1.Algebra Booster

Matrices and Determinants 7.65
a b c
= pqr b c a
c a b
Hence, the result.
25. We have,
2
2bc - a
2
c
2
b
2
c
2
2ac - b
2
b
b a 2ab-
c
2 2 2
a b c – a c b
= b c a ¥ – b a c
c a b – c b a
a b c a c b
=- b c a ¥ b a c
c a b c b a
a b c a b c
= b c a ¥ b c a
c a b c a b
a b c
= b c a
c a b
= (a 3 + b 3 + c 3 – 3abc) 2
26. We have,
– bc
2
b + bc
2
c + bc
2
a + ac – ac
2
c + ac
2
a + ab
2
b + ab – ab
2
2 2
- abc ab + abc ac + abc
1 2 2
= abc a b + abc - abc bc + abc
a 2 c abc b 2
+ c + abc - abc
- bc ab + ac ac + ab
abc
= ab + bc - ac bc + ab
abc ac + bc bc + ac - ab
- bc ab + ac ac + ab
= ab + bc - ac bc + ab
ac + bc bc + ac -ab
ab + bc + ca ab + bc + ca ab + bc + ca
= ab + bc - ac bc + ab
ac + bc bc + ac -ab
(R 1
Æ R 1
+ R 2
+ R 3
)
1 1 1
= ( ab + bc + ca)
ab + bc - ac bc + ab
ac + bc bc + ac -ab
= (ab + bc + ca)
1 0 0
¥ ab + bc -( ab + bc + ca) 0
ac + bc 0 -( ab + bc + ca)
-( ab + bc + ca) 0
= ( ab + bc + ca)
0 -( ab + bc + ca)
= (ab + bc + ca) 3
Hence, the result.
27. Since the given system of equations has non-trivial solutions,
so
D = 0
a
2
a
3
( a + 1)
fi b
2
b
3
( b + 1) = 0
c
2
c
3
( c + 1)
a
2
a
3
a a
2
a 1
fi b
2
b
3
b + b
2
b 1 = 0
c
2
c
3
c c
2
c 1
1 a
2
a a
2
a 1
fi abc1 b
2
b + b
2
b 1 = 0
1 c
2
c c
2
c 1
1 a
2
a 1 a
2
a
fi abc1 b
2
b + 1 b
2
b = 0
1 c
2
c 1 c
2
c
1 a
2
a
fi 1 b
2
b ( abc + 1) = 0
1 c
2
c
fi (a – b)(b – c)(c – a)(abc + 1) = 0
fi (abc + 1) = 0 ( a π b π c)
fi (abc + 1) = 0
28. We have,
2 2 2
( b + c)
a a
2
b
2
( c + a)
2
b
c c ( a + b)
2 2 2
2 2 2 2 2
( b + c) a - ( b + c) a - ( b + c)
2 2 2
= b ( c + a) -b
0
c 0 ( a + b)
- c
2 2 2
2
ÊC2ÆC2-C1ˆ
Á
ËC ÆC -C
˜
¯
3 3 1
( b + c) ( a -b - c) ( a -b-
c)
2
= ( a + b + c) 2
b ( c + a -b) 0
2
c 0 ( a + b - c)