DETERMINANTS
55 2.3
Special Determinants :
(1)
Symmetric determinant.
a h g h b f = abc + 2fgh – af2 – bg2 – ch2 g f c
In this the elements aij = aji or the elements situated at equal distance from the diagonal are equal both in
(2)
magnitude and sign.
Skew symmetric determinant of odd order.
0 b c b 0 a = 0. c a 0
All the diagonal elements are zero and aij = – aji or the elements situated at equal distance from the diagonal
(3)
Circulant
are equal in magnitude but opposite in sign. Its value is zero. a b c b c a = – (a3 + b3 + c3 – 3abc) c a b
The three rows or columns are the cyclic arrangement of the letters a, b, c, i.e. a, b, c ; b, c, a and c, a, (4)
b respectively.
Factors of three important determinants :
(i)
1 1 1 a b c = (a – b) (b – c) (c – a) a2 b2 c 2
1 1 1 (ii) a b c =(a – b) (b – c) (c – a) (a + b + c) a3 b3 c 3
1 1 1 (iii) a 2 b 2 c 2 = (a – b) (b – c) (c – a) (ab + bc + ca) a3 b3 c 3
www.thinkiit.in
DETERMINANTS
57 Ex.3
Solve the equation
Sol.
Let
=
x+α x x x x x+α = 0, (a 0). x x x+α
x x x x x x x x x
= (x + a)
xa x x x x x x xa – x x x + x x x 2
2
2
2
= (x + a) [(x + a) – x ] – x [x (x + a) – x ] + x [x – x (x + a)] 3 2 2 2 = (x + a) – x (x + a) – ax – ax 3 3 2 = (x + a) – x – 3ax 3 2 2 3 3 2 = x + 3ax + 3a x + a – x – 3ax 2
= 3a x + a Equating it to zero, we have 2
3
3a x + a = 0
Ex.4
3
x=–
a 3
If a, b and c are real numbers, and b+c c+ a a+ b = c+ a a+ b b+c = 0 a+ b b+c c+ a
Show that either a + b + c = 0 or a = b = c
Sol.
Here,
2 (a b c ) c a a b = 2 (a b c ) a b b c 2 (a b c ) b c c a
(C1 C1 + C2 + C3)
1 c a ab = 2 (a + b + c) 1 a b b c 1 bc ca 1 c a ab = 2 (a + b + c) 0 b c c a 0 ba c b
(R2 R2 – R1; R3 R3 – R1)
bc c a = 2 (a + b + c) × 1 b a c b = 2 (a + b + c) [(b – c) (c – b) – (b – a) (c – a)] 2 2 2 = 2 (a + b + c) [bc – c – b + bc – bc + ac + ab – a ] 2 2 2 = 2 (a + b + c) (ab + bc + ca – a – b – c ) Equating to zero, we have 2
2
2
2 (a + b + c) (ab + bc + ca – a – b – c ) = 0 2 2 2 Either a + b + c = 0 or ab + bc + ca = a + b + c Either a + b + c = 0 or a = b = c
www.thinkiit.in
DETERMINANTS
59 Ex.8
x + 2 x + 3 x + 2a If a, b, c are in AP, prove that : x + 3 x + 4 x + 2b = 0 x + 4 x + 5 x + 2c
Sol.
Let
x 2 x 3 x 2a = x 3 x 4 x 2b x 4 x 5 x 2c
=
x 2 x 3 x 2a 1 1 2(b a) 2 2 2(c a)
=
x2 x3 x 2a 1 1 2(b a) 0 0 2(c a) 4(b a)
= [2 (c – a) – 4 (b – a)]
(R2 R2 – R1; R3 R3 – R1)
(R3 R3 – 2r2)
x2 x3 1 1
= [2c – 2a – 4b + 4a] (x + 2 – x – 3) = [2c + 2a – 4b] (–1) = (4b – 2a – 2c) Since, a, b, c are in AP, 2b = a + c 4b = 2a + 2c Hence by (1), = 0
Ex.9
(b + c)2 a2 a2 3 Prove that = 2abc (a + b + c) b2 (c + a)2 b2 c2 c2 (a + b)2
Sol.
Replacing C2 by C2 – C1 and C3 by C3 – C1, we have
=
(b c )2 a2 (b c )2 a2 (b c )2 b2 ( c a )2 b 2 0 2 c 0 ( a b )2 c 2
b2 c 2 2bc a b c a b c b2 c ab 0 c2 0 abc
Replacing R1 by R1 – R2 – R3, we have = (a + b + c)
Replacing C2 by C3 +
2
2bc 2c 2b b2 c a b 0 c2 0 abc
1 1 C and C3 by C3 + C , we have b 1 c 1
2bc = (a + b + c)
2
b
2
c2
0 ca
0 b2 c
c2 b
ab
2
= 2bc (a + b + c) [(a + c) (a + b) – bc] 2 2 = 2bc (a + b + c) (a + ab + bc – bc) 3 = 2abc (a + b + c)
www.thinkiit.in
(....1)
=
(a
+
b
+
c)
2
DETERMINANTS
61
UNSOLVED PROBLEMS EXERCISE – I Q.1
2 5 4 3 If A = 2 1 and B = 2 5 , verify that |AB| = |A| |B|
Q.2
b2c 2 bc b c Without expanding evaluate the determinant c 2a2 ca c a a2b2 ab a b
Q.3
1 a a2 bc 2 Without expanding evaluate the determinant 1 b b ac 2 1 c c ab
Q.4
(a x a x )2 (a x a x )2 1 (a y a y )2 (a y a y )2 1 Without expanding evaluate the determinant (a2 a z )2 (a z a z )2 1
Q.5
sin cos sin( ) Without expanding evaluate the determinant sin cos sin( ) sin cos sin( )
Q.6
Prove that
1 a 1 1 1 1 1 1 1 b 1 = abc 1 1 = abc + ab + bc + ca a b c 1 1 1 c
Q.7
Prove that
a b 2c a b c b c 2a b = 2(a + b + c)3 c a c a 2b
Q.8
(b c )2 a2 bc Prove that (c a)2 b2 ca = (a2 + b2 + c2) (a + b + c) (b – c) (c – a) (a – b) (a b)2 c 2 ab
Q.9
1 a2 b2 2ab 2ab 2ab 1 a2 b2 2a Show that = (1 + a2 + b2)2 2 2 2b 2a 1 a b
Q.10
Solve :
Q.11
x x2 1 x3 2 3 If x y z and y y 1 y = 0, then prove that 1 + xyz = 0 2 3 z z 1 z
3x 8 3 3 =0 3 3x 8 3 3 3 3x 8
Q.13
log a p 1 If a, b, c are all positive and are pth, qth and rth terms respectively of a G.P. then prove that log b q 1 = 0 log c r 1 Find the area of a triangle with vertices : (–3, 5), (3, –6), (7, 2)
Q.14
Find the value of so that the points given below are collinear (, 2 – 2), (– + 1, 2) and (–4, –, 6 – 2)
Q.15
Using determinants, find the area of the triangle, whose vertices are (–2, 4), (2, –6) and (5, 4). Are the given points collinear ?
Q.12
www.thinkiit.in
DETERMINANTS
63 Q.11
Using the properties of determinants, prove that
(a)
Q.12
a b c bc c a ab qr r p pq = 2 p q r x y z yz zx xy
[C.B.S.E. 2006]
(b)
3a a b a c ab 3b c b = 3(a + b + c) (ab + bc + ca) ac bc 3c
Using the properties of determinants, prove that
(a)
1 bc bc(b c ) 1 ca ca(c a) = 0. 1 ab ab(a b)
[C.B.S.E. 2007]
x x 2 yz y y 2 zx = (x – y) (y – z) (z – x) (xy + yz + zx) z z 2 xy
(b)
Q.13
x x2 1 x3 If x, y, z are different and y y 2 1 y 3 = 0, show that xyz = –1. z z 2 1 z3
Q.14
a b c If a, b and c are all positive and distinct, show that = b c a has a negative value. c a b
[C.B.S.E. 2008]
[C.B.S.E. 2008] 3x 8 3 3 = 0. 3 3x 8 3 3 3 3x 8
Q.15
Solve for x :
Q.16
Using the properties of determinants.
Q.17
[C.B.S.E. 2008]
[C.B.S.E. 2008]
(a)
x x 2 1 ax 3 y y 2 1 ay 3 = (1 + axyz) (x – y) (y – z) (z – x) z z 2 1 az 3
(b)
1 a2 b2 2ab 2b 2ab 1 a2 b2 2a = (1 + a2 + b2)3 2b 2a 1 a2 b2
(c)
1 x x2 x 2 1 x = (1 – x3)2 x x2 1
(d)
1 a2 bc a3 1 b2 ca b3 = – (a – b) (b – c) (c – a) (a2 + b2 + c2). 1 c 2 ab c 3
Using properties of determinants, prove the following : [C.B.S.E. 2009]
(a)
1 a2 b2 2ab 2b 2ab 1 a2 b2 2a = (1 + a2 + b2)3 2 2 2b 2a 1 a b
(c)
1 1 p 1 p q 2 3 2p 1 3p 2q = 1 3 6 3p 1 6p 3q
(d)
a 2 1 ab ac (b) ba b 2 1 bc = 1 + a 2 + b2 + c 2 2 ca cb c 1
a b c a b b c c a = a3 + b3 + c3 – 3abc bc c a ab
www.thinkiit.in
