1.Algebra Booster

7.62 Algebra Booster
fi
Ê a ˆ Ê b ˆ r
Á + 1 + + 1 + = 2
Ë p -a ˜
¯
Á
Ëq -b ˜
¯ r -c
fi p + q + r = 2
p -a q -b r -c
fi E = 2
13. We have,
sin q cos q sin q
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq + cos q + sin q +
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
sin q cos q sin q
Ê2pˆ Ê2pˆ Ê4pˆ
= 2sinqcosÁ 2cosqcos 2sin 2qcos
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cosÁq - ˜ sin Áq
-
Ë
˜
3 ¯ Ë 3 ¯ Ë
˜
3 ¯
sin q cos q sin q
= –sin q – cos q –sin 2q
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
Á
Ë
˜
3 ¯
sin q cos q sin q
= 0 0 0
Ê 2pˆ Ê 2pˆ Ê 4pˆ
sin Áq - cos q - sin q -
Ë
˜
3 ¯
Á
Ë
˜ Á ˜
3 ¯ Ë 3 ¯
(R 2
Æ R 2
+ R 1
)
= 0
14. The given determinant is
ax -by - c bx + ay cx + a
bx + ay - ax + by - c cy + b
cx + a cy + b -ax - by + c
Applying C 2
Æ aC 1
+ bC 2
+ cC 3
, we get
2 2 2
( )
1 (
2 2 2 )
a
2 2 2
a + b + c x bx + ay cx + a
a + b + c x - ax + by - c cy + b
( a + b + c ) cy + b -ax - by + c
x bx + ay cx + a
1
y - ax + by - c cy + b
a
1 cy + b -ax - by + c
ÊC2ÆC2-bC1ˆ
Applying Á
ËC ÆC -cC
˜ , we get
¯
3 3 1
x ay a
1
y -ax -c b
a
1 cy -ax -by
Applying R 1
Æ (xR 1
+ yR 2
+ R 3
), we get
x
1
ax
2 2
+ y + 1 0 0
y -ax - c b
1 cy -ax -by
1 (
2 2 1)(( )
2
x y ax acx abxy bcy bcy )
= + + + + + -
ax
1 (
2 2 1)(( )
2
= x + y + ax + acx + abxy )
ax
= (x 2 + y 2 + 1) (ax + c + by)
= (x 2 + y 2 + 1) (ax + by + c)
If the given determinant is zero, then
(ax + by + c) = 0, ((x 2 + y 2 + 1) π 0)
Thus, (ax + by + c) = 0 represents a straight line.
15. We have,
Êa b cˆÊa b cˆ
A T A = Áb Á
c a˜Áb ˜Á
c a˜
˜
Ëc a b¯Ëc a b¯
Ê
2 2 2
a + b + c ab + bc + ca ab + bc + caˆ
Á
2 2 2
˜
= Áab + bc + ca a + b + c ab + bc + ca˜
Á
2 2 2˜
Ëab + bc + ac ab + bc + ca a + b + c ¯
Êa b bˆ
= Áb Á
a b˜
˜
Ëb b a¯
where a 2 + b 2 + c 2 = a, ab + bc + ca = b.
Since AA T = I, so, a 2 + b 2 + c 2 = 1 and ab + bc + ca = 0.
Now,
a 3 + b 3 + c 3 – 3abc
= (a + b + c) (a 2 + b 2 + c 2 – ab – bc – ca)
= (a + b + c)(1 – 0)
= (a + b + c)
Also, (a + b + c) 2 = (a 2 + b 2 + c 2 + 2(ab + bc + ca))
fi (a + b + c) 2 = 1 + 2.0 = 1
fi (a + b + c) = 1, since a, b, c are all positive
Thus,
a 3 + b 3 + c 3 – 3abc = 1
fi a 3 + b 3 + c 3 – 3abc + 1 = 3 + 1 = 4.
16. We have,
1 1 1
a a( a + d) ( a + d)( a + 2 d)
D=
1 1 1
a + d ( a + d)( a + 2 d) ( a + 2 d)( a + 3 d)
1 1 1
a + 2 d ( a + 2 d)( a + 3 d) ( a + 3 d)( a + 4 d)
1
=
2 3 2
a( a + d) ( a + 2 d) ( a + 3 d) ( a + 4 d)
( a + d)( a + 2 d) a + 2d a
¥ ( a + 2 d)( a + 3 d) ( a + 3 d) ( a + d)
( a + 3 d)( a + 4 d) ( a + 4 d) ( a + 2 d)