Determinant (Answers)
1.
(a)
1 1 1
0 0 1
2 0
= 2×2 = 4
D = 2× 3 1 1 = 2× 2 0 1 = 2×
3 1
4 2 1
3 1 1
(b)
C1
(c)
Use Sarrus rule to expand the given determinant,
(d)
R 3 → R1 + R 2 + R 3
D
=
and
C2
are in ratio, therefore
D = 0.
D = 2abc .
b+c
c+a
a−b
b−c
a
b
b+c a−b a
= (a + b + c) c + a b − c b
2( a + b + c )
0
a+b+c
2
0
1
= – (a + b + c) (a2 + b2 + c2 – bc – ca – ab)
2.
(e)
4 3 1
4 3 1
C 3 → C 3 − C1
R 3 → R 3 − R1
1 6
1 6 1
1 6 0=
= − 23
D
=
=
3 −5
7 −2 1
3 −5 0
(f)
By Sarrus rule of expansion, result follows.
(a)
a b c 0 1 1
L.H.S. = p q r × 1 0 1 = R.H.S.
x y z 1 1 0
R1 → R1 − R 2
(b)
0 a − b a 2 − b2
=
D
0 b − c b2 − c2 =
R2 → R2 − R3
1
c2
c
a − b a 2 − b2
b−c b −c
2
= (a – b) (b – c) {(b + c) – (a + b)}
1 a
(c)
D
a3
3
= 1 b b
1 c c3
(a – b) (b2 – c2) – (b – c)(a2 – b2)
= (a – b) (b – c) (c – a)
C 3 → C 3 + (ab + bc + ca )C 2 − abcC1
=
=
2
1 a a 2 (a + b + c)
1 b b 2 (a + b + c)
1 c c 2 (a + b + c)
= (a + b + c) × determinant in 2(b)
= (a + b + c) (a – b) (b – c) (c – a)
(d)
D
R 3 → R 3 + ( x + y + z)R 1 − R 2
=
x
x2
y
y2
z
z2
yz + zx + xy yz + zx + xy yz + zx + xy
= (yz + z x + xy) ( x – y ) ( y – z ) ( z – x )
by
(i)
taking out (yz + z x + xy) in R3 ;
(ii)
interchange R2
(iii) result in
(e)
D
with
R3
and then
R1
with
R2 ;
2 (c) where a = x , b = y , c = z .
=
–8abc + 2(b + c)(c + a)(a + b) + 2a(b + c)2 + 2b(c + a)2 + 2c(a + b)2
=
2(b + c)(c + a)(a + b) + 2a(b + c)2 –8abc + 2b (c2 + 2ca + a2) + 2c(a2 + 2ab + b2)
=
2(b + c)(c + a)(a + b) + 2a(b + c)2 + 2bc2 + 2ba2 + 2ca2 + 2cb2
=
2(b + c)(c + a)(a + b) + 2a(b + c)2 + 2a2(b + c) + 2bc(b + c)
=
2(b + c)(c + a)(a + b) + 2(b + c){a(b + c) + a2 + 2bc}
=
2(b + c)(c + a)(a + b) + 2(b + c)(c + a)(a + b)
=
4(b + c)(c + a)(a + b)
1
3.
(a)
(b)
4.
D
1
ω
ω2
ω
ω2
1
ω2
1
ω
D
R1 → R1 + R 2 + R 3
=
1 + ω + ω2
ω
1 + ω + ω2
ω2
1 + ω + ω2
1
ω2
1 − ω3 0
1 = 0 ,since 1 + ω + ω 2 =
= =0.
1− ω ω
ω
= 1 + ω6 + ω6 – ω4 – ω2 – ω6
= 1 + 1 + 1 - ω - ω2 – 1
= 3 – (ω2 + ω + 1) = 3 – 0
=3
R 3 → R 3 + 2R 2
=
( ω3 = 1 )
a3
b3
c3
a3
b3
c3
a
b
c
= (a + b + c) a
b
c = − (a + b + c) 1 b b 3
1
1
1
a+b+c a+b+c a+b+c
1 a a3
1 c c3
= –(a + b + c)2 (a – b) (b – c) (c – a)
5.
R → R1 + R 2 + R 3
D 1
=
a+b+c a+b+c a+b+c
1
1
= (a + b + c ) 2 b b − c − a
2b
b−c−a
2b
2c
C 2 → C 2 − C1 1
0
=
2 b − (a + b + c)
C 3 → C 3 − C1 2c
c−a−b
2c
0
0
2c
= (a + b + c)
− (a + b + c)
0
c−a−b
2c
− (a + b + c)
0
1
2b
0
− (a + b + c)
= (a + b + c)3
6.
D
C 2 → C 2 + C1
=
a a + b + c a2
a 1 a2
D
C 2 → C 2 − 2C 1 − 2 C 3
=
a2
b2
c2
, by 2 (b)
− (a 2 + b 2 + c 2 ) bc
a 2 1 bc
− (a 2 + b 2 + c 2 ) ab
c2
− (a 2 + b 2 + c 2 ) ca = −(a 2 + b 2 + c 2 ) b 2 1 ca
a 2 1 1/ a
= −abc(a + b + c ) b
2
2
2
c
2
2
a3
1 1 / b = −(a + b + c ) b
2
2
2
1 1/ c
c
= (a2 + b2 + c2) (a + b + c) (a – b)(b – c)(c – a),
8.
a2
b a + b + c b 2 = (a + b + c) b 1 b 2 = − (a + b + c) 1 b b 2
c a + b + c c2
c 1 c2
1 c c2
= –(a + b + c) (a – b) (b – c) (c – a)
7.
1 a
3
3
1 ab
a 1
b 1 = (a + b + c
2
2
c 1
1
2
)a
a
3
1
1
b
c
3
c3
b
by 2(c)
abc
abc
abc
a
b
c
( bc)(ca )(ab)
L.H.S. =
bc(c − b ) ca (a − c ) ab( b − a ) =
c−b a −c b−a
abc
abc
2
2
2
b c
c a
a b
b
c
a
1
a b c
R2 → R2 + R3
abc c a b = abc (a3 + b3 + c3 – 3abc)
=
b c a
9.
1 −x
1 −x −x −x
−x
−x
R i → R i + R1
0 r1 + x a + x a + x
1 r1
a
a
D( x ) =
(i = 1,2,3)
0 b + x r2 + x a + x
1 b r2
a
=
0 b + x b + x r3 + x
1 b
b r3
1 a
= D(0) + x 1 r2
1 b
a
r1 1 a
r1 a 1
a + x b 1 a + x b r2 1 ,
r3
b 1 r3
b
by expansion in R1 .
b 1
2
1
r
= D ( 0) + x 1
b
b
1
a
r2
b
1
a
a
r3
0
1
= D(0) + x ∆ …. (*)
1
1
By substitution, it can be seen that
D(-b) = (r1 – b)(r2 – b)(r3 – b) = f(b)
and
D(-a) = (r1 – a)(r2 – a)(r3 – a) = f(a)
By (*),
f(b) = D(-b) = D(0) – b ∆
…. (1)
f(a) = D(-a) = D(0) – a ∆
…. (2)
Solving (1) and (2) for D(0), result follows.
10.
R → R1 + R 2
D ( a , b, c ) 1
=
Therefore
11.
Writing
D=
ab + bc + ca ab + bc + ca ab + bc + ca
bc
ca
ab
= 0 , since R1 and R3 are in ratio.
1
1
a, b, c, (b – c), (c – a), (a – b)
a = sin x ,
a
2ap
b = siny ,
b
2bq
are factors of
c = sinz
c
2cr
1
;
= abc
a (3 − 4 a 2 ) b (3 − 4 b 2 ) c ( 3 − 4 c 2 )
R2 → R2 − R3
=
abc
R 3 → R 3 → R1
1
2p
0
2( q − r )
3 − 4a 2
0.
p = cos x ,
q = cos y ,
1
2p
1
2q
1
2r
3 − 4a 2
3 − 4b 2
3 − 4c 2
0
2( r − p) = 8abc
4( c 2 − b 2 ) 4( a 2 − c 2 )
r = cos z
q−r
r−p
(q − r )(q + r ) ( r − p)( r + p)
= 8abc(q – r)(r – p)(p – q)
= 8 sin x sin y sin z (cos y – cos z)(cos z – cos x)(cos x – cos y)
12.
Method 1
dD
dθ
∴
− sin(θ + x ) sin(θ + x ) 1 cos(θ + x ) cos(θ + x ) 1 cos(θ + x ) sin(θ + x ) 0
= − sin(θ + y) sin(θ + y) 1 + cos(θ + y) cos(θ + y) 1 + cos(θ + y ) sin(θ + y) 0 = 0
− sin(θ + z )
sin(θ + z ) 1
cos(θ + z )
cos(θ + z ) 1
cos(θ + z )
sin(θ + z ) 0
D is independent of θ .
Method 2
x+y x−y
x+y x−y
− 2 sin 
 sin 
 0
 sin 
 2 cos
 2   2 
 2   2 
cos(θ + x ) sin(θ + x ) 1 R 1 → R 1 − R 2
y+z y−z
y−z y−z
cos(θ + y) sin(θ + y) 1
− 2 sin 
=
 sin 
 2 cos
 sin 
 0
 2   2 
 2   2 
cos(θ + z ) sin(θ + z ) 1 R 2 → R 2 − R 3
cos(θ + z )
sin(θ + z )
1
x+y
x+y
sin 
 cos

x−y  y−z  2 
 2  = −4 sin  x − y  sin  y − z  sin  z − x 
= −4 sin 
 sin 


 
 

 2   2  sin  y + z  cos y + z 
 2   2   2 




 2 
 2 
which is independent of
θ.
3
13.
Let
f(z) = given determinant.
f( (1 + a2) sin (π/6) ) = 0,
by substituting into the given determinant,
R1
and
R3 are equal.
f( (1 + a2) sin (π/3) ) = 0,
by substituting into the given determinant,
C1
and
C3 are in ratio.
2
f( (1 + a ) x )
is a determinant of degree
3.
It can be easily seen from the determinant that :
Let α be the third root,
∴ α=−
14.
∆=
=
1
2
−
3
2
− 2 cos x
Sum of roots = α + sin (π/6) + sin (π/3) = 0
.
− 2a cos x
1
1
− 2 cos x
a2
cos( n − 1) x − 2 cos nx cos x cos( n + 1) x
sin( n − 1) x − 2 sin nx cos x sin( n + 1) x
− 2a cos x
1
1
a2
cos( n − 1) x − [cos(n + 1) x + cos( n − 1) x ] cos( n + 1) x
sin( n − 1) x
− [sin( n + 1) + sin( n − 1) x ] sin( n + 1) x
1
1 − 2a cos x + a 2
C 2 → C 2 + C1 + C 3
1
cos( n − 1) x
0
=
− 2 cos x
sin( n − 1) x
0
=
=
15.
x2 – term = 0.
Coefficient of
a2
cos(n + 1) x
sin( n − 1) x
1 − 2a cos x + a 2 cos( n − 1) x cos( n + 1) x
sin( n − 1) x sin( n − 1) x
2 cos x
1 − 2a cos x + a 2
2 cos x
sin 2 x = (1 − 2a cos x + a 2 )sin x
1 cos A cos A 1 − sin A sin A 1 − sin A cos A 1 cos A sin A
L.H.S. = 1 cos B cos B + 1 − sin B sin B + 1 − sin B cos B + 1 cos B sin B
1 cos C
cos C
1 − sin C sin C
1 − sin C cos C
1 cos C
sin C
= 0 + 0 + 0 + R.H.S. = R.H.S.
16.
∆=
sin B
sin C
cos B cos C
−
sin A
sin C
cos A cos C
+
sin A
sin B
cos A cos B
= sin (A – B) + sin (B – C) + sin (C – A)
B+C B−C
C+A C−A
2 cos
2 cos
 sin 

 sin 






  2 
2
2
2
∆
=
sin A sin B − sin C sin C − sin A =
B+C B−C
C+A C−A
 sin 
 − 2 sin 
 sin 

C 3 → C 3 − C1 cos A cos B − cos C cos C − cos A − 2 sin 
 2   2 
 2   2 
C 2 → C 2 − C3
1
0
0
B+C
C+A
cos
 cos

B−C C−A
 2 
 2  = −4 sin  B − C  sin  C − A  sin  A − B 
= −4 sin 
 sin 


 
 

 2   2  sin  B + C  sin  C + A 
 2   2   2 




 2 
 2 
17.
Expansion on both sides gives result.
18.
D = first given determinant
|A| |B| = |AB|
for
2 × 2 determinants.
= D′ (the transpose) = (b – c)(c – a)(a – b) , by 2(b)
Therefore, second given determinant
= DD′ = [(b – c)(c – a)(a – b)]2
4
19.
Interchanging
C2
and
C3
gives the first result.
Let the first given determinant be
b c −a −c −b
a
D2 = c a b c
b c a b
then
D.
2bc − a 2
b2
c2
b2
c2
2ab − c 2
a2
a2
2ca − b 2
a =
c
b
a
= last determinant, by interchanging
20.
and
C2
C3 , then
R2
and
R3 .
b c a
a b c =0.
The other factor is
0 0 0
a
b cb c a
2ab
c a b a b c = bc + a 2
b c a 0 0 0 ab + b 2
 1

s7 =  α4
s 9  α 6
bc + a 2
ab + c 2
2bc
2ac
ab + c 2
1  1 α 2

γ 4  1 β 2
γ 6 1 γ 2
s0
s2
21.
s4
s6
s6
s8
22.
D3 = D3′ = D1 × D2 = D1 × 0 = 0 .
23.
s3
ac + b 2
1
β4
β6
C3 → C3 − C2 1
=
A
a+b
0
c−a
C 2 → C 2 − C1 ab
1
α3 

3
β  = α4
γ 3  α 6
0
a−b
=
b (c − a ) c(a − b )
2ab
ac + b 2
bc + a 2
=
ab + b 2
bc + a 2
2 bc
ab + c 2
ab + c 2 = 0
2ac
1
1 1 α2
α3
β4
β6
γ 4 1 β2
γ6 1 γ2
β3
γ3
R2 ↔ R3
1 1
c−a
a−b
= (c − a )(a − b )
b c
b (c − a ) c(a − b )
= – (c – a)(a – b)(b – c)
x2

X =  y2
 z2

x 1

y 1
z 1
det B = (det x) (det A) = [-(x – y)(y – z)(z – x)][-(a – b)(b – c)(c – a)] =
24.
(a)
A H G aA + hH + gG aH + hB + gF aG + hF + gC ∆ 0
∆ × H B F = hA + bH + fG hH + bF + fF hG + bF + fC = 0 ∆
G
F
C
Therefore,
(b)
gA + fH + cG
∴
F
C
cancel
if ∆ = 0,
both sides = 0
g
0
∆
gG + fF + cC
on both sides,
0
0
0
0 = ∆3
∆
result follows.
and result follows.
0
0 = a∆2
∆
∆ (BC – F2) = a∆2 .
Result follows for both
Consider
gH + fB + cF
if ∆ ≠ 0,
1 H G a 0
∆× 0 B F = h ∆
0
25.
(x – y)(y – z)(z – x)(a – b)(b – c)(c – a)
A 0 G
∆× H 0 F .
∆≠0.
Similar to 24 (b).
G 1 C
5
26.
Multiply
R1
by
a,
R2
by
b,
R3
a 2 + abcx bc
a
a2
a
b b + abcx ca = 0
c c 2 + abcx ab
⇒
2
by
2
b b
c c2
c
bc
x=
1
sin 2 θ
0 1 + sin 2 θ
0
0
Use
sin 2 θ
sin 2 θ
ab + bc + ca
abc
cos 2 θ
cos 2 θ
4 sin 2θ
4 sin 2θ
1 + cos 2 θ
4 sin 2θ
2
cos θ 1 + 4 sin 2θ
Ri → Ri – R1 ,
0
0
=0
i = 2, 3, 4 .
1 sin 2 θ cos 2 θ 4 sin 2θ
1
0
0
−1
−1
−1
1
0
0
1
1 + sin 2 θ + cos 2 θ + 4 sin 2θ sin 2 θ cos 2 θ 4 sin 2θ
0
1
0
0
C →
Ci
=0 1
=0
i =1
0
0
1
0
=
0
0
0
1
4
∑
1 + sin2 θ + cos2 θ + 4sin 2θ = 0
Therefore
,2θ = nπ + (-1)n (-π/6)
2 + 4 sin 2θ = 0
θ=
28.
Method
x
2(c).
(a – b)(b – c)(c – a) (ab + bc + ca) – abcx (a – b)(b – c)(c – a) = 0.
Therefore
27.
C3 → C3 – (a + b + c) C1 + (ab + bc + ca)C2 ,
2(b).
The first determinant can be evaluated by
We get
a 1 bc
ca + abcx b 1 ca = 0
ab
c 1 ab
In the second determinant in the last equation, use
then evaluate by
in the given equation, we get
x2
b b2
c c2
 π
nπ + ( −1) n  −  ,
 12 
2
1
where
n
is an integer.
1
a3
x x2
a3 − b
a3 c
x3
x x2 1
b3 = 0
c3
b2
c2
⇒
1 x
x2
a 3 b b 2 1 − xbc 1 b b 2 = 0
c c2 1
1 c c2
a3 (x – b)(b – c)(c – x) – xbc (x – b)(b – c)(c – x) = 0
(b – c) (x – b) (c – x)(a3 – xbc) = 0
x = b , c,
Method
a3
bc
2
It can be seen that
x = b, x = c
Degree of the equation
From the determinant,
a
and
let
α
be the third root.
Coefficient of
x – term
= b2c
Coefficient of
constant term
= - a 3b2c
= α bc
Product of roots
x = b , c,
=3
are roots of the given equation by Factor Theorem.
3
= - (- a3b2c)/( b2c),
solve for
α
we have:
3
bc
6
29.
∆ = a3 + 2b3 – 3ab2 = (a + 2b)(a – b)2 = 7(x + 1)(x – 2)
Therefore
a
30.
b c
,
2
a
x = -1
or
b ca
2.
c b
s
p p
c a b = c a bb a c = p s p
b c a
b c a c b a p p s
a
b c
2
a
b ca
b c
x
z
y
c a b = c a bc a b = y x z
b c a
b c a b c a z y x
31.
OA = x 1i + y1 j + z1k ,
Area of ABC =
1
2
OB = x 2 i + y 2 j + z 2 k ,
AB × AC =
1
2
1 x1
1 x2
y1
y2
1 x3
y3
OC = x 3 i + y 3 j + z 3k
Area of parallelogram formed by OB and OC as adjacent sides = OB × OC
x1
Volume of parallelepiped = OA • OB × OC = x 2
y1
y2
z1
z2
x3
y3
z3
(
32.
Area of triangle formed by
By
33.
)
(x , y) , (x1 , y1) , (x2 , y2) = 0
no. 31 , result followed.
Li : Ai + Bi y + Ci = 0,
i = 1, 2, 3
⇒
L3
⇒
(x, y) = (∆x / ∆ , ∆y / ∆ )
⇒
Coefficient determinant = 0
are concurrent
passes through the intersection point of
satisfies
L1 and
L2
L3
The condition is not sufficient,
Consider
3
parallel lines which are not concurrent, the coefficient determinant = 0
7