θ cos2θ − sin2θ
1+ tan2θ = cos2θ + sin2θ

3) tan2q = 2tanθ cos2θ − sin2θ
1− tan2θ = cos2θ

Proof: 1) = sin 2q = sin (q + q ) cos2θ + sin2θ
= sin q cos q + cos q sin q cos2θ

= 2 sin q cos q .. . …(1) = 1− csoins22θθ
= 2sinθ cosθ sin 2θ
1 + cos2θ
1

2sinq cosq = 11+− ttaann22θθ …….(4)
= sin2 q + cos2 q

2sinθ cosθ / cos2 θ From (1), (2), (3) and (4) we get
(sin2 θ + cos2 θ / cos2 θ
= cos2q = cos2q − sin2q = 2cos2q - 1

2sinθ cosθ = 1 − 2 sin2q
= sin2 θ = 1− tan2θ
cos2 θ +1
1+ tan2θ

2tanθ 3) tan2q = 2tanθ
= 1+ tan2θ . . . . .(2) 1+ tan2θ
From (1) and (2)
Note that the substitution 2q = t transforms
2tanθ sin2q = 2sinq.cosq into sint = 2sin t .cos t .
sin 2q = 2sinqcosq = 1+ tan2θ
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2) cos2q = cos (q + q) Simillarly,

= cosq cosq - sinq sinq cos2q = cos2q - sin2q and cost = cos2 t - sin2 t
22
= cos2q - sin2q …….(1)
= cos2q - (1 - cos2q) 2 tanθ 2 tan t
= cos2q - 1 + cos2q 1− tan2θ 2
= 2cos2q - 1 ……. (2) tan2q = and tant = 1- tan2 t

2

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q 2t = 3tanθ − tan3θ
Also if tan 2 = t then sinq = 1- t2 1− 3tan2θ

1−t2 and tanq = 2t ∴ tan3q = 3tanθ − tan3θ
and cosq = 1+ t2 1- t2 1− 3tan2θ

3.3.2 Trigonometric functions of triple angle
(3q)

Theorem : 1) For any angle q SOLVED EXAMPLES

1) sin3q = 3sinq - 4sin3q Ex. 1) Prove that 1 + tanq tan q = sec q
2
2) cos3q = 4cos3q -3cosq
Solution :
3) tan3q = 3tanθ − tan3θ
1− 3tan2θ L.H.S = 1 + tan q tan ( q )
2
Proof:
sin θ
1) sin3q = sin (2q + q) sinθ 2
cosθ θ
= sin2qcosq + cos2qsinq = 1 + .

cos
2
= 2sinqcosq.cosq + (1- 2 sin2q) sinq

= 2sinqcos2q + sinq − 2sin3q 2sin θ cos θ .sin θ

= 2sinq (1- sin2q) + sinq - 2sin3q = 1 + 22 2
cosθ cos θ
= 2sinq - 2sin3q + sinq - 2sin3q 2

=3sinq - 4sin3q 2sin2 θ

∴ sin3q = 3sinq - 4sin3q = 1 + 2
cosθ

(Activity) 1− cosθ
= 1 + cosθ
2) cos3q = cos (2q + q)

3) tan3q = tan (2q + q) cosθ +1− cosθ
= cosθ
tan 2θ + tanθ
= 1− tan 2θ tanθ 1
= cosθ = secq = R.H.S
 2tanθ  + tanθ
=  1− tan2θ  Ex. 2) Prove that tan 20° tan 40° tan 60° tan 80°
=3
1 2tanθ  Solution :
− 1 − tan2θ  tanθ L.H.S.
= tan 20° tan 40° tan 60° tan 80°
( )2tanθ + tanθ 1− tan2θ = tan 20° tan 40°. 3 tan 80°
= 3 tan20° tan (60° - 20°) tan (60° + 20°)
1− tan2θ

( ) = 1− tan2θ − 2 tan2 θ

1− tan2θ

= 2tanθ + tanθ − tan3θ
1− 3tan2θ

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= 3 tan20°. tan60ο − tan20ο . tan60ο + tan20ο = −3cos3θ + 3cosθ 3sinθ − 3sin3θ
1+ tan60ο tan20ο 1− tan60ο tan20ο cosθ + sinθ

3 − tan20ο . 3 + tan20ο 3cosθ (1− cos2θ ) 3sinθ (1− sin2θ )
1+ 3tan20ο 1− 3tan20ο
= 3 tan20°. = cosθ + sinθ



= 3 tan20° . 3 − tan2 20ο = 3 sin2q + 3 cos2q
1− 3tan2 20ο = 3(sin2q + cos2 q)
= 3(1) = 3 = R.H.S.

3tan20ο − tan3 20ο
= 3 1− 3tan2 20ο

= 3 tan [3 (20)]° Ex. 5) Prove that tan5A + tan3A =4cos2A.cos 4A
= 3 tan 60° tan5A − tan3A
= 3 . 3 = 3 = R. H. S
Soln. : L.H.S. = ttaann55AA + tan3A
− tan3A

Ex. 3) Prove that 2cosec2x + cosecx = secx. cot (x/2) sin5A + sin3A
= csoins55AA − csoins33AA
Solution : L.H.S. = 2cosec2x + cosecx
cos5A cos3A
= sin22x + si1nx
sin5Acos3A + cos5Asin3A
2
= 2sinx cosx + 1 = cos5 A cos3 A
sinx sin5Acos3A − cos5Asin3A

1+ cosx cos5 A cos3 A
= sinx cosx
sin5Acos3A + cos5Asin3A
2cos2 ( x / 2) = sin5Acos3A − cos5Asin3A
= 2sin ( x / 2)cos ( x / 2)cosx = ssiinn82AA = 2sins4inA2cAos4A
= csoins((xx // 22)) . co1sx
= 2.2sin2 A cos 2 A cos 4 A
= cot (x/2) . secx = R. H. S. sin2 A

= 4 cos2 A cos 4A = R. H. S.

Ex. 4) Prove that cos3θ − cos3θ + sin3θ + sin3θ Ex. 6) Show that , where (cos q + isin q)3
= 3 cosθ sinθ = cos 3q + i sin 3q, where i2 = -1.

Solution : L.H.S. Solution : L.H.S.= [cosq + i sinq]3

= cos3θ − cos3θ + sin3θ + sin3θ = cos3 q + 3 i cos2 q sin q + 3 i2 cosq sin2 q + i3 sin3q
cosθ sinθ = cos3 q + 3i ( 1 - sin2 q) sinq - 3cosq sin2 q - i sin3q

= cos3θ − 4cos3θ − 3cosθ  + sin3θ + 3sinθ − 4sin3θ  = cos3 q + 3i sin q - 3isin3q- 3cosq (1- cos2 q) - isin3q
cosθ sinθ = cos3 q + 3isin q - 3isin3q - 3cosq + 3cos3 q - isin3q

45

= [4 cos3 q − 3cosq] + i [3sin q - 4sin3q] = tan  π + A 
= cos3q + isin3q  4 

= R.H.S. = R.H.S.

Ex. 7) Show that 4sinq cos3q - 4cosq sin3q Ex. 9) Find sin x , cos x , tan x
= sin4q 2 2 2

Solution: if tan x = 4 , x lies in II quadrant.
3
L.H.S = 4sinq cos3q - 4cosq sin3q
= 4sinq cosq [cos2 - sin2q] Solution : we know that 1 + tan2q = sec2q
= 2. (2sinq cosq ) (cos2 q - sin2q)
= 2. sin2q . cos2q sec2x = 1 + ( - 4 )2 = 1 + 16 = 9 +16 = 25
= sin 4q 3 9 9 9
= R.H.S.
sec x = ±5
3

But x lies in II quadrant.

Ex. 8) Show that 1+ sin2A = tan  π + A  ∴ secx is negative.
1− sin2A  4
∴ sec x = − 5 ∴ cos x = − 3
Solution : 3 5

L.H.S. = 1+ sin2A sin x = 1− cos2x = 1− (− 3)2 = 1− 9 = ±4
1− sin2A 5 25 5

= ssiinn22 A + cos 2 A + 2sinAcosA 4 [∴ x lies in II quadrant]
A + cos 2 A − 2sinAcosA ∴ sin x = 5
But sin x =
(sinA + cosA)2 1− cosx 1 −  − 3 
= (cosA − sinA)2 2 2=  5 

sinA + cosA 2
= cosA − sinA
= 5+3 = 4 2
2×5 5= 5

cosA + sinA cos x = 1+ cosx = 1 −  − 3 
= coscAo−sAsinA 2 2  5 

cosA 2

1+ sinA = 5−3 1= 1
= 1− csoinsAA 2×5 = 55

cosA sin x 4
= 11+− ttaannAA
tan x = 2 = 5 = 4×5 = 4 =2
2 cos x 1 51

2 5

= tan π4 +tanA π ∴ sin x = 2 , cos x = 1 , tan x = 2
1− tan 4 2 5 25 2
π [∴ 1 = tan ]
4 tanA

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Ex. 10) Find the value of tan π = 1 [3 + cos 2x + 2cos2x cos  π − 2π  ]
8 2  3 

Solution : let x= π ∴ 2x = π = 1 [3 + cos 2x - 2cos2x cos π ]
8 4 2 3

2 tan x 1
we have tan 2 x = 1- tan2x = 2 [ 3 + cos 2x - cos2x ]

π 2tan π 3
4 8 = 2 = R. H. S.
∴ tan = 1− tan2 π

8 2y Ex. 12) Find sin π
∴ 1 = 1− y2 10
let y = tan π
8 Solution : πc = 18°
10
∴ 1 - y2 = 2y
Let, q = 18°, 2q = 36°, 3q = 54°

∴ y2 +2y - 1 = 0 We have 2q + 3q = 90°
2q = 90° - 3q
∴ y = −2 + 2 2 = - 1 ± 2 ∴ sin 2q = sin (90° - 3q)
2sin q.cos q = cos 3 q
π 2 π ∴ 2sin q.cos q = 4.cos3 q - 3 cosq
8 8 2sin q = 4.cos2 q - 3
Since lies in I quadrant y = tan positive 2sin q = 4 (1 - sin2q) - 3
2sin q = 4 - 4sin2q - 3
∴ tan π = 2 - 1
8

Ex. 11) Prove that

cos2 x + cos2  x + π  + cos2  π − 2π  = 3
 3  3  2

Solution : L.H.S. 4sin2q + 2sinq - 1 = 0

= cos2 x + cos2  x + π  + cos2  π − 2π  ∴ sin q = −2 ± 4 + (4)(4)(1)
 3   3  2(4)

1+ cos2x 1 + cos 2  x + π  = −2 ± 2 5
2 3  2(4)
= + +
2
−1± 5
1+ cos 2  x − 2π  ∴ sin q = 4
 3 

2 −1+ 5
∴ sin q = 4
1 2π
= 2 [3 + cos 2x + cos  2x + 3  [∵ q is an acute angle]
 

+ cos  2x − 2π  −1 + 5
 3  4
∴ sin 18° =

= 1 [3 + cos 2x + 2cos2x cos 2π ] ∴ sin π = −1 + 5
2 3 10 4

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EXERCISE 3.3 xv) 2cos4x +1 = (2 cosx -1) (2 cos2x -1)
2cosx +1
1) Find values of : i) sin π ii) cos π 3
8 8
xvi) cos2x + cos2 (x +120°) + cos2 (x-120°) = 2

2) Find sin 2x, cos 2x, tan 2x if secx = −13 , xvii) 2cosec 2x + cosecx = sec x cot  x
 2 
π 5
2
<x<π  x + π   π − 2π 
 3   3
xviii) 4cosx cos + cos2  = cos3x

3) Prove the following: 2

i) 1− cos2θ = tan2q xix) sinx tan  x  + 2cosx =  x 
1+ cos2θ  2   2 
1 + tan2

ii) (sin3x +sinx)sin x + (cos3x-cosx) cosx = 0 Let's :Learn

iii) (cosx+cosy)2+ (sinx -siny)2 = 4cos2 ( x + y) 3.4 Factorization formulae:

2 Formulae for expressing sums and differences
of trigonometric functions as products of sine and
iv) (cosx-cosy)2+(sinx -siny) 2 = 4sin2 ( x − y) cosine functions are called factorization formulae.
Formulae to express products in terms of sums and
2
v) tanx + cotx = 2 cosec2x differences are called defactorization formulae.

vi) cosx + sinx − cosx − sinx = 2tan2x 3.4.1 Formulae for conversion of sum or
cosx − sinx cosx + sinx difference into product.

vii) 2 2 2 2cos8x = 2 cosx

viii) 16 sinq cosq cos2q cos4q cos8q = sin16q Theorem: 9) For any angles C and D,

ix) sin3x + cos3x = 2cot2x 1) sin C + sin D = 2 sin C+D cos C−D
 2   2 

cosx sinx C+D  C − D 
 2   2 
cot  x  −1 2) sin C - sin D = 2 cos sin
x)  2 
cosx =
1+ sinx  x   C + D   C − D 
 2  3) cos C + cos D = 2 cos  2  cos  2 
cot +1

 θ   θ  4) cos C - cos D = -2 sin  C + D  sin C−D
xi)  2   2   2   2 
tan  θ  + cot  θ 
cot  2  − tan  2  = secq  C + D   D−C 
 2   2 
= 2 sin sin

xii) 1 − 1 = cot 2A Proof : We know that
tan3A − tanA cot3A − cotA Let, A = C + D and B = C − D
sin680
22
xiii) cos7° cos 14° cos28° cos 56° = 16cos830 ∴ A+B = C and A–B = D

( ) ( )xiv) sin2 −160ο

sin2 70ο
sin 180ο −θ = sec2 20° using these values in equations
+ 48

sinθ

sin (A + B ) + sin (A + B) = 2sin AcosB SOLVED EXAMPLES
sin (A - B) + sin (A - B) = 2 cos A sinB

we get Ex. 1) Prove the following :
i) sin 40° - cos 70° = 3 cos 80°
sin C + sin D = 2 sin C+D  cos C−D ii) cos 40° + cos 50° + cos 70° + cos 80°
 2   2  = cos 20° + cos 10°

sin C - sin D = 2 cos C+D  sin C−D
 2   2 

Simillarly Solution :

the equations, i) LHS = sin 40° - cos 70°
cos ( A + B ) = cos A cos B - sinAsin B ...... (3)
= sin (90° - 50°) - cos70°

cos ( A - B ) = cos A cos B + sinAsin B ...... (4) = cos 50° - cos70°

gives, C+D  C−D = - 2 sin 60°sin ( - 10°)
 2   2 
cosC + cos D = 2 cos cos

∴ cosC - cos D = - 2 sin C+D  sin C−D  = 2 sin 60°sin 10°
 2   2  = 2× 3 cos 80 = 3 cos 80°

∴ sin (−θ) = sinθ 2
= R.H.S

L.H.S.

∴ −sin C−D  = sin  −  C − D   = cos 40° + cos 50° + cos 70° + cos 80°
 2    2  

 D − C  = (cos 80° + cos 40°) + (cos 70° + cos 50°)
 2 
= sin  80 + 40   80 − 40   70 + 50   70 − 50 
 2   2   2   2 
= 2 cos cos + 2cos cos

∴ cosC - cos D = 2 sin C+D  sin  D−C  = 2cos 60° cos 20° + 2cos60° cos10°
 2   2 

= 2 cos60° (cos 20° + cos 10°)

3.4.2 Formulae for conversion of product in to =2 1 (cos20° + cos 10°)
sum or difference : 2
For any angles A and B
1) 2sin A cos B = sin (A + B) + sin (A - B) = cos20° + cos 10° = R. H. S.
2) 2cos A sin B = sin (A + B) - sin (A - B)
3) 2cos A cos B = cos (A + B) + cos (A - B) Ex. 2) Express the following as sum or difference
4) 2sin A sin B = cos (A - B) - cos (A + B) of two trigonometric function:

i) 2 sin4q cos 2q
Solution : = 2 sin 4q cos 2q
= sin (4q + 2q) + sin (4q - 2q)
= sin 6q + sin 2q

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ii) 4 sin  A+ B  sin  A−B Ex. 4) Prove that following.
 2   2 
i) cos (7x − 5y) + cos (7 y − 5x) = cot (x + y)
 A+ B A− B   A+ B A− B  sin (7x − 5y) + sin (7 y − 5x)
= 2 ×[cos  2 − 2  − cos  2 + 2 

= 2[cos B − cos A] Solution : L.H.S.

= 2cosB − 2cos A cos (7x − 5y) + cos (7 y − 5x)
= sin (7x − 5y) + sin (7 y − 5x)

Ex. 3) Show that 2 cos  7x − 5y + 7 y − 5x  cos  7 x − 5 y − 7 y + 5x 
 2   2 
sin8x + sin2x =
i) cos2x − cos8x = cos3x  7 x − 5 y + 7 y − 5x   7x − 5 y − 7 y + 5x 
2sin  2  cos  2 

Solution : L.H.S cos ( x + y) cos (6x − 6y)
= sin8x + sin2x = sin ( x + y) cos (6x − 6y)

cos2x − cos8x cos ( x + y)

2 sin  8x + 2x  cos  8x − 2x  = sin(x + y = cot (x + y) = R. H. S.
 2   2 
=
 2 x + 8x   8x − 2x 
2sin  2  sin  2  ii) sin6q + sin4q - sin2q = 4cosq sin2q cos3q

2sin 5x cos 3x Solution : L.H.S.
2sin 5x sin 3x
= = sin6q + sin4q - sin2q

= cot3x  6θ + 4θ   6θ − 4θ 
 2   2 
= R. H. S. = 2sin cos - 2sinq cosq

sin2 ∝ +sin2β tan (∝ +β ) = 2sin5qcosq - 2sinqcosq
ii) sin2 ∝ −sin2β = tan (∝ −β )
= 2 cosq [sin5q -sinq]
Solution :
sin2 ∝ +sin2β = 2 cosq 2 cos  5θ +θ  sin  5θ −θ 
 2   2 
L.H.S. = sin2 ∝ −sin2β
= 2 cosq.2cos3qsin2q
2 sin  2 ∝ +2β  cos  2 ∝ −2β  = 4 cosqsin2qcos3q
 2   2 
= = R.H.S.
 2 ∝ +2β   2 ∝ −2β 
2cos  2  sin  2 

= sin (∝ +β ) . cos (∝ −β ) cos3xsin9x − sinxcos5x
cos (∝ +β ) sin (∝ −β ) iii) cosxcos5x − sin3xsin9x = tan8x

= tan (a + b ) . cot (a - b ) Solution : L.H.S.
cos3xsin9x − sinxcos5x
tan (∝ +β )
= tan (∝ −β ) = cosxcos5x − sin3xsin9x

= R. H.S. 2cos3xsin9x − 2sinxcos5x
= 2cosxcos5x − 2sin3xsin9x
50

[sin (3x + 9x) − sin (3x − 9x) ]−[ sin ( x + 5x) + sin ( x − 5x)] EXERCISE 3.4
= cos ( x + 5x) + cos ( x − 5x) − cos (9x − 3x) − cos (3x + 9x)

sin12x − sin (−6x) − sin6x − sin (−4x) 1) Express the following as a sum or difference
= cos6x + cos (−4x) − cos6x + cos12x of two trigonometric function.

sin12x + sin 6x − sin6x + sin4x i) 2sin 4x cos 2x
cos6x + cos 4x − cos6x + cos12x
= ii) 2sin 2π cos π
3 2
sin12x + sin4x
= cos12x + cos4x iii) 2cos4q cos2q

 12x + 4x   12 − 4 x  iv) 2cos35° cos75°
 2   2 
2 sin cos 2) Prove the following :

=  12x + 4x  cos( 12 x− 4 x 
 2  2 
2cos i) sin2x + sin2 y = tan ( x + y)
sin2x − sin2 y tan ( x − y)

sin8x ii) sin6x + sin 4x - sin2x = 4cosx sin2x cos3x
= cos8x = tan8x = R.H.S.

1 sinx − sin3x + sin5x − sin7x
iv) cos 20° cos 40° cos 60° cos 80° = 16 iii) cosx − cos3x − cos5x + cos7x = cot2x
Solution : L.H.S. iv) sin18° cos39° + sin6° cos15° = sin24° cos33°

= cos 20° cos 40° cos 60° cos 80° 1
v) cos20° cos40° cos60° cos80° = 16
= cos 20° . cos 40° . 1 . cos 80°
2 3
vi) sin20° sin40° sin60° sin80° = 16
= 1 [cos 20° cos 40° cos 80°]
2 Let's :Learn
1
= 4 [cos(20°+40°) + cos (20°- 40°)] cos 80° 3.5 Trigonometric functions of angles of a
triangle
1
= 4 [cos60°) + cos (- 20°)] cos 80° Notation: In D ABC; m ∠ BAC = A,
m ∠ ABC = B, m ∠ ACB = C
= 1 [ 1 cos80° + cos 20° cos 80°] ∴A+B+C= p
4 2
Result 1) In D ABC, A + B + C = p
11 ∴A+B= p-C
= 4 [ 2 cos80° + cos 20° cos 80°] ∴ sin (A + B) = sin (p - C)
∴ sin (A + B) = sin C
1 11
= 8 cos80° + 4 . 2 . 2cos 20° cos 80° Simillarly:

= 1 cos80° + 1 [cos (20 + 80 )+cos (20 - 80)] sin (B + C) = sin A and
8 8

11
= 8 cos80° + 8 [cos 100° + cos (- 60)°]

1 11
= 8 [cos80° + [cos 180° - 80°)] + 8 × 2

1 11
= 8 [cos80° - cos 80°] + 16 = 16

= R. H.S. sin (C + B) = sin B
51

Result 2) In DABC, A+B+C = p ii) In DABC, A+B+C = p ∴ B + C = p - B
∴ B + C = p - C
∴ cos (B +C) = cos (p - A) ∴ cos  A + C  = cos ( π - B ) = sin B
∴ cos (B + C) = - cos A  2  2 2 2

∴ cos  A+C  = sin B
 2  2

Simillarly: and Verify.
cos (A + B) = - cos C
cos (C + A) = - cos B 1) cos  A+ B  = sin C
 2  2

Result 3) for any DABC  B +C  A
 2  2
C 2) cos = sin
2
i) sin  A+ B  = cos
 2 

sin  B +C  = cos A SOLVED EXAMPLES
 2  2

C+ A B Ex. 1) In DABC prove that
sin  2  = cos 2
i) sin2A + sin2B - sin2C = 4cosA cosB sinC

 A + B  C Solution : L.H.S. = sin2A + sin2B - sin2C
 2  2
ii) cos = sin  2A+ 2B   2A− 2B 
 2   2 
= 2sin cos - sin2C

cos  B+C  = sin A = 2sin(A + B) cos (A- B) -2sinC cosC
 2  2

C+ A B = 2sin (p-C) cos (A-B) - 2sinC cos [p - (A+B )]
 2  = sin 2
cos = 2sinC cos (A - B) + 2sinC cos(A+ B)

Proof. = 2sinC [cos (A - B) + cos(A+ B)

i) In DABC, A+B+C = p ∴ A +B = p - C = 2 sinC.2cos  A−B+ A+B cos  A−B− A−B
 2   2 
 A+ B  π −C π C
∴  2  = 2 = 2 - 2
= 4 sinC cosA cosB

 A+ B  π C C = 4 cosA cosB sinC
 2  2 2 ) = cos 2
sin = sin ( - = R.H.S.

sin  A + B  = cos C ABC
 2  2 ii) cosA+cosB + cosC =1+ 4sin 2 sin 2 sin 2
Solution : L.H.S = cosA + cosB + cosC
Verify.

 B+C  A = 2 cos  A+B cos  A − B  + 1 - 2 sin2 C
 2   2  2
1) sin  2  = cos 2

2) cos  B +C  = sin B = 2 cos π − C  cos  A− B  + 1 - 2 sin2 C
 2  2  2 2   2  2

52

= 1+ 2 sin C cos  A−B - 2 sin2 C = cosC.2sin  A−B+ A+ B  sin  A+ B− A+ B 
2  2  2  2   2 

C cos  A − B  C  = 2cosC sin A sin B
2  2  2 
= 1+ 2 sin − sin = 2 sin A sin B cos C

= R. H.S.

= 1+ 2 sin C [cos  A − B  − sin  π − A + B ]
2  2   2 2
iv) cotA cot B + cot B cot + cot C cotA = 1
= 1+ 2 sin C [cos  A − B  − cos  A+ B ] Solution : In DABC, A + B + C = p
2  2   2 ∴ A+B =p-C
∴ tan ( A+ B ) = tan ( p - C)
= 1+ 2 sin C .2sin  A − B + A + B  sin  A + B − A + B 
2  4   4  tanA + tanB
∴ 1− tanA tanB = tan ( p - C)

C AB ∴ tanA + tan B = - tanC + tanA tanB tanC
= 1+ 4 sin 2 sin 2 sin 2
∴ tanA + tan B + tanC = tanA tanB tanC
A B C
= 1+ 4 sin 2 sin 2 sin 2 ∴ 1+1+1 = 1 .1 .1
cotA cotB cotC cotA cot B cotC
= R.H.S.

∴ cot A cot B + cotB cotC + cotC cotA = 1

iii) sin2A + sin2B - sin2C = 2 sinA sinB cosC

Solution : L.H.S. = sin2A + sin2B - sin2C AB BC C A
v) tan 2 tan 2 + tan 2 tan 2 +tan 2 tan 2 =1
= 1− cos2A + 1− cos2B − sin2C
22 Solution : In DABC, A + B + C = p

1 ∴ A + B = p - C ∴ A+B = π −C = π - C
= 2 [2 - cos2A - cos2B] - sin2C 2 2 2 2

1 AB πC
= 1 - 2 [cos2A + cos2B] - sin2c tan ( 2 + 2 ) = tan ( 2 - 2 )

= 1- 1 . 2 cos  2 A + 2B  cos  2 A − 2B  - sin2C tan A + tan B
2  2   2 
∴ 2 2 = cot C
tan A tan B 2
= 1 - sin2C - cos( A + B ) + cos (A- B) 1−
22

= cos2C - cos [ p - C ] cos (A- B) tan A + tan B 1

= cos2C + cosC cos (A- B) ∴2 2 = tan C
2
1− tan A tan B
22
= cos C [cosC + cos (A- B)]
A BC AB
∴ [tan 2 + tan 2 ] tan 2 =1 - tan 2 tan 2
= cos C [cos[ p - (A+B)] + cos (A- B)]

= cos C [- cos (A+B ) + cos (A- B)] ∴ tan A C +tan B tan C =1-tan A tan B
2 tan 2 2 2 2 2

= cos C [cos (A-B) - cos (A+ B)] ∴ tan A tan C +tan B tan C +tan A tan B =1
2 2 2 2 2 2
53

vi) cosA − cosB + cosC +1 = cot A cot C C 2sin B cos A
cosA + cosB + cosC −1 2 2 2 22
= cot . 2sin A sin B

Solution : L.H.S. = cosA − cosB + cosC +1 22
cosA + cosB + cosC −1
cos A
[cosA − cosB] + [1+ cosC] = cot C 2
= [cosA + cos B] − [1− cos C] 2
sin A
2

2 sin  A + B  sin  B − A  + 2cos2 C = cot C cot A
 2   2  2 2 2
=
 A + B   A − B   C 
2 cos  2  cos  2  +  −2sin 2 2  = R.H.S.

2 sin  π − c  sin  B − A  + 2cos2 C
 2 2   2  2
= EXERCISE 3.5

2 cos  π − c  cos  A − B  +  −2sin2 C 
 2 2   2   2 
In DABC, A + B + C = p show that

2 cos C sin  B − A  + 2cos2 C 1) cos2A + cos2 B + cos 2 C
2  2  2 = -1- 4 cos A cos B cos C
=
C  A + B  − 2sin2 C
2 sin 2 cos  2  2
2) sin A + sin B + sin C
cos C [sin  B − A  + cos C] A BC
2  2  2
= 4 cos 2 cos 2 cos 2

= C  A − B  C 3) cos A + cos B - cosC
2  2  2 A BC
sin [cos − sin
= 4 cos 2 cos 2 sin 2 - 1

sin  A + B  + sin  B − A 
 2   2 
C 4) sin2 A + sin2 B + sin2C = 2 + 2cosAcosBcosC
= cot 2
[cos  A− B  − sin (π − A+ B )] ABC
 2  2 2 5) sin2 2 + sin2 2 - sin2 2

sin  A + B  + sin  B − A  A BC
 2   2  = 1- 2cos 2 cos 2 sin 2

C
= cot 2 A B C A B C
cos  A−B  − cos ( A+ B ) 6) cot 2 +cot 2 +cot 2 =cot 2 cot 2 cot 2
 2  2

 A + B B −A   A + B B − A  7) tan2A + tan2B + tan2C = tan2A tan2B tan2C
 2 2   2 2 
+ −

C 2 sin cos 8) cos2 A +cos2 B - cos2C = 1- 2sinA sinB cosC
2 22
= cot .
 A − B A + B   A + B A − B 
 2 + 2   2 − 2 

2 sin sin
22

54

Let's Remember 2 tanθ
14) sin2q = 2sinqcosq = 1+ tan2 θ

cos2q = cos2q - sin2q = 2cos2q - 1

1) cos ( A - B) = cosA cosB + sinA sinB = 1 -2 sin2q = 1− tan2θ
1+ tan2θ
2) cos ( A + B) = cosA cosB - sinA sinB
2tanθ
3) sin ( A + B) = sinA cosB + cosA sinB tan 2q = 1− tan2θ

4) sin ( A - B) = sinA cosB - cosA sinB 15) sin 3q = 3 sin q - 4 sin3 q

5) cos π −θ  = sinq , cos π −θ  sin 3q = 4cos3q - 3cosq
 2   2  tan3q = 3tanθ − tan3θ

= sinq, tan π −θ  = cotq 1− 3tan2θ
 2 
2 tan θ

π   π  16) sinq = 2sin θ cos θ = 2
 2   2  22 1+ tan2 θ
6) sin +θ = cosq, cos +θ

2

= sinq , tan π + θ  = - cotq , cosq = cos2 θ - sin2 θ = 2cos2 θ - 1
 2 
22 2

7) tan ( A + B) = tanA + tanB = 1 - 2sin2 θ 1− tan2 θ
1− tan AtanB 2 2
= θ

1+ tan2 2

8) tan ( A - B) = tanA − tanB 2tan θ
1+ tan AtanB
tanq = 2
1− tan2 θ
9) sin (p - q) = sinq, cos (p - q) 2
= - cos q, tan (p - q) = -tanq
17) 1 + cosq = 2cos2 θ , 1 - cosq = 2sin2 θ
10) sin (p + q) = - sinq, cos (p + q) 22
= - cos q tan (p+ q) = tanq
1 + cos2q = 2cos2q, 1 - cos2q = 2sin2q

11) sin  3π −θ  = - cosq, cos  3π −θ  18) sin C + sinD = 2sin C+D cos C −D 
 2   2   2   2 

= sinq, tan  3π −θ  = cot q sin C - sinD = 2cos C + D  sin  C − D 
 2   2   2 

 3π   3π  cos C + cosD = 2cos  C + D  cos C − D 
 2   2   2   2 
12) sin +θ = - cos q, cos +θ

 3π +θ  cos C − cosD = −2sin C +D  sin  C − D 
 2   2   2 
= sinq, tan = - cot q

13) sin (2p - q) = - sinq, cos (2p - q) cos C − cosD = 2 sin  C + D  sin  D −C 
 2   2 
=cosq, tan (2p- q) = - tanq

55

19) 2 sinA cosB = sin (A + B) + sin (A − B) vii) tan15° = tan π = 3 −1 = cot75° = cot 5π
2 cosA sinB = sin (A + B) − sin (A − B) 12 3 +1 12
2 cosA cosB = cos (A + B) + cos (A − B)
2 sinA sinB = cos (A − B) − cos (A + B) viii) tan75° = tan π = 2 + 3
12

20) For ∆ABC, = 3 +1= cot15° = cot π
3 −1 12

sin (A + B) = sinC, sin (B + C) = sinA ix) tan(22.5°) = tan π
sin (A + C) = sinB 8

cos(A + B) = − cosC, cos (B + C) = − cosA = 2 - 1 = cot67.5 = cot 3π
cos(A + C) = − cosB 8

sin  A+B C sin  B+C  x) tan(67.5°) = tan 3π
 2  = cos 2 ,  2  8
π
= 2 + 1 = cot (22..5°) = cot 8

= cos A , sin  A+C  = cos B
2  2  2

cos  A+ B  = sin C , cos  B+C  MISCELLANEOUS EXERCISE - 3
 2  2  2 

= sin A cos  A+C  = sin B I) Select correct option from the given
2,  2  2 alternatives.

Activity : 1) The value of sin (n+1) Asin (n+2) A +cos
Verify the following. (n+1) A cos (n+2) A is equal to
A) sin A B) cosA C) − cos A D) sin2A

i) sin18° = sin π = 5 −1 = cos 72° = cos 2π 2) If tan A − tan B = x and cot B − cotA = y
10 4 5 then cot (A − B ) = …

ii) cos36° = cos π = 5 +1 = sin 54° = sin  3π  A) 1 − 1 B) 1 − 1
5 4  10  y x x y

iii) sin72° = sin 2π = 10 + 2 5 = cos18° = cos π C) 1 + 1 D) xy
5 4 10 x y x−y

iv) sin36° = sin 2π = 10 − 2 5 = cos54° = cos 3π 3) If sinq = nsin (q + 2 a) then tan (q + a)
5 4 10 is equal to

v) sin15° = sin π = 3 −1 = cos75° = cos 5π A) 1+ n tan a B) 1− n tan a
12 2 2 12 2−n 1+ n

vi) cos15° = cos π = 3 +1 sin75° = sin 5π C) tan a D) 1+ n tan a
12 2 = 12 56 1− n

2

4) The value of cosθ is equal to…… 10) The numerical value of tan200 tan800 cot500
1+ sinθ is equal to…..
A) 3 B) 1 C) 2 3 D) 1
A) tan θ −π  B) tan  − π − θ 
 2 4   4 2  3 23

C) tan π −θ  D) tan  π + θ  II) Prove the following.
 4 2   4 2 
1) tan20° tan80° cot50° = 3

5) The value of cosA cos (60° − A)cos (60° +A) 2) If sin a sinβ − cos a cos β + 1= 0
then prove cot a tan β = −1
is equal to…..

A) 1 cos 3A B) cos 3A 3) cos 2π cos 4π cos 8π cos1165π = 1
2 D) 4cos3A 15 15 15 16

C) 1 cos 3A
4
π 3π )1+ cos 5π   7π  1
4) (1+ cos 8 ) (1+ cos 8 8   1 + cos 8  = 8

6) The value of

sin π sin 3π sin 5π sin 7π sin 9π sin 11π sin 13π is .... 5) cos12°+ cos 84° + cos 156° + cos132° = − 1
14 14 14 14 14 14 14 2

A) 1 B) 1 11 6) cos  π + x  + cos  π − x  = 2 cos x
16 64 C) 128 D) 256  4   4 

7) If a + β + ϰ = π then the value of 7) sin5x − 2sin3x + sinx = tanx
sin2 a + sin2 β-sin2 ϰ is equal to….. cos5x − cosx

A) 2sina B) 2sina cos β sinϰ 8) sin2 6x − sin2 4x = sin2x sin10x

C) 2sin a sinβcosϰ D) 2sin a sinβsinϰ 9) cos2 2 x − cos2 6 x = sin4x sin8x

8) Let 0 < A, B < π satisfying the equation 10) cot4x ( sin5x + sin3x) = cot x (sin5 x − sin3x)
2
11) cos9x − cos5x = − sin2x
3sin2 A +2sin2 B =1 and 3sin2 A − 2sin2 B = 0

then A+2B is equal to.... sin17x − sin3x cos10x

A) π B) π C) π D) 2π 12) If sin 2A = λsin 2 B then prove that
2 4
tan ( A + B) = λ +1
tan ( A − B) λ −1

9) In∆ABC if cot A cot B cotC > 0 13) 2cos2A +1 = tan(60° + A) tan ( 60° − A)
then the triangle is…. 2cos2A −1

A) Acute angled B) right angled 14) tan A+tan (60° + A)+tan (120° + A)= 3tan3A
C) obtuse angled 15) 3tan610° - 27 tan410° + 33tan210° = 1

D) isosceles right angled

57

16) cosec 48° + cosec 96° + 23) sin18° = 5 −1
cosec 192° + cosec 384° = 0 4

17) 3(sinx - cosx)4 + 6(sin x + cosx)2 + 24) cos36° = 5 +1
4 (sin6 x + cos6 x) = 13 4

18) tan A + 2 tan2A + 4tan4A + 8cot 8A = cotA 25) sin36° = 10 2 5

3π 4
2
19) If A + B + C = then cos2A + πc = 1 2− 2
26) sin 2
cos2B + cos2C = 1 − 4sinAsinBsinC
8

20) In any triangle ABC, sin A − cosB = cosC 27) tan π = 2 -1
then ∠ B = π/2 8

28) tan6° tan42° tan66° tan78° = 1

21) tan3x + cot3x = secxcosecx−2sinx cosx 29) sin47° + sin 61° − sin11° − sin25° = cos7°
1+ tan2 x 1+ cot2 x
30) 3 cosec20° − sec20° = 4

22) sin20° sin40° sin80° = 3 31) In ∆ABC, ∠ C = 2π then prove that
8 3
3
cos2 A + cos2 B − cosAcosB = 4

58

4 Determinants and Matrices

 Let's Study a b 1st row
c d 2nd row
· Definition and Expansion of Determinants
· Minors and Co-factors of determinants 1st 2nd
· Properties of Determinants
· Applications of Determinants column column
· Introduction and types of Matrices
· Operations on Matrices The value of the determinant ab
· Properties of related matrices ad – bc. c d is

4.1 Introduction SOLVED EXAMPLES

We have learnt to solve simultaneous 7 9 cosθ sin θ
equations in two variables using determinants. Ex. Evaluate i) −4 3 ii) − sinθ cosθ
We will now learn more about the determinants
because they are useful in Engineering 4 i where i2 = –1 iv) log 2 log 2
applications, and Economics, etc. iii) −2i 7 4 4

The concept of a determinant was 2 4
discussed by the German Mathematician G.W.
Leibnitz (1676-1714) and Cramer (1750) Solution :
developed the rule for solving linear equations
using determinants. 79
i) −4 3 = 7 × 3 − (−4) × 9 = 21 + 36 = 57

Let's Recall ii) cosθ sinθ = cos2θ − (−sin2θ)
− sinθ cosθ
4.1.1 Value of a Determinant
In standard X we have studied a method of = cos2θ + sin2θ = 1

solving simultaneous equations in two unknowns 4i = 4 × 7 − (−2i) × i = 28 + 2i2
using determinants of order two. In this chapter, iii) −2i 7 = 28 + 2(−1) [Qi2 = −1]
we shall study determinants of order three. = 28 − 2 = 26

ab
The representation c d is defined as the

determinant of order two. Numbers a, b, c, d iv) log 2 log 2 = 4×log4 2 − 2×log42
are called elements of the determinant. In this 4 4
arrangement, there are two rows and two columns.
2 4 = log424 − log422

59

= log416 − log44 a11 a12 a13
= 2 log44 − log44 D = a21 a22 a23
= 2 × 1 − 1 = 2 − 1 = 1
a31 a32 a33

The determinant can be expanded as follows:

Let's Understand D = a11 a22 a23 − a12 a21 a23 + a13 a21 a22
a32 a33 a31 a33 a31 a32
4.1.2 Determinant of order 3
Definition - A determinant of order 3 is SOLVED EXAMPLES

a square arrangement of 9 elements enclosed
between two vertical bars. The elements are
arranged in 3 rows and 3 columns as given below.

a11 a12 a13 R1 R1 are the rows Evaluate:
a21 a22 a23 R2 Cj are the column
a31 a32 a33 R3 3 -4 5
i) 1 1 - 2
C1 C2 C3
2 3 -1
Here aij represents the element in ith row and
jth column of the determinant. secθ tanθ 0

e.g. a31 represents the element in 3rd row and ii) tanθ secθ 0
1st column.
00 1
In general, we denote determinant by Capital
Letters or by ∆ (delta). 2-i 3 -1
iii) 3 2 - i 0 where i = -1
We can wirte the rows and columns
separately. e.g. here the 2nd row is [a21 a22 a23] 2 -1 2-i
and 3rd column is
Solution :

a13  3 -4 5
a23 
 1 -2 1 -2
i) 1 1 - 2 = 3 3 -1 − (−4) 2 -1
a33  2 3 -1

Expansion of Determinant 11
+5 2 3
We will find the value or expansion of a 3x3
determinant.We give here the expansion by the = 3(−1+6) + 4(−1+4) + 5(3−2)
1st row of the determinant D.
= 3×5+4×3+5×1
There are six ways of expanding a
determinant of order 3, corresponding to each = 15+12+5
of three rows (R1, R2, R3) and three columns
(C1, C2, C3). = 32

60

secθ tanθ 0 Definitions
ii) tanθ secθ 0
1 The minor of aij - It is defined as the
00 determinant obtained by eliminating the ith row
and jth column of A. That is the row and the
= secθ secθ 10 − ta n θ ta 0nθ 10 column that contain the element aij are omitted.
0 We denote the minor of the element aij by Mij

In case of above determinant A

+0 tan θ secθ a22 a23
0 0 Minor of a11 = M11 = a32 a33 = a22. a33 − a32 .a23

= secθ (secθ – 0) − tanθ (tanθ – 0) + 0 Minor of a12 = M12 = a21 a23 = a21. a33 − a31. a23
= sec2θ − tan2θ a31 a33
=1
Minor of a13 = M13 = a21 a22 = a21. a32 − a31. a22
a31 a32

2−i 3 −1 Similarly we can find minors of other
iii) 3 2 − i 0 elements.
2−i
2 −1 Cofactor of aij -

= (2 − i)[(2 − i)2 − 0] − 3[3(2 − i) − 0] cofactor of aij = (−1)i+j minor of aij = Cij
− 1[ − 3 − 2(2 − i)] ∴ Cofactor of element aij = Cij = (−1)i+j Mij
The same definition can also be given for
= (2 − i)3 − 9(2 − i) + 3 + 2(2 − i)
elements in 2×2 determinant. Thus in a b
= 8 − 12i + 6i2 − i3 − 18 + 9i + 3 + 4 − 2i cd

= 8 − 12i + 6(−1) + i − 18 + 9i + 3 + 4 − 2i The minor of a is d.
(since i2= −1) The minor of b is c.
The minor of c is b.
= 8 − 6 − 18+ 7 − 12i + 9i − 2i + i The minor of d is a
= − 9 − 4i

SOLVED EXAMPLES

Let's Learn Ex. 1) Find Minors and Cofactors of the
elements of determinant
4.1.3 Minors and Cofactors of elements of
determinants i) 2 −3
47
a11 a12 a13
Let A = a21 a22 a23 be a given determinant. Solution : Here a11 a12 = 2 −3
a21 a22 4 7
a31 a32 a33

M11 = 7
C11 = (−1)1+1 M11 = (−1)1+1 .7 = 7

61

M12 = 4 M23 = 1 2 = −1 −10 = −11
C11 = (−1)1+1 M12 = (−1)1+2 .4 = −4 5 -1

M21 = −3 C23 = (−1) 2+3 M23 = (−1) 2+3 . (−11) = 11
C21 = (−1)1+1 M21 = (−1)2 +1 . (−3) = 3
2 -3
M22 = 2 M31 = 0 4 = 8 − 0= 8
C22 = (−1)1+1 M22 = (−1)2+2 . 2= 2 C31 = (−1) 3+1 M31 = (−1) 3+1 . 8 = 8

1 2 -3 1 -3
ii) -2 0 4 M32 = -2 4 = 4 − 6 = −2
C32 = (−1)2 3+2 M32 = (−1) 3+2 . (−2) = 2
5 -1 3

Solution : 12
M33 = -2 0 = 0+4 = 4
a11 a12 a13 1 2 - 3
C33 = (−1) 3+3 M33 = (−1) 3+3 . 4 = 4
Here a21 a22 a23 = -2 0 4
Expansion of determinant by using Minor
a31 a32 a33 5 -1 3 and cofactors of any row/column

04 a11 a12 a13
M11 = -1 3 = 0 + 4 = 4 A = a21 a22 a23

C11 = (−1)1+1 M11 = (−1)1+1 . 4 = 4 a31 a32 a33
-2 4
= a11C11 + a12C12 + a13C13 (By 1st row)
M12 = 5 3 = −6 −20 = −26
= a12C12 + a22C22 + a32C32 (By 2nd column)
C12 = (−1)1+2 M12 = (−1)1+2 .(−26) = 26
Ex. 2) Find value of x if
-2 0 x -1 2
M13 = 5 = 2 − 0 = 2
-1 i) 2x 1 - 3 = − 10
3 -4 5
C13 = (−1)1+3 M13 = (−1)1+3 . 2 = 2
∴ x (5 − 12) − (−1) (10x + 9) + 2(−8x − 3)
2 -3 = −10
M21 = -1 3 = 6 − 3 = 3
C21 = (−1) 2+1 M21 = (−1) 2+1 . 3 = −3 ∴ x(−7) + (10x + 9) − 16x − 6 = −10

1 -3 ∴ −7x + 10x − 16x + 9 − 6 + 10 = 0
M22 = 5 3 = 3+15 = 18 ∴ 10x − 23x +13 =0
C22 = (−1) 2+2 M22 = (−1)1+1 . 18 = 18 ∴ 13x = 13
∴ x = 1

62

x32 = 2(0 + 6) − 5(0−2) – 1(3−2)
ii) x x 1 = 9
= 2(6) − 5(−2) – 1(1)
1 01 = 12 + 10 − 1
= 22 – 1
∴ x (x−0) − 3(x−1) + 2(0−x) =9 = 21
∴ x2 − 3x + 3 − 2x = 9
∴ x2 − 5x +3 =9 Interpretation: From (a) and (b) it is seen that
∴ x2 − 5x − 6 = 0 the expansion of determinant by both ways gives
∴ (x−6) (x+1) =0 the same value.
∴ x−6 = 0 or x+1 = 0
∴ x = 6 or x = −1 EXERCISE 4.1

1 −1 2 by Q.1) Find the value of determinant
Ex. 3) Find the value of −2 3 5
2 −4 2i 3 3 −4 5
−2 0 −1 7 −15 4 −i iii) 1 1 − 2

expanding along a) 2nd row b) 3rd column and i) ii)
Interprete the result.
23 1

a) Expansion along the 2nd row ahg
iv) h b f
= a21 c21 + a22 c22 + a23 c23
gfc
−1 2 12
= –2(−1)2+1 0 −1 + 3(−1)2+2 −2 −1
Q.2) Find the value of x if

+ 5(−1)2+3 1 −1 x2 − x +1 x +1 x −1 2
−2 0
i) x +1 x +1 = 0 ii) 2x 1 − 3 = 29
= 2(+1 −0) + 3(–1+4) − 5(0−2) 3 −4 5
= 2(1) + 3(3) − 5(−2)
= 2 + 9 + 10 Q.3 Find x and y if 4i i3 2i
= 21 i2 = −1 1 3i2 4 = x+iy where
5 −3 i

b) Expansion along 3rd coloumn

= a13 c13 + a23 c23 + a33 c33 Q.4) Find the minor and cofactor of element of
the determinant
= −2 3 + 5(−1)2+3 1 −1 +
2(−1)1+3 −2 0 −2 0 2 −1 3
D = 1 2 −1
1 −1
− 1(−1)3+3 −2 3 57 2

63

2 -3 5 = a1 (b2 c3 – b3c2) – a2(b1c3 + c1b3) +
Q.5) Evaluate A = 6 0 4 Also find minor a3(b1c2 – c1b2)

1 5 -7 = a1(b2 c3 – b3c2) – b1(a2c3 – a3c2) +
c1(a2b3 – b2a3) ---------- (ii)
and cofactor of elements in the 2nd row of
determinant and verify From (i) and (ii) D= D1
a) − a21.M21 + a22.M22 − a23.M23 = value of A
b) a21 C21 + a22 C22 + a23.C23 = value of A Ex.
1 2 -1
where M21, M22 , M23 are minor of a21 , a22, a23
and C21, C22, C23 are cofactor of a21 , a22, a23 Let A = 3 -1 2
02 1
Q.6) Find the value of determinant expanding
along third column -1 2 3 2 3 -1
= 1 2 1 – 2 0 1 – 1 0 2
-1 1 2
-2 3 - 4 = 1(–1–4) – 2 (3–0) – 1(6–0)
= –5 – 6 – 6
-3 4 0 = –17 ………………………….. (i)

4.2 Properties of Determinants by interchanging rows and columns of A we get
determinant A1
In the previous section we have learnt how
to expand the determinant. Now we will study 1 30 -1 2 2 2
some properties of determinants. They will help 1
us to evaluate the determinant more easily. A1 = 2 -1 2 = 1 2 1 -3 -1
-1 2 1

∙ Let’s Verify… 2 -1
+ 0 -1 2
Property 1 - The value of determinant
remains unchanged if its rows are turned into = 1(–1–4) – 3 (2+2) +0
columns and columns are turned into rows. = –5 –12
= –17 ………………………………...(ii)
Verification: ∴ A = A1 from (i) and (ii)

a1 b1 c1 Property 2 - If any two rows (or columns) of
Let D = a2 b2 c2 the determinant are interchanged then the value
of determinant changes its sign.
a3 b3 c3
The operation Ri ↔ Rj change the sign of the
= a1 .(b2 c3 − b3 c2) – b1 (a2c3 – a3c2) + determinant.

c1(a2b3 – a3b2 ) Note : We denote the interchange of rows by
---------- (i) Ri ↔ Rj and interchange of columns by Ci ↔ Cj.

a1 a2 a3 Property 3 - If any two rows (or columns)
of a determinant are identical then the value of
Let D1 = b1 b2 b3 determinant is zero.
c1 c2 c3

= a1 (b2 c3 – b3 c2) – a2 (b1c3 – c1b3) +
a3(b1c2 − c1b2)

64

R1 ↔ R2 D = D1 R1 → R1 + kR3
then D1 = −D ....... (property 2) ...... (I)
But R1 = R2 hence D1 = D .......... (II) A1 = a1 + ka3 b1 + kb3 c1 + kc3
∴ adding I and II a2 b2 c2
a3 b3 c3
2D1 = 0 ⇒ D1 = 0
Simplifying A1, using the previous properties,
i.e. D = 0 we get A1 = A.

Property 4 - If each element of a row (or 1 23
a column) of determinant is multiplied by a Ex. : Let B = -1 2 0 = 1(2–0) – 2(–1–0) +
constant k then the value of the new determinant
is k times the value of given determinant. 1 21
3(–2–2) = 2 + 2 – 12
The operation Ri → kRi gives multiple of the = 4–12 = –8 ----------(i)
determinant by k.
1 23
Remark i) Using this property we can take Now, B = -1 2 0
out any common factor from any one row (or any
one column) of the given determinant 1 21

ii) If corresponding elements of any two rows R1 → R1 + 2R2
(or columns) of determinant are proportional
(in the same ratio) then the value of the 1+ 2(−1) 2 + 2(2) 3 + 2(0)
determinant is zero. B1 = −1 2 0

Property 5 - If each element of a row 1 21
(or column) is expressed as the sum of two
numbers then the determinant can be expressed -1 6 3
as sum of two determinants B1 = -1 2 0 = –1(2–0) – 6(–1–0) + 3(–2–2)

For example, 1 21

a1 + x1 b1 + y1 c1 + z1 a1 b1 c1 x1 y1 z1 = –2 + 6 – 12 = 6 – 14 = –8 ----(ii)
b2 c2 = a2 b1 c1 + a2 b2 c2
a2 b3 c3 a3 b3 c3 a3 b3 c3 From (i) and (ii) B = B1
a3 Remark : If more than one operation from
above are done, make sure that these operations
Property 6 - If a constant multiple of all are completed one at a time. Else there can be
elements of any row (or column) is added to mistake in calculation.
the corresponding elements of any other row
(or column ) then the value of new determinant Main diagonal of determinant : The main
so obtained is the same as that of the original diagonal (principal diagonal) of determinant A is
determinant. The operation Ri ↔ Ri + kRj does collection of entries aij where i = j
not change the value of the determinant.
OR
Verification
Main diagonal of determinant : The set of
a1 b1 c1 elements ( a11 , a22, a33 , ---- ann) is called the main
A = a2 b2 c2 diagonal of the determinant A.

a3 b3 c3

65

a11 a12 a13 1 23
e.g. D = a21 a22 a23 here a11, a22 , a33 are = 100 505 606 707 by using property

a31 a32 a33 1 23
element of main diagonal
= 100 × 0 (R1 and R3 are identical)
Property 7 - (Triangle property) - If each = 0
element of a determinant above or below the main
diagonal is zero then the value of the determinant 312 313 314
is equal to product of its diagonal elements. ii) 315 316 317 = 0

that is 318 319 320

a1 b1 c1 a1 0 0 312 313 314
0 b2 c2 = a2 b2 0 = a1b2c3 L.H.S. = 315 316 317
0 0 c3 a3 b3 c3
318 319 320
Remark : If all elements in any row or any
column of a determinant are zeros then the value C2→ C2 – C1
of the determinant is zero.
312 1 314
= 315 1 317

318 1 320

SOLVED EXAMPLES C3→ C3 – C1

Ex. 1) Show that 312 1 2
= 315 1 2
101 202 303
i) 505 606 707 = 0 318 1 2

12 3 take 2 common from C3

101 202 303 312 1 1
LHS = 505 606 707 = 2 315 1 1

12 3 318 1 1

R1 → R1 – R3 = 2 (0) ( C2 and C3 are identical)

100 200 300 1 a bc 1 a a2
= 505 606 707 Ex. 2) Prove that 1 b ca = 1 b b2

12 3 1 c ab 1 c c2

100×1 100× 2 100× 3 1 a bc
L.H.S. = 1 b ca
= 505 606 707
1 c ab

12 3 R1 →aR1

66

a a2 abc xyz
= 1 1 b ca Ex. 3) If -x y z = k.xyz then find the value

a 1 c ab x - y z of k

R2 → bR2 Sloution :

a a2 abc xyz
= 1 × 1 b b2 abc L .H.S. = -x y z

a b 1 c ab x -y z

R3 → cR3 R2→R2 +R1

= 1a 1 1 a a2 abc x yz
b c = 0 2 y 2z
× × b b2 abc
c c2 abc x -y z

a a2 1 R3 → R3 + R1
= a1bc × abc b b2 1
xyz
c c2 1 = 0 2 y 2z

2x 0 2z

(taking abc common from C3) xyz
= 2 × 2 0 y z taking (2 common
a a2 1
= b b2 1 x0z
from R2 and R3)
c c2 1 = 4[x(yz) – y(0 – xz)+z(0 – xy)]
= 4[xyz + xyz – xyz]
C1 ↔C3 = 4xyz
From given condition
1 a2 a L.H.S. = R.H.S.
= (–1) 1 b2 b 4xyz = k xyz
∴ k = 4
1 c2 c
EXERCISE 4.2
C2 ↔C3
Q.1) Without expanding evaluate the following
1 a a2 determinants.
= (–1)(–1) 1 b b2

1 c c2

1 a a2 1 a b+c 23 4 2 7 65
= 1 b b2 = R.H.S.
i) 1 b c + a ii) 5 6 8 iii) 3 8 75
1 c c2
1 c a + b 6x 9x 12x 5 9 86

67

x+y y+z z+x 4.3 APPLICATIONS OF DETERMINANTS
Q.2) Prove that z + x x + y y + z 4.3.1 Cramer’s Rule

y+z z+x x+y In linear algebra Cramer’s rule is an explicit
formula for the solution of a system of linear
xyz equations in many variables. In previous class
=2 z x y we studied this with two variables. Our goal here
is to expand the application of Cramer’s rule to
yzx three equations in three variables (unknowns) .
Variables are usually denoted by x, y and z.
Q.3) Using properties of determinant show that
Theorem - Consider the following three linear
a+b a b equations in variables three x, y, z.
i) a a + c c = 4abc
a1x + b1y + c1z = d1
b c b+c a2x + b2y +c2z = d2
a3x + b3y + c3z = d3
1 logx y logx z Here ai , bi , ci and di are constants.
ii) log y x 1 log y z = 0 The solution of this system of equations is

logz x logz y 1 Dx Dy Dz
x= D ,y= D ,z= D
Q.5) Solve the following equations. provided D≠0 where

x + 2 x + 6 x −1
i) x + 6 x −1 x + 2 = 0

x −1 x + 2 x + 6

x -1 x x - 2 a1 b1 c1 d1 b1 c1
ii) 0 x - 2 x - 3 = 0 D = a2 b2 c2 Dx = d2 b2 c2

0 0 x-3 a3 b3 c3 d3 b3 c3

4+x 4−x 4−x a1 d1 c1 a1 b1 d1
Q.6) If 4 − x 4 + x 4 − x = 0 then find the Dy = a2 d2 c2 Dz = a2 b2 d2

4−x 4−x 4+x a3 d3 c3 a3 b3 d3

values of x Remark :

Q.7) Without expanding determinants show that 1) You will find the proof of the Cramer’s Rule
in QR code.

13 6 233 121 2) If D = 0 then there is no unique solution for
the given system of equations.
6 1 4 + 4 2 1 2 = 10 3 1 7

3 7 12 1 7 6 3 26

68

SOLVED EXAMPLES Ex. 2) By using Cramer’s rule solve the following
linear equations.
Ex. 1) Solve the following equation by using x +y – z = 1, 8x +3y – 6z = 1, –4x – y + 3z = 1
Cramer’s rule.
Solution : Given equations are
x+y+z = 6, x–y+z = 2, x+2y–z = 2 x + y – z = 1
8x + 3y – 6z = 1
Solution : Given equations are –4x – y + 3z =1
x+y+z = 6 x–y+z = 2 x+2y–z = 2
1 1 -1
11 1 D = 8 3 - 6
D = 1 -1 1
-4 1 3
1 2 -1
= 1(9–6) – 1(24–24) –1(–8+12)
= 1(1–2) – 1(–1–1) + 1(2+1) = 3 + 0 – 4
= –1 + 2 + 3 = –1
= –1 + 5
= 4 1 1 -1
Dx = 1 3 - 6
61 1
Dx = 2 -1 1 1 -1 3

2 2 -1 = 1(9–6) – 1(3+6) –1 (–1–3)
= 3 – 9 + 4
= 6(1–2) – 1(–2–2) + 1(4+2) = –2
= –6 + 4 + 6
= 4 1 1 -1
Dy = 8 3 - 6
16 1
Dy = 1 2 1 -4 1 3

1 2 -1 = 1(3+6) – 1(24–24) – 1(8+4)
= 9 – 0 – 12
= 1(–2–2) – 6(–1–1) + 1(2–2) = –3
= –4 + 12 + 0
1 11
= 8 Dz = 8 3 1

1 16 -4 -1 1
Dz = 1 -1 2
= 1(3+1) – 1(8+4) + 1(–8+12)
1 22 = 4 – 12 + 4
= 8 – 12
= 1(–2–4) – 1(2–2) + 6(2+1) = –4
= –6 + 0 + 18
= 12 ∴x= Dx -2 = 2, y = Dy = -3 = 3 and
D= -1 D -1
Dx 4 Dy 8 Dz
∴x= D= 4 = 1, y = D = 4 =2 and z D = Dz -4
D -1
∴z= = =4

12 = 3 are solutions of given equation. ∴x =2, y = 3, z = 4 are the solutions of the given
4 equations.

69

Ex. 3) Solve the following equations by using ∴ p = Dp = -5 = 5
determinant D -3 3 ,

1+ 1 +1 = –2 , 1 − 2 +1 =3 , 2 − 1 +3 = –1 Dq 5 -5
x y z x y z x y 2 q = q = -3 = 3 ,

Solution : Dr 6
r = D = -3 = –2
Put 1 =p 1 =q 1 =r
x y z 3
1 5 ∴ x = 5 ,
∴ Equations are ∴ x =p= 3
p + q + r = –2
p – 2q + r = 3 1 -5 -3
2p – q + 3r = –1 ∴ y = q = 3 ∴ y = 5 ,

1 11 1 -1
D = 1 - 2 1 ∴ z = r = –2 ∴ z = 2

2 -1 3 ∴x= 3 ,y= -3 ,z= -1 are the solutions of
5 5 2
= 1(–6+1) – 1(3–2) + 1(–1+4)
= –5 – 1 + 3 the equations.
= –3
Ex. 4) The cost of 2 books, 6 notebooks and
-2 1 1 3 pens is Rs.120. The cost of 3 books,
Dp = 3 - 2 1 4 notebooks and 2 pens is Rs.105. while the
cost of 5 books, 7 notebooks and 4 pens is
-1 -1 3 Rs.183. Using this information find the cost
of 1 book, 1 notebook and 1 pen.
= –2(–6+1) – 1(9+1) + 1(–3–2)
= 10 – 10 – 5 Solution : Let Rs. x, Rs. y and Rs. z be the cost of
= –5 one book, one notebook and one pen respectively.
Then by given information we have,
1 -2 1
Dq = 1 3 1 2x + 6y + 3z = 120

2 -1 3 3x + 4y + 2z = 105

= 1(9+1) + 2(3-2) + 1(–1–6) 5x + 7y + 4z = 183
= 10 + 2 – 7
263
= 5 D = 3 4 2

1 1 -2 574
Dr = 1 - 2 3
= 2(16–14) – 6(12–10) + 3(21–20)
2 -1 -1
= 2(2) – 6(2) + 3(1)
= 1(2+3) – 1(–1–6) – 2(–1+4)
= 5 + 7 –6 = 4 – 12 + 3
= 6
= 7 – 12

= –5

70

120 6 3 40 6 3 4.3.2 Consistency of three equations in two
Dx = 105 4 2 = 3 35 4 2 variables

183 7 4 61 7 4 Consider the system of three linear equations
in two variables x and y
= 3 [40(16–14) – 6(140–122) + 3(245–244)]
= 3[40(2) – 6(18) + 3(1)] a1x+ b1y +c1 = 0 (I)
= 3[80 –108 +3]
= 3[ 83 – 108] a2x+ b2y +c2 = 0 (II)
= 3[– 25] = –75
a3x+ b3y +c3 = 0 (III)
2 120 3 2 40 3
Dy = 3 105 2 = 3 3 35 2 These three equations are said to be
consistent if they have a common solution.
5 183 4 5 61 4
Theorem : The necessary condition for the
= 3[2(140–122) – 40(12–10) + 3(183–175)]
= 3[ 2(18) – 40(2) + 3(8)] equation a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0,
= 3[ 36–80+24] a3x + b3y + c3 = 0 to be consistent is
= 3[ 60–80]
= 3[–20] a1 b1 c1
= –60 a 2 b 2 c 2 = 0
a3 b3 c3

Proof : Consider the system of three linear
equations in two variables x and y.

2 6 40 2 6 120 a1x + b1y + c1 = 0 (I)
Dz = 3 4 35 = 3 3 4 105 a2x + b2y + c2 = 0
a3x + b3y + c3 = 0
5 7 61 5 7 183

= 3[2(244–245) – 6(183–175) + 40(21–20)] We shall now obtain the necessary condition
= 3[2(–1) – 6(8) + 40(1)] for the system (I) be consistent.
= 3[–2 – 48 + 40 ]
= 3[–50+40] Consider the solution of the equations
= 3[–10]
= –30 a2x+b2y = – c2
a3x+b3y = – c3

∴x= Dx = -75 = 15, y = Dy = -60 = 12, If a2 b2 ≠ 0 then by Cramer’s Rule the system
D -5 D -5 a3 b3

of two unknowns, we have

Dz -30 -c2 b2 a2 -c2
z = D = -5 = 6
x = -c3 b3 a3 -c3 put these
a2 b2 , y = a2 b2

∴ Rs.15, Rs. 12, Rs. 6 are the costs of one book, a3 b3 a3 b3
one notebook and one pen respectively.
values in equation a1x + b1y + c1 = 0 then

71

-c2 b2 a2 -c2 x + y − 2 = 0, 2x +3y − 5 = 0, 3x − 2y − 1 = 0

a1 -c3 b3 + b1 a3 -c3 + c1 = 0 a1 b1 c1 1 1 -2
a2 b2 a2 b2
a2 b2 c2 = 2 3 -5
a3 b3 a3 b3 a3 b3 c3 3 -2 -1

i.e a1 -c2 b2 + b1 a2 -c2 + c1 a2 b2 = 0 = 1(–3–10) –1(–2+15) –2(–4–9)
-c3 b3 a3 -c3 a3 b3
= –13 –13 + 26 = 0
i.e − a1 c2 b2 − b1 a2 c2 + c1 a2 b2 = 0 ∴ Given equations are consistent.
c3 b3 a3 c3 a3 b3
ii) x + 2y − 3 = 0, 7x + 4y − 11 = 0,
b2 c2 a2 c2 a2 b2 = 0 2x +3y + 1 = 0
b3 c3 a3 c3 a3 b3
i.e a1 − b1 + c1 a1 b1 c1 1 2 -3

a1 b1 c1 a2 b2 c2 = 7 4 -11
i.e a2 b2 c2 = 0
a3 b3 c3 23 1
a3 b3 c3
= 1(4+33) –2(7+22) –3(21–8)
Note : This is a necessary condition that the
equations are consistant. The above = 37 –58 39 = 37 –97
condition of consistency in general is not
sufficient. = −60 ≠ 0
∴Given system of equations is not consistent.

SOLVED EXAMPLES iii) x + y = 1, 2x + 2y = 2, 3x + 3y = 5
Solution : x + y = 1, 2x + 2y = 2, 3x + 3y = 5 are
Ex. 1) Verify the consistency of following given equations.
equations
Check the condition of consistency.
2x+2y = −2, x + y = −1, 3x + 3y = −5
1 1 -1 1 1 -1
Solution : By condition of consistency consider 2 2 -2 = 2 1 1 -1
3 3 -5 3 3 -5
aa12 bb12 cc12 = 2 2 2
a3 b3 c3 1 1 1 = 2(0) = 0 (R1 and R2 are identical)
3 3 5 Let us examine further.

= 2(5−3) − 2(5−3) + 2(3−3) = 4 − 4 + 0 = 0 Note that lines given by the equations
x + y = 1 and 3x + 3y = 5 are parallel to each
But the equations have no common Solution. other. They do not have a common solution, so
(why?) equations are not consistent.

Ex. 2) Examine the consistency of following Ex. 3) Find the value of k if the following
equations. equations are consistent.

i) x + y = 2, 2x +3y = 5, 3x − 2y = 1 7x – ky = 4 , 2x + 5y = 9 and 2x + y = 8

Solution : Write the given equation in standard Solution : Given equations are
form.
7x – ky – 4 = 0, 2x + 5y –9 = 0,

72

2x + y – 8 = 0 are consistent Area of triangle ABC = 1 PQ.[AP+CQ]
2
7 -k 4
∴ 2 5 -9 = 0 + 1 QR.[QC+BR]– 1 PR.[AP+BR]
22
2 1 -8

∴ 7(–40+9) + k(–16+18) −4(2−10) = 0 = 1 (y1+y3) (x3–x1) + 1 (y2+y3) (x2– x3)
2 2
∴ 7(–31) +2k –4(–8) = 0
– 1 (y1+y2) (x2–x1)
∴ –217 + 2k + 32 = 0 –185 + 2k = 0 2

∴ 2k = 185 ∴ k = 185 = 1 [y1.x3 –x1y1 + x3y3–x1y3+x2y2−x3y2 +x2y3
2 2

4.3.3 Area of triangle and Collinearity of –x3y3−x2y1+x1y1–x2y2+x1y1]
three points.
= 1 [ y1x3–x1y3–x3y2+x2y3–x2y1+x1y2]
2

Theorem : If A(x1 , y1) , B(x2 , y2) and C(x3 , y3) = 1 [x1(y2–y3) – x2(y1–y3) + x3(y1–y2)]
are vertices of triangle ABC then the area of 2
triangle is
=1 x1 x2 x3 1 x1 y1 1
2 2
y1 y2 y3 = x2 y2 1
1 1 1 x3 y3 1
x1 y1 1
1 x2 y2 1 Remark:

2 x3 y3 1 i) Area is a positive quantity. Hence we
always take the absolute value of a determinant.
Proof : Consider a triangle ABC in Cartesian
coordinate system. Draw AP, CQ and BR ii) If area is given, consider both positive and
perpendicular to the X axis negative values of the determinant for calculation
of unknown co-ordinates.

iii) If area of a triangle is zero then the given
three points are collinear.

SOLVED EXAMPLES

Ex. 1) Find the area of the triangle whose vertices
are A(−2, −3) , B(3, 2) and C(−1, −8)

Fig. 4.1 Solution : Given (x1, y1) = (−2, −3), (x2, y2) =
(−2, −3), and (x3, y3) = (−1 , −8)
From the figure,
Area of ∆ABC = Area of trapezium PACQ +Area We know that area of triangle
of trapezium QCBR –Area of trapezium PABR
x1 y1 1 -2 -3 1
= 1 x2
y2 1 = 1 3 2 1
2 x3 y3 1 2 -1 -8 1

73

= 1 [–2(2+8)+3(3+1)+1(–24+2)] Solution : Given (x1, y1) ≡ (3, 7), (x2, y2) ≡ (4, –3)
2 and (x3, y3) ≡ (5, –13)

= 1 [−20+12−22] Area of ∆ = 1 x1 y1 1 3 71
2 2 x2 y2 1 = 1 4 −3 1
x3 y3 12 5 −13 1
= 1 [−42+12] = 1 [–30] = –15
22 = 1 [3(–3+13) – 7(4–5) + 1(−52+15)]
2
Area is positive.
= 1 [30+7–37] = 1 [37–37] = 0
∴ Area of triangle = 15 square unit 22

This gives the area of the triangle ABC A(∆ABC) = 0 ∴ A, B, C are collinear points
in that order of the vertices. If we consider the
same triangle as ACB, then triangle is considered Ex. 4) Show that the following points are collinear
in opposite orientation. The area then is 15 by determinant method.
sq. units. This also agrees with the rule that
interchanging 2nd and 3rd rows changes the sign A(2, 5), B(5, 7), C(8, 9)
of the determinant.

Ex. 2) If the area of triangle with vertices Solution : Given A ≡ (x1, y1) = (2, 5) ,
P(−3, 0), Q(3, 0) and R(0, K) is 9 square B ≡ (x2, y2) ≡ (5, 7), C ≡ (x3, y3) ≡ (8,9)
unit then find the value of k.
x1 y1 1
Solution : Given (x1, y1) ≡ (–3, 0), (x2, y2) If x2 y2 1 = 0 then, A, B, C are collinear
≡ (3, 0) and (x3, y3) ≡ (0, k) and area of ∆ is 9 sq. y3 1
unit. x3

We know that area of ∆ = 1 x1 y1 1 2 51
2 x2 y2 1 ∴ 5 7 1 = 2(7–9) –5(5−8) +1(45–56)
x3 y3 1
8 91

∴ ±9 = 1 −3 0 1 = –4 +15 − 11 = –15 +15 = 0
2 ∴ A, B, C are collinear.
3 0 1 ( Area is positive but the

0 k1

determinant can be of either sign) EXERCISE 4.3

∴ ±9 = 1 [–3(0 – k) + 1(3k – 0)] Q.1) Solve the following linear equations by
2 using Cramer’s Rule.
i) x+y+z = 6, x–y+z = 2, x+2y–z = 2
∴ ±9 = 1 [3 × 3k] ∴ ±9 = 3k ∴ k = ±3
2 ii) x+y−2z = –10,
2x+y–3z = –19, 4x+6y+z = 2
Ex. 3) Find the area of triangle whose vertices are
A(3, 7) B(4, −3) and C(5, –13). Interpret iii) x+z = 1, y+z = 1, x+y = 4
your answer.

74

iv) −2 − 1 − 3 = 3 , 2 − 3 + 1 = –13 MISCELLANEOUS EXERCISE - 4 (A)

x yz xyz

and 2 − 3 = –11 (I) Select the correct option from the given
x z alternatives.

Q.2) The sum of three numbers is 15. If the ab a+b
second number is subtracted from the sum b+c
of first and third numbers then we get 5. Q.1 The determinant D = b c
When the third number is subtracted from 0
the sum of twice the first number and the a+b b+c
second number, we get 4. Find the three
numbers. = 0 lf

A) a,b, c are in A.P.

B) a , b, c are in G.P.

Q.3) Examine the consistency of the following C) a, b, c are in H.P.
equations.
D) α is root of ax2 + 2bx + c = 0
i) 2x−y+3 = 0, 3x+y−2=0, 11x+2y−3 = 0
ii) 2x+3y−4=0, x+2y=3, 3x+4y+5 = 0 xk xk+2 xk+3
iii) x+2y−3 =0, 7x+4y−11=0, 2x+4y−6= 0 Q.2 If yk yk+2 yk+3 = (x–y) (y–z) (z–x)

Q.4) Find k if the following equations are zk zk+2 zk+3
consistent.
i) 2x+3y-2=0, 2x+4y−k=0, x−2y+3k =0 ( 1 + 1 + 1z ) then
x y
ii) kx +3y+1=0, x+2y+1=0, x+y=0
A) k= –3 B) k = –1 C) k = 1 D) k = 3

Q.5) Find the area of triangle whose vertices are sinθ .cosφ sinθ .sinφ cosθ
Q.3 Let D = cosθ .cosφ cosθ .sinφ −sinθ then
i) A(5,8), B(5,0) C(1,0)
ii) P( 3 , 1), Q(4, 2), R(4, −1 ) −sinθ .sinφ sinθ .cosφ 0

22 A) D is independent of θ
iii) M(0, 5), N(−2, 3), T(1, −4)

Q.6) Find the area of quadrilateral whose B) D is independent of φ
vertices are C) D is a constant
D) dD at θ =π/2 is equal to 0
A(−3, 1), B(−2, −2), C(3,−1), D(1,4)
d
Q.7) Find the value of k, if the area of triangle
whose vertices are P(k, 0), Q(2, 2), R(4, Q.4 The value of a for which system of equation
3) is 3 sq.unit a3x + (a + 1)3y + (a + 2)3z = 0 ax + (a +1) y
2 + (a + 2) z = 0 and x + y + z = 0 has non zero
Soln. is
Q.8) Examine the collinearity of the following
set of points A) 0 B) –1 C) 1 D) 2
i) A(3, −1), B(0, −3), C(12, 5)

ii) P(3, −5), Q(6, 1), R(4, 2)
iii) L(0, 1 ), M (2, −1), N(−4, 7 )

22

75

b+c c+a a+b C) Determinant is number associated to
r+ p square matrix
Q.5 q + r z+x p+q =
y+z x+ y D) None of these

cba bac abc (II) Answer the following questions.

A) 2 r q p B) 2 q p r C) 2 p q r 2 −5 7 1 −3 12

zyx yxz x yz Q.1) Evaluate i) 5 2 1 ii) 0 2 −4

acb 902 97 2
D) 2 p r q
Q.2) Evaluate determinant along second column
xzy
1 −1 2
Q.6 The system 3x – y + 4z = 3, x + 2y –3z 3 2 −2
= –2 and 6x + 5y + λz = –3 has at least one
Solution when 0 1 −2

A) λ = –5 B) λ = 5 23 5
Q.3) Evaluate i) 400 600 1000
C) λ = 3 D) λ = –13
48 47 18

x37 101 102 103
ii) 106 107 108 by using properties
Q.7 If x = –9 is a root of 2 x 2 = 0 has
123
76x

other two roots are Q.4) Find minor and cofactor of elements of the
determinant.
A) 2, –7 B) –2, 7 C) 2, 7 D) –2, –7
−1 0 4 1 −1 2
6i −3i 1 i) −2 1 3 ii) 3 0 −2
Q.8 If 4 3i −1 = x + iy then
0 −4 2 1 0 3
20 3 i
Q.5) Find the value of x if
A) x = 3 , y = 1 B) x = 1 , y = 3
C) x = 0 , y = 3 D) x = 0 , y = 0 1 4 20 1 2x 4x

Q.9 If A(0,0), B(1,3) and C(k,0) are vertices of i) 1 −2 −5 = 0 ii) 1 4 16 = 0
triangle ABC whose area is 3 sq.units then
value of k is 1 2x 5x2 11 1

A) 2 B) –3 C) 3 or −3 D) –2 or +2 Q.6) By using properties of determinant prove

Q.10 Which of the following is correct x+y y+z z+x
A) Determinant is square matrix that z x y = 0
B) Determinant is number associated to
matrix 111

76

Q.7) Without expanding determinant show that iii) (k−2)x +(k−1)y =17 ,
(k−1)x+ (k−2)y = 18 and x + y = 5
b + c bc b2c2 xa yb zc
Q.11) Find the area of triangle whose vertices are
i) c + a ca c2a2 = 0 ii) a2 b2 c2 i) A(−1,2), B(2,4), C(0,0)

a + b ab a2b2 111 ii) P(3,6), Q(−1,3), R (2,−1)

xyz iii) L(1,1), M (−2,2), N (5,4)
= a b c
Q.12) Find the value of k
bc ca ab i) If area of triangle is 4 square unit and
vertices are P(k, 0), Q(4, 0), R(0, 2)
lmn nfw ii) If area of triangle is 33/2 square unit and
vertices are L (3,−5), M(−2,k), N (1,4)
iii) e d f = l e u
Q.13) Find the area of quadrilateral whose
uvw md v vertices are A (0, −4), B(4, 0) , C(−4, 0),
D (0, 4)
0 ab
iv) −a 0 c = 0 Q.14) An amount of ` 5000 is put into three
investments at the rate of interest of 6% , 7%
−b −c 0 and 8% per annum respectively. The total
annual income is ` 350. If the combined
a11 income from the first two investments is
Q.8) If 1 b 1 =0 then show that ` 70 more than the income from the third.
Find the amount of each investment.
11c
Q.15) Show that the lines x−y=6, 4x−3y=20 and
1 + 1 + 1 = 1 6x+5y+8=0 are concurrent .Also find the
1− a 1−b 1− c point of concurrence

Q.9) Solve the following linear equations by Q.16) Show that the following points are collinear
Cramer’s Rule. by determinant

i) 2x−y+z = 1, x+2y+3z = 8, 3x+y−4z =1 a) L (2,5), M(5,7), N(8,9)
b) P(5,1), Q(1,−1), R(11,4)
ii) 1 + 1 = 3 , 1 + 1 = 5 , 1 + 1 = 4
x y 2 y z 6 z x 3

iii) 2x+3y+3z=5 , x−2y+z = – 4 , Further Use of Determinants
3x– y– 2z=3
1) To find the volume of parallelepiped and
iv) x–y+2z=7 , 3x+4y–5z=5 , 2x–y+3z=12 tetrahedron by vector method

Q.10) Find the value of k if the following equation 2) To state the condition for the equation
are consistent. ax2+2hxy+by2 +2gx+2fy+c =0 representing a
pair of straight lines.
i) (k+1)x+(k−1)y+(k−1) = 0
(k−1)x+(k+1)y+(k−1) = 0 3) To find the shortest distance between two
(k−1)x+(k−1)y+(k+1) = 0 skew lines.

ii) 3x+y−2=0 kx+2y−3=0 and 2x−y = 3 4) Test for intersection of two line in three
dimensional geometry.

77

5) To find cross product of two vectors and B is a matrix having 3 rows and 2 columns.
scalar triple product of vectors The order of B is 3×2. There are 6 elements in
matrix B.
6) Formation of differential equation by
eliminating arbitrary constant. 1+ i 8 
iii) C =  i −3i , C is a matrix of order 2×2.

Let's Learn −1 9 2 

4.4 Introduction to Matrices : iv) D =  3 0 −3 , D is a matrix of order 2×3.

The theory of matrices was developed by
a Mathematician Arthur Cayley. Matrices are In general a matrix of order m x n is represented
useful in expressing numerical information
in compact form. They are effectively used by
in expressing different operators. Hence in
Economics, Statistics and Computer science they  a11 a12 … a1j … a1n 
are essential.  
 a 21 a 22 … a2j … a 2n 
Definition : A rectangular arrangement of mn
numbers in m rows and n columns, enclosed in A = [ aij]mxn =  a 31 a 32 … a3j … a3n 
[ ] or ( ) is called a matrix of order m by n.  … … … … … … 
 
A matrix by itself does not have a value or any  
special meaning.  a i1 ai2 … a ij … a in 

Order of the matrix is denoted by m × n, read as am1 am2 … amj … amn 
m by n.
Here aij = An element in ith row and jth column.
Each member of the matrix is called an element
of the matrix. 2 −3 9  a11 a12 a13 
Ex.In matrixA= 1 −7 = a21 a22 
Matrices are generally denoted by A, B, C ,… and 0 a 23 
their elements are denoted by aij, bij, cij, … etc.
e.g. aij is the element in ith row and jth column of 4 −2 1  a31 a32 a33 
the matrix.
a11 = 2, a12 = –3, a13 = 9, a21 = 1, a22 = 0, a23 =
2 −3 9  –7, a31 = 4, a32 = –2, a33 = 1
For example, i) A = 1 0 −7 Here a32 = – 2
4.4.1 Types of Matrices :
4 −2 1 
1) Row Matrix : A matrix having only one
row is called as a row matrix. It is of order
1 x n, Where n ≥ 1.

Ex. i) [–1 2]1×2 ii) [0 –3 5]1×3

A is a matrix having 3 rows and 3 columns. The 2) Column Matrix : A matrix having only one
order of A is 3×3, read as three by three. There column is called as a column matrix. It is of
are 9 elements in matrix A. order m x 1, Where m ≥ 1.

−1 −5 1 5
Ex. i) 02x1 ii) −9
ii) B =  2 0  −33x1
 
 6 9 

78

Note : Single element matrix is row matrix 5) Diagonal Matrix : A square matrix in which
as well as column matrix. e.g. [5]1x1 every non-diagonal element is zero, is called
a diagonal matrix.

3) Zero or Null matrix : A matrix in which 5 0 0
every element is zero is called as a zero or Ex. i) A = 0 0 0 = diag (5,0,9)
null matrix. It is denoted by O.
0 0 93x3
0 0 0
Ex. i) O = 0 0 0 −1 0 
ii) B =  0 −52x2
0 0 03x3

0 0 −1 0 0 
ii) O = 0 0  
iii) C=  0 −2 0 
0 03x2
 0 0 −32x2

4) Square Matrix : A matrix with equal Note: If a11, a22, a33 are diagonal elements of a
number of rows and coloumns is called a diagonal matrix A of order 3, then we write the
square matrix.
matrix A as A = Diag.
 5 −3 i 
 −7
Examples, i) A =  1 0 6) Scalar Matrix : A diagonal matrix in which
all the diagonal elements are same, is called
2i −8 9 3x3 as a scalar matrix.

ii) C = −1 0 5 0 0
 1 −5 2x 2 For Ex. i) A = 0 5 0

Note : A matrix of order n×n is also called as 0 0 53x3
square matrix of order n.

Let A = [ aij]nxn be a square matrix of order n then ii) B = −2 0 
 −2  2 x 2
(i) The elements a a11, 22, a33,… aii … ann are called the  0
diagonal elements of matrix A.
7) Unit or Identity Matrix : A scalar matrix in
Note that the diagonal elements are defined which all the diagonal elements are 1(unity) ,
only for a square matrix. is called a Unit Matrix or an Identity Matrix.
An Identity matrix of order n is denoted
(ii) Elements aij, where i ≠ j are called non by In.
diagonal elements of matrix A.
1 0 0 1 0
(iii) Elements aij, where i < j represent elements Ex. i) I3 = 0 1  0 1
above the diagonal. 0 0  ii) I2 =
0
(iv) Elements aij, where i > j represent elements Note: 1
below the diagonal.
1. Every Identity matrix is a Scalar matrix but
Statements iii) and iv) can be verified by observing every scalar matrix need not be Identity
square matrices of different orders. matrix. However a scalar matrix is a scalar
multiple of the identity matrix.

79

2. Every scalar matrix is a diagonal matrix but ii) B = −3 1
every diagonal matrix need not be a scalar  8 2 x 2
matrix.  1

8) Upper Triangular Matrix : A square matrix  2 4 −7
in which every element below the diagonal  5 −1
is zero, is called an upper triangular matrix. iii) C=  4

Matrix A = [ aij]nxn is upper triangular if −7 −1 −33x3
aij = 0 for all i > j.
Note:
4 −1 2
For Ex. i) A = 0 0 3 The scalar matrices are symmetric. A null square
matrix is symmetric.
0 0 93x3
12) Skew-Symmetric Matrix : A square matrix
ii) B = −3 1 A = [ aij]nxn in which aij = − aji, for all i and j,
 8 2 x 2 is called a skew symmetric matrix.
 0
Here for i = j, aii = − aii, 2aii = 0 aii = 0 for
9) Lower Triangular Matrix : A square matrix all i = 1, 2, 3,……n.
in which every element above the diagonal
is zero, is called a lower triangular matrix. In a skew symmetric matrix each diagonal
element is zero.

Matrix A = [ aij]nxn is lower triangular if e.g.  0 5
aij = 0 for all i < j. i) A = −5 02x2

 2 0 0  0 4 −7
For Ex. i) A = −1 1 0 ii) B = −4 0 
5 
−5 1 93x3
 7 −5 0 3x3

 7 0 Note : A null square matrix is also a skew
ii) B = −1 32x2 symmetric.

10) Triangular Matrix : A square matrix is 13) Determinant of a Matrix : Determinant of
called a triangular matrix if it is an upper a matrix is defined only for a square matrix.
triangular or a lower triangular matrix.
If A is a square matrix, then the same arrangement
Note : The diagonal, scalar, unit and null matrices of the elements of A also gives us a determinant,
are also triangular matrices. by replacing square brackets by vertical bars. It is
denoted by |A| or det(A).
11) Symmetric Matrix : A square matrix A =
If A = [ aij]nxn, then is of order n.
[ aij]nxn in which aij = aji, for all i and j, is called
a symmetric matrix.  1 3
Ex. i) If A = −5 42x2
a h g
i) A = h  13
Ex. b f  then |A| = -5 4

g f c3x3

80

 2 −1 3 15) Transpose of a Matrix : The matrix
ii) If B = −4 1 5 obtained by interchanging rows and columns
of matrix A is called Transpose of matrix A.
 7 −5 03x3 It is denoted by A' or AT. If A is matrix of
order m × n, then order of AT is n × m.
2 -1 3
then |B| = -4 1 5 If AT = A' = B then bij = aji

7 -5 0 −1 5 
 −2
14) Singular Matrix : A square matrix A is e.g. i) If A =  3
said to be a singular matrix if |A| = det(A)
= 0, otherwise it is said to be a non-singular  4 7  3×2
matrix.
then AT = −1 3 4
6 3  −2 7 2×3
Ex. i) If A = 8 42x2  5

then |A| = 6 3 = 24–24 = 0. 1 0 −2
84 ii) If B = 8 
−1 2 
Therefore A is a singular matrix.
2 3 4 4 3 5  3×3

ii) If B = 3 4 5 then  1 8 4
4 5 63x3  3
then BT =  0 −1
234
|B| = 3 4 5 −2 2 5 3×3

456 Remark:

1) If A is symmetric then A = AT

2) If B is skew symmetric, then is B = −BT

|B| = 2(24–25)–3(18–20) + 4(15–16) Activity :
= – 2+6–4 = 0
Construct a matrix of order 2 × 2 where the aij th
|B| = 0 Therefore B is a singular matrix.
(i + j)2
 2 −1 3 element is given by aij = 2+i
iii) A = −7 4 5 then
−2 1 63X 3 Solution : Let A =  a11 a12  be the required
a 21 
2 -1 3 a 22  2 X 2
|A| = -7 4 5
matrix.
-2 1 6
Given that aij = (i + j)2 , a11 = (……..)2 = 4 ,
2+i …… + 1 3

|A| = 2(24-5)-(–1)(-42+10)+3(–7+8) a12 = (........)2 = 9 = ......
= 38–32+3 = 9 ........... 3

|A| = 9, As |A| ≠ 0 , A is a non-singular a21 = (2 +1)2 = ........ ,
matrix. 2+2 4

81

a22 = (……….)2 = ........ = 4  2 a 3
2+2 ........ Ex. 3) Find a, b, c if the matrix A = −7 4 5

4 … is a symmetric matrix.  c b 6

∴ A=  3   2 a 3

.....  Solution : Given that A = −7 4 5 is a
..... 4 

symmetric matrix.  c b 6

SOLVED EXAMPLES aij = aji for all i and j
As a12 = a21
x + y y + z z + x ∴ a = –7
 
Ex. 1) Show that the matrix  1 1 1  As a32 = a23
∴ b = 5
is a singular matrix.  z x y 
As a31 = a13
x + y y + z z + x ∴ c = 3
 
Solution : Let A =  1 1 1 

 z x y 

x+y y+z z+x EXERCISE 4.4

∴ |A| = 1 1 1
z xy

Now |A| = (x+y)(y−x)−(y+z)(y−z)+(z+x)(x−z) (1) Construct a matrix A = [aij]3×2 whose elements
= y2 – x2 – y2 + z2 + x2 – z2
=0 aij are given by (i) aij = (i - j)2
5-i
∴ A is a singular matrix.
(ii) aij = i – 3j (iii) aij = (i + j)3
5

−1 −5
 
Ex. 2) If A =  2 0  Find (AT)T. (2) Classify the following matrices as, a row,
a column, a square, a diagonal, a scalar, a
 3 −4  3X 2 unit, an upper triangular, a lower triangular,
a symmetric or a skew-symmetric matrix.
−1 −5
 
Solution : Let A =  2 0  3 −2 4  0 4 7

 3 −4  3X 2 (i) 0 0 −5 (ii) −4 0 −3

∴ AT =  −1 2 3 0 0 0  −7 3 0 
−5 0 −4  2 x 3

−1 −5 5
 
  (iii)  4  (iv) 9 2 −3
 
∴ (AT)T = 2 0 =A −3

 3 −43X2

82

6 0  2 0 0  7 3 1
(v) 0 6  0 (6) If A = −2 −4 1 , Find (AT)T.
(vi)  3 −1 1
3  5 9 1
−7

    1 3 a 
3 0 0  10 −15 27   
(vii) 0  −15  5
5 0  (viii) 0 34  (7) Find a, b, c if  b −5  is a symmetric
 −7
    −4 c 
0 0 1   27 34 5  0 

 3 3 matrix. 

1 0 0 0 0 1 
(ix) 0 1 0 (x) 0 1 0  0 −5i x
 
0 0 1 1 0 0 (8) Find x, y, z if  y 0 z  is a skew

 3 −2 
 2 0
(3) Which of the following matrices are singular
or non singular ? symmetric matrix.

a b c (9) For each of the following matrices, using its
  transpose state whether it is a symmetric, a
(i)  p q r  skew-symmetric or neither.

2a − p 2b − q 2c − r  1 2 −5  2 5 1

5 0 5   3 5 7 (i)  2 −3 4  (ii) −5 4 6
 
−5 4 9  −1 −6 3
(ii) 1 99 100 (iii) −2 1 4
6 99 105  3 2 5
0 1+ 2i i − 2
(iii) −1− 2i 0 
7 −7 
 2 − i
(iv) 7 5 0 
−4 7
(10) Construct the matrix A = [ aij]3×3 where
(4) Find k if the following matrices are singular aij = i−j. State whether A is symmetric or
skew symmetric.
7 3   4 3 1
−2 k   k 1 4.5 Algebra of Matrices :
(i)  (ii)  7 (1) Equality of matrices
(2) Multiplication of a matrix by a scalar
10 9 1 (3) Addition of matrices
(4) Multiplication of two matrices.
k −1 2 3
 1 2 (1) Equality of matrices : Two matrices A and
(iii)  3 B are said to be equal if (i) order of A = order
of B and (ii) corresponding elements of
 1 −2 4 A and B are same, that is if aij=bij for; all i,j
and symbolically written as A=B.
(5) If A = 5 1 −1 , Find (AT)T.
3 2 
0 

83

15 14 (3) Addition of Two matrices : A and B are
Ex. (i) If A 12 102x2 two matrices of same order. Their addition
denoted by A + B is a matrix obtained
15 14 15 14 by adding the corresponding elements of A
B = 10 122x2 and C = 10 122x2 and B.

Here A ≠ B, A ≠ C but B = C by definition of Note: A+B is possible only when Aand B are
equality. of same order.

A+B is of the same order as that of A and B.

2a − b 4  1 4  Thus if A = [ aij]mxn and B = [ bij]mxn then A+B
Ex. (ii) If  2 = −7 a + 3b = [ aij + bij]mxn
 −7

then using definition of equality of matrices, we  2 3 1
Ex. A = −1 −2 02x3 and
have 2a – b = 1 …...(1) and a + 3b = 2 …… (2)

Solving equations (1) and (2), we get a = 5 and −4 3 1 
b= 3 7
B=  5 7 −82x3 Find A+B.

7

(2) Multiplication of a Matrix by a scalar: Solution : By definition of addition,
If A is any matrix and k is a scalar, then
the matrix obtained by multiplying each A+B = 2+( − 4) 3+3 1+1 
element of A by the scalar k is called the  −2+7 0+( − 8)2x3
scalar multiple of the given matrix A and is  −1+5
denoted by kA.
−2 6 2 
=  4 5 −82x3

Thus if A = [aij]mxn and k is any scalar then Note : If A and B are two matrices of the same
kA = [kaij]mxn. order then subtraction of the two matrices is
defined as, A–B = A+(–B), where –B is the
Here the order of matrix A and kA are same. negative of matrix B.

15 3 14 15
and k = 2 ,
Ex. A = 3 2 Ex. If A = 3 2 and B = 2 6,
9
47 3x2 0 5 4
Find A–B. 3x2 3x2

then kA = 3 A Solution : By definition of subtraction,
2

3 15 −1 4 1 −5
−2 + −2 
15 22 A–B = A+(–B) =  3 5  −4 6 
93 
3 2  0 −9
= 2 3 2 = 6 21

47 3x2 2 3x2  −1+1 4 + (−5)  0 −1

= 3 + (−2) −2 + 6  =  1 4 
  
0 + (−4) 5 + (−9) −4 −4

84

Some Results on addition and scalar 10 −6 −6 −21
multiplication : If A, B, C are three matrices    
conformable for addition and α, β are scalars, =  2 0  +  9 −3 
then
−8 −4 −6 6 
(i) A+B = B+A, That is, the matrix addition is
commutative. 10 − 6 −6 − 21  4 −27
   
(ii) (A+B)+C = A+(B+C), That is, the matrix =  2+9 0−3  =  11 −3 
addition is associative.
−8 − 6 −4 + 6  −14 2 
(iii) For matrix A, we have A+O = O+A = A, That
is, a zero matrix is conformable for addition Ex. 2) If A = diag(2, –5, 9), B = diag(–3, 7, –14)
and it is the identity for matrix addition. and C = diag(1, 0, 3), find B−A−C.

(iv) For a matrix A, we have A+(−A) = (−A) +A Solution : B−A−C = B−(A+C)
= O, where O is a zero matrix conformable Now, A+C = diag(2, –5, 9) + diag(1, 0, 3)
with matrix A for addition. Where (−A) is
additive inverse of A. = diag(3, −5, 12)

(v) α(A±B) = αΑ ± αB B−A−C= B–(A+C)
=diag(–3, 7, –14)–diag(3, –5, 12)
(vi) (α ± β)A = αΑ ± βA
−6 0 0 
(vii) α(β·A) = (α·β)·A  
= diag(−6, 12, −26) =  0 12 0 
(viii) OA = O
 0 0 −26

2 3 −1 1 3 2 
Ex. 3) If A = 4 7 5  , B = 4 6 −1 and

SOLVED EXAMPLES 1 −1 6 
C = 0 2 −5 , find the matrix X such that
5 −3 3A−2B+4X = 5C.
  Solution : Since 3A–2B+4X = 5C
Ex. 1) If A =  1 0  and ∴ 4X = 5C–3A+2B

−4 −2 1 −1 6  2 3 −1
∴ 4X = 5 0 2 −5 –3 4 7 5 
2 7
B = −3  1 3 2 
1  , find 2A – 3B. +2 4 6 −1

 2 −2 5 −5 30   −6 −9 3 
= 0 10 −25 + −12 −21 −15
Solution : Let 2A – 3B
2 6 4 
5 −3  2 7 + 8 12 −2
  –3 −3 
= 2  1 0  1 

−4 −2  2 −2

85

 5 − 6 + 2 −5 − 9 + 6 30 + 3 + 4  1
= 0 −12 + 8 10 − 21+12 −25 −15 − 2 By (1) – (2) , 3Y = A – B , ∴ Y = 3 (A – B)

 1 −8 37  1  2 −1 −2 1  4 −2
= −4 1 −42 3 −13   −1  1
∴Y = 3  −  3 −2  = 3 −2 4 
 −7 
−2  4  0 
1 37 
 −2 421
∴X= 11 −8 37  =  4 2  4 −2 
4 −4 1 −42 1 = −323 3 
−1 4 − 4 

3

2x +1 −1 −1 6 4 5   7 
− 
Ex. 4) If  3 4 y +  3 0 = 6 12 ,  3 0 
 

find x and y. From (1) X + Y = A, ∴ X = A – Y,

2x +1 −1 −1 6 4 −2 
Solution : Given  4 y +  0 3 
 3  3 −332 4 
2 −1 
 
4 5  ∴ X =  1 3  – 3
= 6 12 
−3 −2  7
− 
2x 5  4 5   3 0 

∴  6 4 y  = 6 12
 
 4 2 2 −1 
 2 − 3 −1 +  3 
∴ Using definition of equality of matrices,  3   3 5 
we have   5 
X = −13++3273 4 
2x = 4, 4y = 12 ∴ x = 2, y = 3 3− 3  = 3 3
 
 2 
−2 + 0  − −2 
  3
2 −1
 
Ex. 5) If X +Y=  1 3  and

−3 −2

X – 2Y EXERCISE 4.5

−2 1
 −1 then find X ,Y.  2 −3 −1 2
=  3 −2  −4 , B =  2 and
 5  2
 4 (1) If A =

2 −1 −2 1 −6 1   0 3
−1
Solution : Let A =  1 3  and B =  3 −2  4 3
    4 Show that (i) A +B = B+A
−3 −2  4 C =  −1

Let, X + Y = A …. (1), X – 2Y = B ……(2), −2 1
Solving (1) and (2) for X and Y
(ii) (A+B)+C = A+(B+C)

86

1 −2 1 −3  i 2i 2i i 
(2) If A = 5 3  , B = 4 −7 , then find the (8) If A = −3  and B =  −3 , where
2   2

matrix A –2B+6I, where I is the unit matrix -1 = i, findA+B andA–B. Show thatA+B is
of order 2.
a singular. Is A–B a singular ? Justify your
answer.

 1 2 −3  9 −1 2  2x + y −1 1
(3) If A = −3 7 −8 , B = −4 2  (9) Find x and y, if  3 4y 4
5 

 0 −6 1   4 0 −3

then find the matrix C such that A+B+C is a −1 6 4 3 5 5
zero matrix. +  3 0 3 = 6 18 7

1 −2 −1 −2 2a + b 3a − b 2 3 
(10) If = c + 2d  = 4 −1 , find a, b,
 3 −5 , B =  4 2  2c − d 
(4) If A =    and

−6 0   1 5  c and d.

2 4 (11) There are two book shops owned by Suresh
 −4 , find the matrix X such that and Ganesh. Their sales (in Rupees) for
C==  −1 books in three subject – Physics, Chemistry
and Mathematics for two months, July and
−3 6  August 2017 are given by two matrices
A and B.
3A –4B+5X = C.

(5) Solve the following equations for X and Y, if July sales (in Rupees), Physics Chemistry
Mathematics.
 1 −1 0 −1
3X–Y= −1  and X–3Y= 0 −1
1  5600 6750 8500
A = 6650 7055 8905 First Row Suresh/
(6) Find matrices A and B, if Second Row Ganesh

2A–B = 6 −6 0 and August sales (in Rupees), Physics Chemistry
−4 2 1 Mathematics

3 2 8 6650 7055 8905 
A–2B = −2 1 −7 B = 7000 7500 10200 First Row
Suresh/ Second Row Ganesh then,

(7) Simplify, (i) Find the increase in sales in Rupees from
July to August 2017.
 cosθ sinθ 
cosθ −sinθ cosθ + (ii) If both book shops got 10 % profit in the
month of August 2017, find the profit for
sinθ  sinθ −cosθ each book seller in each subject in that
cosθ  month.
sinθ 

87

(4) Algebra of Matrices (continued) 3
Ex.2 : Let A= [1 3 2]1×3 and B = 2 , find AB.
Two Matrices A and B are said to be
conformable for the product AB if the number of 13×1
columns in A is equal to the number of rows in B. Does BA exist? If yes, find it.
i.e. A is of order mxn and B is of order nxp.
Solution : Product AB is defined and order of
In This case the product AB is amatrix AB is 1.
defined as follows.
3
n ∴ AB = [1 3 2] 2 = [1 × 3 + 3 × 2 + 2 × 1]

∑Am×n × Bn×p = Cm×p , where Cij = aikbkj 1
k =1 = [11]1×1

a11 a12 ... a1k ... a1n  Again since number of column of B = number of
a 21  rows of A=1
a22 ... a 2k ... a 2n 
∴ The product BA also is defined and order of
If A = [ aik]m×n = a 31 a32 ... a 3k... a 3n  BA is 3.
a i1 ai2 ... aik ... a in → ith row

am1 am2 ... amk ...amn 

b11 b12 ... b1j ... b1p 
b21 
b31 b22 ... b2j ... b2p  3 3×1 3×3 3× 2
B = [ bkj]n×p = BA = 2 [ 1[13322]1]×31×3= 2×1 2× 3 2× 2
b32 ... b3j ... b3p 

bp1 bp2 ... bnj ... bnp  13×1 1×1 1× 3 1× 2 3×3

↓ 3 9 6
jth column = 2 6 4

then 1 3 23×3
Cij = ai1 b1j + ai2 b2j + ........ + ain bnj

Remark : Here AB and BA both are defined but
AB ≠ BA.

−1 −2 1 2
Ex.3 : A = −3  −1
SOLVED EXAMPLES 2  ,B= −2 
 1 0 3×2 
2×2

b11  Find AB and BA which ever exist.
b21 
Ex.1 : If A = [a11 a12 a13] and B =  Solution : Here A is order of 3 × 2 and B is of
order 2 × 2. By conformability of product, AB is
Find AB. b31  defined but BA is not defined.

Solution : Since number of columns of −1 −2 1 2
A = number of rows of B = 3 ∴ AB = −3  −1 −2
2 
Therefore product AB is defined and its order  1
is 1. (A)1×3 (B)3×1 = (AB)1×1 0 
AB = [a11 × b11 + a12 × b21 + a13 × b31]
−1+ 2 −2 + 4 1 2
= −3 − 2 −6 − 4 = 5 10
2 + 0  1 2
 1+ 0

88

 3 2 −1 , Note :
Ex.4 : Let A = −2 5 
4  2×3 From the above solved numerical Examples, for
the given matrices A and B we note that,
 3 −3
B= −4  i) If AB exists , BA may or may not exist.
2  2×2 ii) If BA exists , AB may or may not exist.
iii) If AB and BA both exist they may not
Find AB and BA which ever exist.
be equal.
Solution : Since number of columns of A ≠
number of rows of B. Product of AB is not 4.6 Properties of Matrix Multiplication :
defined. But number of columns of B = number
of rows of A = 2, the product BA exists, 1) For matrices A and B, matrix
multiplication is not commutative that
 3 −3  3 2 −1 is AB ≠ BA.
∴ BA = −4  −2 
2  5 4  2) For three matrices A,B,C. Matrix
multiplication is associative. That is
 9 + 6 6 −15 −3 −12 (AB)C = A(BC) if orders of matrices
= −12 − 4 −8 +10  are suitable for multiplication.
4+8 

 15 −9 −15 1 2 1 −1 2
e.g. Let A= 4 3 , B = 0 −1 3 ,
= −16 2 12 


−2 1 
 −1
4 −3 −1 3  C=  3
Ex.5 : Let A = 5  and B =  −2 Find
2   4  0 2 

AB and BA which ever exist. 1 2 1 −1 2
Then AB = 4 3 0 −1 3
Solution : Since A and B are two matrix of same
order 2 × 2.

∴ Both the product AB and BA exist and are of 1+ 0 −1− 2 2 + 6 1 −3 8 
same order 2 × 2 = 4 + 0  = 4 17
−4 − 3 8+ 9  −7

4 −3 −1 3  1 −3 8 −2 1
AB = 5 2   4 −2 (AB)C = 4 −7 17  −1
 3 2 

−4 −12 12 + 6 −16 18  0

=  −5 + 8 15 − 4 =  3 11  −2 − 9 + 0 1+ 3 +16  −11 20
 = −8 − 21+ 0 4 + 7 + 34 = −29 45

−1 3  4 −3 −4 +15 3 + 6  …..(1)
BA =  4 −2 5 2  = 16 −10 −12 − 4
−2 1
1 −1 2  −1
11 9  ∴ BC = 0 −1 3  3 2 
=  6 −16
Here AB ≠ BA  0

−2 − 3 + 0 1+1+ 4  −5 6
=  0 +1+ 6 = −3 7
 0−3+0

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1 2 −5 6 5) For any matrix A there exists a null
A(BC) = 4 3 −3 7 matrix O such that a) AO = O and
b) OA = O.
 −5 − 6 6 +14  −11 20
= −20 − 9 24 + 21 = −29 45 ..…(2) 6) The product of two non zero matrices
From (1) and (2), (AB)C = A(BC) can be a zero matrix. That is AB = O
but A ≠ O, B ≠ O

1 0  0 0
e.g. Let A = 2 0 , B = −1 1 ,
3) For three matrices A,B,C, multiplication
is distributive over addition. 1 0  0 0
i) A(B+C)= AB + AC Here A ≠ 0, B ≠ 0 but AB= 2 0 −1 1
(left distributive law)
0 0
ii) (B+C)A= BA + CA That is AB = 0 0 = O
(right distributive law)
7) Positive integral powers of a square
These laws can be verified by examples. matrix A are obtained by repeated
multiplication of A by itself. That is
4) For a given square matrix A there exists A2 = AA, A3 = AAA, ……,
a unit matrix I of the same order as that
of A, such that AI=IA=A. An = AA … n times

I is called Identity matrix for matrix (Activity)
multiplication.

3 −2 −1
e.g. Let A = 2 0 
4  , 1 −5 2 −3
If A = 6 7  , B = 4 8  ,
1 3 2 

1 0 0 Find AB-2I, where I is unit matrix of
and I = 0 1 0 order 2.

0 0 1 1 −5 2 −3
Solution : Given A = 6  , B = 4 
7  8 

3 −2 −1 1 0 0 1 −5 2 −3 ... ...
Then AI = 2 0  0 0
4  1 Consider AB−2I = 6 7  4 8  –2 ... ...
 
1 3 2  0 0 1

3+ 0 + 0 0−2+0 0+ 0−1  ...... −3 − 40 ... 0 
= 2 + 0 + 0 0+0+0 0 + 0 + 4 ∴AB−2I = 12 + 28  –  ...
0+3+0 0 + 0 + 2 ......   0
1+ 0 + 0
 ... −43 ... 0 
= 40 ...  –  0 ...

3 −2 −1
= 2 0 
4  = IA = A … −43

1 3 2  ∴AB−2I = 40 … 


90

SOLVED EXAMPLES 19 15 15  5 15 15
= 15 19 15 – 15 5 15
−2 1
−1 , 15 15 19 15 15 5 
2 
1 −1 2  3 14 0 0 
Ex. 1: If A= 0 −1 3 , B =   
 0  
∴ A2 –5A = 0 14 0 = 14I

show that matrix AB is non singular.  0 0 14

1 −1 2 −2 1  ∴ A2–5A is a scalar matrix.
Solution : let AB = 0 −1 3
 3 −1 3 −2
 2  Ex. 3 : If A= 4 −2 , Find k, so that A2 –kA+2I
 0 = O, where I is an identity matrix and O is null
matrix of order 2.
= −2 − 3 + 0 1+1+ 4  Solution : Given A2 –kA+2I = O
 0 +1+ 6
 0−3+0 3 −2 3 −2
∴Here, A2 = AA = 4 −2 4 −2
−5 6
= −3 7 ,  9 − 8 −6 + 4 1 −2
= 12 − 8 −8 + 4 = 4 −4
-5 6
∴|AB| = -3 7 ∴ A2–kA+2I = O

= –35 + 18 = –17 ≠ 0
∴ matrix AB is nonsingular.

1 3 3 1 −2 3 −2 1 0
Ex. 2 : If A = 3 1 3 prove that A2–5A is a ∴ 4 −4 – k 4 −2 + 2 0 1 = O
scalar matrix. 3 3 1
1 −2 3k −2k  2 0 0 0
Solution : Let A2 = A.A ∴ 4 −4 – 4k  + 0 2 = 0 0
1 3 3 1 3 3 −2k 

= 3 1 3 3 1 3 1− 3k + 2 −2 + 2k  0 0
3 3 1 3 3 1 ∴ −4 + 2k + 2 = 0 0
1+ 9 + 9 3 + 3 + 9 3 + 9 + 3  4 − 4k

= 3 + 3 + 9 9 +1+ 9 9 + 3 + 3 ∴ Using definition of equality of matrices, we
3 + 9 + 3 9 + 3 + 3 9 + 9 +1 have
19 15 15
1 – 3k + 2 = 0 ∴ 3k=3
= 15 19 15
15 15 19 –2 + k = 0 ∴ 2k=2 k=1
19 15 15 1 3 3
4 – 4k = 0 ∴ 4k=4
∴ A2 –5A = 15 19 15 – 5 3 1 3
15 15 19 3 3 1 –4 + 2k + 2=0 ∴ 2k=2

Ex. 4 : Find x and y, if

  6 3 −4 −1
3 −1 2  −04 = [x
[2 0 3]   5 + 2  1 y]

4 −3

91